Price Indexation, Habit Formation, and the Generalized Taylor Principle *
|
|
- Δήλια Κοτζιάς
- 8 χρόνια πριν
- Προβολές:
Transcript
1 Federal Reserve Bank of Dallas Globalization and Monetary Policy Institute Working Paper No Price Indexation Habit Formation and the Generalized Taylor Principle * Saroj Bhattarai Penn State University Jae Won Lee Rutgers University Woong Yong Park University of Hong Kong August 2013 Abstract We prove that the Generalized Taylor Principle under which the nominal interest rate reacts more than one-for-one to inflation in the long run is a necessary and under some extra mild restrictions on parameters sufficient condition for determinacy in a sticky price model with positive steady-state inflation interest rate smoothing in monetary policy partial dynamic price indexation and habit formation in consumption. JEL codes: E31 E52 E58 * Saroj Bhattarai 615 Kern Building University Park PA Sub31@psu.edu. Jae Won Lee 75 Hamilton Street NJ Hall New Brunswick NJ jwlee@econ.rutgers.edu. Woong Yong Park 908 KK Leung Building University of Hong Kong Pokfulam Road Hong Kong. wypark@hku.uk. We are grateful to Thomas Lubik for encouragement and advice and Hantaek Bae for useful comments. The views in this paper are those of the authors and do not necessarily reflect the views of the Federal Reserve Bank of Dallas or the Federal Reserve System.
2 1 Introduction One of the most important guiding principles for practical monetary policy is the Generalized Taylor Principle which asserts that in order to ensure price stability the nominal interest rate needs to respond more than one-for-one to inflation in the long run. Indeed Bullard and Mitra 2002 Woodford 2003 and Lubik and Marzo 2007 show that the Generalized Taylor Principle is a necessary and sufficient condition for a unique stable equilibrium in simple sticky price models when the central bank follows a Taylor rule that is a rule where the nominal interest rate responds to both inflation and output. 1 While these results are highly influential most sticky price models that are taken to the data now routinely feature various propagation mechanisms such as habit formation and price indexation following Christiano Eichenbaum and Evans 2005 and Smets and Wouters To the best of our knowledge the determinacy properties of such models have been studied only numerically. We contribute to the literature by showing analytically that the Generalized Taylor Principle is a necessary and under some extra mild restrictions sufficient condition for determinacy in a more general environment than considered by previous studies. In particular we consider a sticky price model with non-zero steady-state inflation dynamic partial price indexation and habit formation in consumption and in which the central bank follows a Taylor rule where the nominal interest rate is determined by its lag and partially responds to both current inflation and output. 2 As a by-product of our analysis we also characterize analytically the full solution of the model when a unique equilibrium exists. We find that habit formation in consumption and interest rate smoothing in the Taylor rule does not affect the Generalized Taylor Principle while dynamic partial price indexation requires monetary policy to respond to inflation and/or output more strongly to ensure determinacy. This is because dynamic partial price indexation decreases the long-run trade-off between inflation and output in the model while habit formation does not affect the long-run trade-off at all. Moreover interest rate smoothing does not affect the Generalized Taylor Principle as it does not change the extent of long-run impact of interest rates on inflation. Our results can be practically applied in likelihood-based estimation of monetary models to impose parameter restrictions that lead to determinacy or indeterminacy separately. For example in Bhattarai Lee and Park 2013 we estimate a sticky price model under different combinations of monetary and fiscal policy regimes and where each regime including one which features indeterminacy is imposed by making use of the analytical boundary condition 1 Carlstron Fuerst and Ghironi 2006 show that the Taylor principle is a necessary and sufficient condition for determinacy in a two-sector model where the nominal interest rate responds only to inflation. 2 To preserve analytical tractability we do not allow for sticky wages or investment in the model. 2
3 derived here. In particular having an analytical boundary greatly aids in making the posterior simulation stable and helps substantially with convergence. 2 Model The model is based on the prototypical New Keynesian set-up in Woodford detailed exposition of the model is in the appendix. equilibrium conditions and the monetary policy rule which are The Here we present the log-linearized Y t ηy t 1 = E t Y t+y t R t E t π t+1 + d t 1 π t γπ t 1 = β E t π t+1 γπ t + κ ϕy t + 1 ] Y t ηy t 1 + u t 2 R t = ρ R R t ρ R φ π π t + φ Y Y t + ε Rt 3 where Y is output π is inflation and R is the nominal interest rate. 3 The parameter 0 < η < 1 governs habit formation 0 < γ < 1 governs dynamic price indexation 0 < β < 1 is the discount factor κ > 0 is a composite parameter that depends inversely on the extent of price stickiness and ϕ > 0 is the inverse of the Frisch elasticity of labor supply. The model is therefore a generalization of the text-book purely forward looking New Keynesian model. In particular habit formation introduces persistence in the IS equation 1 while dynamic price indexation introduces persistence in the Phillips curve 2. Finally the Taylor rule 3 takes a standard form with interest rate smoothing and has the smoothing parameter 0 < ρ R < 1 and feedback parameters φ π 0 and φ Y 0 on inflation and output respectively. 4 The exogenous shock d t is a normalized preference shock and u t is a normalized markup shock. We assume that they evolve according to an AR1 process as follows d t = ρ d d t 1 + ε dt u t = ρ u u t 1 + ε ut where ε dt and ε ut are i.i.d. and have finite mean and variance. The shock ε Rt which is also i.i.d. and has finite mean and variance captures an unanticipated deviation of monetary policy from the Taylor rule. Since stationary shocks do not matter for determinacy of the model equilibrium we can drop d t u t and ε Rt from the model in most of the derivations and proofs below. 3 Y π and R denote the log deviation of the variables from their respective state state value. To keep the presentation uncluttered we do not use a hat to denote log deviations. Note that in the appendix variables with no hats denote variables in levels not log deviations. 4 Clearly when η γ = 0 and ρ R = 0 the model reduces to a completely forward-looking set-up. 3
4 3 Results We present our results in steps. We first show a condition on the roots of a fifth-order characteristic equation for determinacy of the model. Then we derive a necessary condition and a sufficient condition in terms of the model parameters for a unique stable equilibrium. Finally given that this condition is met we analytically characterize the unique solution of the model. 3.1 Condition on Characteristic Roots for Determinacy To derive a condition for equilibrium determinacy of the model we first collapse the three equations 1-3 into a single equation for Y t and its leads and lags and then use the factorization method with the lag operator. 5 The method boils down to finding a condition about the roots of a univariate characteristic equation. It is essentially equivalent to the standard method that uses the eigenvalue decomposition but turns out to be easier to apply in our case since we can make use of some properties of a high-order polynomial. Because of the complicated lag structure introduced by habit formation in consumption dynamic price indexation and interest rate smoothing in monetary policy we use equations in different time periods to eliminate π t and R t and their leads and lags. After a series of algebraic operations we obtain L 5 + a 4 L 4 + a 3 L 3 + a 2 L 2 + a 1 L 1 + a 0 Et 1 Y t 2 = E t 1 w t 1 4 where L is the lag operator w t 1 = β 1 ρ d ρ R ρ d γ 1 βρ d d t 1 β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 uu t 1 and a 4 = 1 + β 1 + η + γ + ρ R + κβ 1 ϕ + 1 ] + 1 ρ R φ Y κ 1 β a 3 =β 1 + η + γ + ρ R 1 + β 1 + ηγ + ηρ R + γρ R + 1 ρ R κβ 1 φ π ϕ βγ φ 1 η Y κ 1 + ρ R 1 ρ R ϕ η 1 η 1 ρ R 1 η ] η + γ + ρ R β 1 + ηγ + ηρ R + γρ R 1 + β 1 + ηγρ R a 2 = + 1 ρ R κβ 1 η φ π + φ Y κ 1 γ + ρ R η 1 η 1 ρ R 5 See Sargent 1987 or Hamilton 1994 for a detailed presentation of the factorization method. 1 η 4
5 a 1 =ηγβ 1 + ρ R β 1 η + γ + ηγ + βηγ a 0 = ηγρ R β 1. Note that a 4 a 2 a 0 < 0 and a 3 a 1 > 0. 6 Let λ 1 λ 2 λ 3 λ 4 and λ 5 be the five roots of the characteristic equation for the left hand side of 4 f z z 5 + a 4 z 4 + a 3 z 3 + a 2 z 2 + a 1 z + a 0 = 0 5 where λ 1 λ 2 λ 3 λ 4 λ 5. Then 4 can be written as 1 λ 1 L 1 λ 2 L 1 λ 3 L L 1 λ 4 L 1 λ 5 Et 1 Y t+1 = E t 1 w t 1. 6 The condition for 4 to have a unique stable solution for E t 1 Y t+1 is therefore that f z = 0 has three roots inside the unit circle and two roots outside the unit circle λ 1 λ 2 λ 3 < 1 < λ 4 λ 5. 7 Under the condition 7 we can derive a unique solution for E t 1 Y t+1 and use it to solve for E t 1 Y t which can be used in turn to solve for endogenous variables such as Y t π t and R t. If there are more than three roots inside the unit circle 4 is not determinate and has multiple solutions of E t 1 Y t+1 which results in equilibrium indeterminacy. If there are less than three roots inside the unit circle there does not exist any stable solution of E t 1 Y t Generalized Taylor Principle We now translate the condition 7 for equilibrium determinacy into a condition with respect to the parameters of the model. We first derive a necessary condition which turns out to be the Generalized Taylor Principle and then show that this is also sufficient for equilibrium determinacy under a weak additional assumption Necessary Condition for Determinacy A necessary condition for 7 is f 1 > 0 since f z is a fifth-order polynomial and f z < 0 for real z 0. Otherwise there exist no root two roots or four roots inside the unit circle. 6 The detailed derivation is presented in the appendix. Note that L 1 is the forward operator. The lag operator and forward operator apply to the time subscript of a variable but not on the time period in which the expectation of the same variable is taken. 5
6 Note that f 1 = 1 + a 4 + a 3 + a 2 + a 1 + a 0 = 1 ρ R κβ 1 ϕ + 1 and therefore f 1 > 0 is equivalent to φ π Sufficient Condition for Determinacy φ π + ] 1 γ 1 β φ Y 1 κ ϕ γ 1 β φ Y > 1. 8 κ ϕ + 1 It turns out that a mild restriction on parameters is needed for 8 to be sufficient for equilibrium determinacy. There exist some parameter values for which 8 is met but the model does not have a unique stable equilibrium. 7 We instead derive a sufficient condition for equilibrium determinacy and then show that this sufficient condition is only slightly stronger than 8. Using an exhaustive grid search on the parameter space we find that the difference between the two conditions is practically unimportant. Let g z = a 3 z 3. A stronger version of the Rouché Theorem by Glicksberg 1976 states that if the strict inequality f z g z < f z + g z 9 holds on the unit circle C = {z : z = 1} and f z and g z have no zeros on C then f z = 0 and g z = 0 have the same number of roots inside C. 8 Here each root is counted as many times as its multiplicity. Choose z C. Then there exists ω 0 2π] such that z = e iω = cos ω + i sin ω and it follows that f z g z = e i5ω + a 4 e i4ω + a 2 e i2ω + a 1 e iω + a 0 = e i3ω e i5ω + a 4 e i4ω + a 2 e i2ω + a 1 e iω + a 0 { 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω] 2 = + 1 a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω] 2 } 1/2 7 These parameter values are practically not relevant as we discuss in detail later. Moreover numerically we find that they lead to an explosive solution. 8 While f and g are real-valued we extend the domain of f and g to include all the complex numbers for the proof. Note however that the roots of f and g with the extended domain are the same as the roots of f and g with the original domain. 6
7 and f z + g z = a3 e i3ω + e i5ω + a 4 e i4ω + a 3 e i3ω + a 2 e i2ω + a 1 e iω + a 0 = a 3 + e i3ω e i5ω + a 4 e i4ω + a 3 e i3ω + a 2 e i2ω + a 1 e iω + a 0 { 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + a 3 ] 2 = a a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω] 2 } 1/2. First suppose that 1 a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω 0. Then f z 0. Now define h µ z µa 3 + { 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + µa 3 ] a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω] 2 } 1/2 for µ 0 1]. Observe that h µ z µ since a 3 > 0 and = a ] 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + µa 3 a 3 > 0 h µ z µa 3 0 < 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + µa 3 h µ z µa 3 < 1. Therefore h µ z is strictly increasing in µ over 0 1] which implies that the inequality 9 holds since h 0 z = f z g z and h 1 z = f z + g z. Now suppose that 1 a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω = Then we assume that 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + a 3 > It follows that since a 3 > 0 the inequality 9 holds. Also under this assumption f z 0. It is obvious that g z does not have zeros on C. Therefore according to the stronger version of the Rouché Theorem f z = 0 has exactly three roots inside the unit circle as g z = 0 has three roots inside the unit circle. This concludes the proof that the condition 7
8 that 11 holds for any ω satisfying 10 is sufficient for equilibrium determinacy. Note that 11 does not have to hold for ω that does not satisfy 10. Let us denote this sufficient condition by 10 11]. Note that there is the following relationship between the conditions found so far 10 11] 7 8 where 7 is the necessary and sufficient condition for equilibrium determinacy. 9 We now summarize our main result in the following proposition. Proposition The Generalized Taylor Principle. Under the standard assumption on the domain of the parameters a necessary condition for equilibrium determinacy of the model 1-3 is For any ω 0 2π] such that φ π + 1 γ 1 β φ Y > κ ϕ a 1 sin 2ω + a 4 a 2 sin ω a 0 sin 3ω = 0 13 assume that 1 + a 1 cos 2ω + a 4 + a 2 cos ω + a 0 cos 3ω + a 3 > Then the condition 12 is both necessary and sufficient for equilibrium determinacy. It is easy to show that when η = 0 γ = 0 and ρ R = 0 that is when the model 1-3 is purely forward-looking the sufficient condition 13 14] is equivalent to the necessary condition 12. That is 12 is necessary and sufficient. In general the sufficient condition 13 14] is only slightly stronger than the necessary condition 12. Using an exhaustive grid search we find that 13 14] is practically equivalent to 12 in that those parameter values that meet 12 but not 13 14] are not relevant. In particular when a condition that β is greater than either of η γ or ρ R which is not restrictive at all is further assumed 12 is found to be necessary and sufficient for equilibrium determinacy We can derive a sufficient condition that fz = 0 has only one root or five roots inside the unit circle by defining gz = a 1 z or gz = z 5 respectively. The sufficient conditions are mutually exclusive of each other and also with 10 11]. A parameter value that satisfies either of these sufficient conditions is an example of the discrepancy between 10 11] and 8. In an exhaustive grid search we find that there are no parameter values under 8 that produce five roots inside the unit circle. See Section for details. 10 We discuss the relationship between the sufficient condition 13 14] and the necessary condition 12 and the grid search in Section
9 3.2.3 Economic Intuition Economic intuition implied by the condition 12 in the proposition above is well known. Suppose that the endogenous variables are stable. Then the Phillips curve 2 implies the following long-run relationship between inflation and output dy = 1 γ 1 β dπ 15 κ ϕ + 1 where dπ and dy are the sizes of a permanent change in inflation and output respectively. Combining this with the long-run relationship implied by the Taylor rule 3 leads to dr = φ π + ] 1 γ 1 β φ Y dπ 16 κ ϕ + 1 where dr is the size of a permanent change in the nominal interest rate. 11 Note that the condition 12 is exactly given by the term in the brackets in 16 being greater than 1 and implies that when it is fulfilled the nominal interest rate reacts to a rise in inflation by more than one-for-one in the long run. Thus the real interest rate eventually rises when inflation rises which works to counteract the increase in inflation and stabilizes the economy. This property is referred to as the Generalized Taylor Principle in the literature. Therefore our main result is indeed that the Generalized Taylor Principle 12 is both necessary and sufficient for the existence of a unique stable equilibrium in our model except for some parameter values that are ruled out by the sufficient condition 13 14]. Moreover note that the habit formation parameter η does not directly influence condition 12 since it does not affect the long-term inflation and output gap trade-off in the model via 2. Exactly for this reason η does not appear in 15. Our results overall generalize those in Bullard and Mitra 2002 Woodford 2003 and Lubik and Marzo 2007 who consider a purely forward-looking New Keynesian model. When γ = 0 12 indeed simplifies to the condition for a unique equilibrium shown in these papers. With partial dynamic price indexation that is 0 < γ < 1 our proposition shows that ceteris paribus the Taylor rule feedback coefficients φ π and/or φ Y have to be larger to ensure a determinate equilibrium. This is because dynamic inflation indexation reduces the long-run trade-off between inflation and output in the model as shown clearly by Note that the interest rate smoothing parameter does not appear since we consider permanent changes. 12 In fact if one were to allow for complete price indexation then the long-run trade-off between inflation and output would disappear completely as discussed in Woodford In such a case the condition for determinacy would simply be φ π > 1. Note that even with non-zero steady-state inflation unlike in Coibion and Gorodnichenko 2011 the Generalized Taylor Principle is necessary and sufficient to ensure determinacy because we allow for partial dynamic price indexation as well as partial indexation to steady-state inflation. 9
10 3.2.4 The Sufficient Condition and the Generalized Taylor Principle Using an exhaustive grid search on the parameter space we find that the sufficient condition 13 14] is practically equivalent to the Generalized Taylor Principle That is for most of the parameter values the two conditions are met simultaneously. Note that 13 can be rewritten as 4a 0 cos 2 ω a 1 cos ω + a 4 a 2 + a 0 = 0 17 for ω such that sin ω 0. In most of the cases there does not exist any ω that invokes 17 since the quadratic equation in terms of cos ω does not have a real root on 0 1]. Even though there exists such ω solving is often true under 12. Also for the values of ω such that sin ω = 0 14 always holds given 12. Next those parameter values that satisfy 12 but not 13 14] are not practically relevant as they are quite extreme and implausible. In particular using the grid search we find that 12 is necessary and sufficient under an extra condition that β > η γ or ρ R. This extra condition rules out these irrelevant and extreme parameter values. In addition we note that in the grid search we do not find any parameter values that meet 12 but produce multiple stable equilibria. So we conjecture that equilibrium indeterminacy is ruled out under 12. In particular parameter values with β < η γ and ρ R that are sometimes found to violate the sufficient condition 13 14] while satisfying 12 generate an explosive equilibrium. Our conjecture is that such parameter values make the model dynamics too persistent leading to an explosive solution and that 13 14] rules such solution out. Lastly using the grid search we find 13 14] to be necessary and sufficient for equilibrium determinacy generally. That is any parameter values that result in equilibrium determinacy are found to satisfy 13 14]. 3.3 Model Solution Finally we present the complete solution to the model under equilibrium determinacy. 14 can rewrite 6 as We L 1 λ 4 L 1 λ 5 Et 1 Y t+1 λ 1 + λ 2 + λ 3 Y t + λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 Y t 1 13 We provide technical details of the grid search in the appendix. 14 For the detailed derivation see the appendix. λ 1 λ 2 λ 3 Y t 2 ] = E t 1 w t 1 10
11 which is solved as E t 1 Y t+1 = λ 1 + λ 2 + λ 3 Y t λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 Y t 1 + λ 1 λ 2 λ 3 Y t ρ d ρ R ρ d γ 1 βρ d β λ 4 ρ d λ 5 ρ d d t 1 1 ρ R φ π + ρ R ρ u ] ρ 2 u u t 1. β λ 4 ρ u λ 5 ρ u Note that ε Rt 1 does not appear in the solution of E t 1 Y t+1 since it is independent over time. The two-step ahead expectation of output E t 1 Y t+1 is uniquely determined and stable. We can further find a unique solution for E t 1 Y t and finally Y t π t and R t using 18. The solution for Y t is Y t = ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ Y 1 + Φ Y 1 Y t 1 + ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ π 1 + Φ π 1 π t 1 + ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ R 1 + Φ R 1 R t 1 + ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ d0 + Φ d0 d t + ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ u0 + Φ u0 u t + ΦY0 Φ 1 π0 Ψ 1 π0ψ Y0 Φπ0 Ψ 1 π0ψ εr 0 + Φ εr 0 ε Rt where Φ Y0 = Ψ 1 π0 ΨY0 Φ 1 Y1 Φ ] Y0 + Ψ Y 1 + Ψ R 1 1 ρ R φ Y + κβ 1 ϕ + 1 Φ π0 = β βγ Ψ 1 π0 ΨY0 Φ 1 Y1 Φ π0 + Ψ π 1 + Ψ R 1 1 ρ R φ π η Φ Y 1 = κβ 1 Ψ 1 π0ψ Y0 Φ 1 Y1 Φ Y 1 Φ π 1 = β 1 γ Ψ 1 π0ψ Y0 Φ 1 Y1 Φ π 1 Φ R 1 = Ψ 1 π0ψ R 1 ρ R Φ d0 = Ψ 1 π0 ΨY0 Φ 1 Y1 Φ d0 + Ψ d0 ρ d Φ u0 = β 1 Ψ 1 π0 ΨY0 Φ 1 Y1 Φ u0 + Ψ u0 ρ u Φ εr 0 = Ψ 1 π0ψ R 1 and Ψ π0 = Φ 1 Y1 Φ π0 1 ρ R φ π β βγ ] 11
12 Ψ Y0 = 1 Φ 1 Y1 Φ Y0 + η + 1 ρ R φ Y + κβ 1 ϕ + 1 ] Ψ Y 1 = η Φ 1 Y1 Φ Y 1 κβ 1 η Ψ π 1 = Φ 1 Y1 Φ π 1 + β 1 γ Ψ R 1 = ρ R Ψ d0 = Φ 1 Y1 Φ d0 1 Ψ u0 = Φ 1 Y1 Φ u0 + β 1 Ψ εr 0 = and Φ Y1 =1 λ 1 + λ 2 + λ 3 + η + ρ R + 1 ρ R φ Y + κβ 1 ϕ + 1 ] Φ Y0 =η + ρ R λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 + ρ R η + 1 ρ R φ π κβ 1 ϕ + 1 β βγ κ ϕ + 1 η κβ 1 ] + ρ R κβ 1 ϕ + 1 Φ π0 = β 1 β βγ 2 γ ] β βγ 1 ρ R φ π + ρ R η η Φ Y 1 = λ 1 λ 2 λ 3 ρ R η 1 ρ R φ π κβ 1 + β βγ κ η ρ R κβ 1 Φ π 1 = 1 ρ R φ π β 1 γ β 2 γ 1 + βγ + ρ R β 1 γ Φ d0 = ρ d ρ R 1 + ρ d γ 1 βρ d β λ 4 ρ d λ 5 ρ d Φ u0 = 1 ρ R β 1 φ π β 1 ρ u + β βγ ] + ρ R β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 u. β λ 4 ρ u λ 5 ρ u Inflation π t is solved for as π t =Ψ 1 π0ψ Y0 Y t + Ψ 1 π0ψ Y 1 Y t 1 + Ψ 1 π0ψ π 1 π t 1 + Ψ 1 π0ψ R 1 R t 1 + Ψ 1 π0ψ d0 d t + Ψ 1 π0ψ u0 u t + Ψ 1 π0ψ εr 0ε Rt which can be further solved to remove Y t on the right hand side. The interest rate R t is then simply determined by the Taylor rule 3. 12
13 4 Conclusion We show analytically that the generalized Taylor Principle under which the nominal interest rate reacts more than one-for-one to inflation in the long-run is a necessary and sufficient condition for determinacy in a sticky price model with non-zero steady-state inflation partial dynamic price indexation and habit formation in consumption. References 1] Bhattarai Saroj Jae W. Lee and Woong Y. Park Policy Regimes Policy Shifts and U.S. Business Cycles Unpublished. 2] Bullard James and Kaushik Mitra Learning About Monetary Policy Rules Journal of Monetary Economics ] Carlstrom Charles T Timothy S Fuerst and Fabio Ghironi Does it Matter for Equilibrium Determinacy What Price Index the Central Bank Targets? Journal of Economic Theory ] Christiano Lawrence J. Martin Eichenbaum and Charles L. Evans Nominal Rigidities and the Dynamic Effects of a Shock to Monetary Policy Journal of Political Economy ] Coibion Olivier and Yuriy Gorodnichenko Monetary Policy Trend Inflation and the Great Moderation: An Alternative Interpretation American Economic Review ] Glicksberg Irving A Remark on Rouché Theorem American Mathematical Monthly ] Hamilton James Time Series Analysis Princeton NJ: Princeton University Press. 8] Lubik Thomas and Massimiliano Marzo An Inventory of Simple Monetary Policy Rules in a New Keynesian Macroeconomic Model International Review of Economics and Finance ] Sargent Thomas Macroeconomic Theory New York NY: Academic Press. 10] Smets Frank R. and Raf Wouters Shocks and Frictions in U.S. Business Cycles: A Bayesian DSGE Approach American Economic Review
14 11] Woodford Michael Interest and Prices: Foundations of a Theory of Monetary Policy Princeton NJ: Princeton University Press. 14
15 Appendix A Model A.1 Households There is a continuum of households in the unit interval. Each household specializes in the supply of a particular type of labor. A household that supplies labor of type-j maximizes the utility function: E 0 β t H δ t log C j j t t ηc t ϕ t=0 1+ϕ where C j t is consumption of household j C t is aggregate consumption and H j t denotes the hours of type-j labor services. The parameters β ϕ and η are respectively the discount factor the inverse of the Frisch elasticity of labor supply and the degree of external habit formation while δ t represents an intertemporal preference shock that follows: Household j s flow budget constraint is: δ t = δ ρ δ t 1 expε δt. ] P t C j t + B j t + E t Q tt+1 V j t+1 = W t jh j t + V j t + R t 1 B j t 1 + Π t where P t is the price level B j t is the amount of one-period risk-less nominal bond held by household j R t is the interest rate on the bond W t j is the competitive nominal wage rate for type-j labor and Π t denotes profits of intermediate firms. In addition to the government bond households trade at time t one-period state-contingent nominal securities V j t+1 at price Q tt+1 and hence fully insure against idiosyncratic risk. A.2 Firms The final good Y t which is consumed by the government and households is produced by perfectly competitive 1 firms assembling intermediate goods Y t i with the technology Y t = 0 Y ti θt 1 θ t θ t 1 θ t di where θ t denotes time-varying elasticity of substitution between intermediate goods that follows θ t = θ 1 ρ θ θ ρ θ t 1 expε θt with the steady-state value θ. The corresponding price index for the final consumption good is P t = P ti 1 θt 1 θ t di where Pt i is the price of the intermediate good i. The optimal demand for Y t i is given by Y t i = P t i/p t θt Y t. Monopolistically competitive firms produce intermediate goods using the production function: Y t i = H t i where H t i denotes the hours of type-i labor employed by firm i. We do not include a productivity shock for simplicity. A firm resets its price optimally with probability 1 α every period. Firms that do not optimize adjust 15
16 their price according to the simple partial dynamic indexation rule: P t i = P t 1 iπ γ t 1 π1 γ where γ measures the extent of indexation and π is the steady-state value of the gross inflation rate π t P t /P t 1. All optimizing firms choose a common price P t profits: where E t k=0 X tk A.3 Monetary policy { α k Q tt+k Pt X tk W ] t+ki Y t+k i A t+k to maximize the present discounted value of future π t π t+1 π t+k 1 γ π 1 γk k 1 1 k = 0 The central bank sets the nominal interest rate according to a Taylor-type rule: ρr R πt t R = Rt 1 ] 1 ρr φπ Yt R φ Y exp ε Rt 19 π which features smoothing through the dependence on the lag and systematic responses of interest rates to output and deviation of inflation from the steady-state π. The steady-state value of R t is R and ε Rt is the non-systematic monetary policy shock that is i.i.d.. A.4 Equilibrium Equilibrium is characterized by the prices and quantities that satisfy the households and firms optimality conditions the monetary policy rule and the clearing conditions for the product labor and asset markets: 1 0 C j t dj = Y t H t j = H j t 1 0 V j t dj=0 and 1 0 B j t dj=b t = 0. Note that C j t = C t due to the complete market assumption. We use approximation methods to solve for equilibrium: we obtain a first-order approximation to the equilibrium conditions around the non-stochastic steady state. The approximation leads to the equations in the text. We reparameterize the shocks so that where κ = 1 αβ 1 α / α 1 + ϕθ]. d t = 1 ρ δ δ t u t = κ 1 θ 1 θ t 16
17 B Derivations B.1 Derivation of the Characteristic Equation We collapse the following three equations into a single equation with respect to Y t and its leads and lags Y t ηy t 1 = E t Y t+y t R t E t π t+1 + d t 20 π t γπ t 1 = β E t π t+1 γπ t + κ ϕy t + 1 ] Y t ηy t 1 + u t 21 R t = ρ R R t ρ R φ π π t + φ Y Y t ] + ε Rt. 22 First push 20 one period ahead take E t and subtract 20 multiplied by ρ R from it as E t Y t+y t ρ R Y t ηy t 1 = E t Y t+2 ηe t Y t+1 ρ R E t Y t+y t E t R t+1 ρ R R t + E t π t+2 ρ R E t π t+1 + E t d t+1 ρ R d t = E t Y t+2 ηe t Y t+1 ρ R E t Y t+y t 1 ρ R φ π E t π t+1 + φ Y E t Y t+1 + E t π t+2 ρ R E t π t+1 + ρ d ρ R d t 23 where we used 22 to eliminate R t. Now lag 23 one period multiply γ to it and subtract it from the expectation of 23 given information in period t 1 as E t 1 Y t+e t 1 Y t ρ R E t 1 Y t ηy t 1 γ E t 1 Y t ηy t 1 ρ R Y t Y t 2 ] = E t 1 Y t+2 ηe t 1 Y t+1 ρ R E t 1 Y t+e t 1 Y t γ E t 1 Y t+e t 1 Y t ρ R E t 1 Y t ηy t 1 ] 1 ρ R φ π E t 1 π t+1 γe t 1 π t 1 ρ R φ Y E t 1 Y t+1 γe t 1 Y t + E t 1 π t+2 γe t 1 π t+1 ρ R E t 1 π t+1 γe t 1 π t ] + ρ d ρ R ρ d γ d t 1. Then push 24 one period ahead take E t 1 multiply β and subtract it from 24 to obtain E t 1 Y t+e t 1 Y t ρ R E t 1 Y t ηy t 1 γ E t 1 Y t ηy t 1 ρ R Y t Y t 2 ] β {E t 1 Y t+2 ηe t 1 Y t+1 ρ R E t 1 Y t+y t γ E t 1 Y t+e t 1 Y t ρ R E t 1 Y t ηy t 1 ]} = E t 1 Y t+2 ηe t 1 Y t+1 ρ R E t 1 Y t+e t 1 Y t γ E t 1 Y t+e t 1 Y t ρ R E t 1 Y t ηy t 1 ] β {E t 1 Y t+3 ηe t 1 Y t+2 ρ R E t 1 Y t+2 ηe t 1 Y t+1 γ E t 1 Y t+2 ηe t 1 Y t+1 ρ R E t 1 Y t+e t 1 Y t ]} 1 ρ R φ π E t 1 π t+1 γe t 1 π t β E t 1 π t+2 γe t 1 π t+1 ] 1 ρ R φ Y E t 1 Y t+1 γe t 1 Y t β E t 1 Y t+2 γe t 1 Y t+1 ] + E t 1 π t+2 γe t 1 π t+1 β E t 1 π t+3 γe t 1 π t+2 ] ρ R E t 1 π t+1 γe t 1 π t β E t 1 π t+2 γe t 1 π t+1 ] + ρ d ρ R ρ d γ d t 1 βe t 1 d t
18 But note that from 21 E t 1 π t+1 γe t 1 π t β E t 1 π t+2 γe t 1 π t+1 = κ ϕe t 1 Y t ] E t 1Y t+e t 1 Y t +E t 1 u t+1 and E t 1 π t+2 γe t 1 π t+1 β E t 1 π t+3 γe t 1 π t+2 = κ ϕe t 1 Y t ] E t 1Y t+2 ηe t 1 Y t+1 +E t 1 u t+2 which can be plugged into 25 to eliminate π t and its leads and lags. After arranging terms we finally obtain L 5 + a 4 L 4 + a 3 L 3 + a 2 L 2 + a 1 L 1 + a 0 Et 1 Y t 2 = E t 1 w t 1 where L is the lag operator w t 1 = β 1 ρ d ρ R ρ d γ 1 βρ d d t 1 β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 uu t 1 and a 4 = 1 + β 1 + η + γ + ρ R + κβ 1 ϕ + 1 ] + 1 ρ R φ Y κ 1 β a 3 =β 1 + η + γ + ρ R 1 + β 1 + ηγ + ηρ R + γρ R + 1 ρ R κβ 1 φ π ϕ η βγ φ Y κ 1 + ρ R 1 ρ R ϕ η + 1 η 1 ρ R 1 η η + γ + ρ R β 1 + ηγ + ηρ R + γρ R 1 + β 1 + ηγρ R a 2 = + 1 ρ R κβ 1 φ η π 1 η + φ Y κ 1 γ + ρ R η 1 ρ R 1 η a 1 =ηγβ 1 + ρ R β 1 η + γ + ηγ + βηγ a 0 = ηγρ R. ] B.2 Solution of 4 for Output Expectations The expectational difference equation 4 can be written as 1 λ 1 L 1 λ 2 L 1 λ 3 L L 1 λ 4 L 1 λ 5 Et 1 Y t+1 = E t 1 w t 1 where λ i s i = are the roots of the characteristic equation 4 and λ 1 λ 2 λ 3 < 1 < λ 4 λ 5. Note that L 1 λ 4 Et 1 Y t+1 λ 1 + λ 2 + λ 3 Y t + λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 Y t 1 λ 1 λ 2 λ 3 Y t 2 ] = L 1 1 λ 5 Et 1 w t 1 = λ 1 5 λ s 5 E t 1w t 1+s. s=0 18
19 Since E t 1 w t 1+s = β 1 ρ d ρ R ρ d γ 1 βρ d E t 1 d t 1+s β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 ue t 1 u t 1+s for s 1 it follows that s=0 = β 1 ρ d ρ R ρ d γ 1 βρ d ρ s dd t 1 β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 uρ s uu t 1 λ s 5 E t 1w t 1+s =β 1 ρ d ρ R ρ d γ 1 βρ d λ 1 5 ρ s d dt 1 β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 u λ 1 5 ρ s u ut 1 s=0 = ρ d ρ R ρ d γ 1 βρ d β 1 λ 1 5 ρ d t 1 1 ρ R φ π + ρ R ρ u ] ρ 2 u d β 1 λ 1 5 ρ u t 1 u and thus L 1 λ 4 Et 1 Y t+1 λ 1 + λ 2 + λ 3 Y t + λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 Y t 1 λ 1 λ 2 λ 3 Y t 2 ] = ρ d ρ R ρ d γ 1 βρ d β λ 5 ρ d d t ρ R φ π + ρ R ρ u ] ρ 2 u u t 1. β λ 5 ρ u By inverting L 1 λ 4 and solving the equation in the same way we can show that E t 1 Y t+1 = λ 1 + λ 2 + λ 3 E t 1 Y t λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 Y t 1 + λ 1 λ 2 λ 3 Y t ξ d d t 1 + ξ u u t 1 27 s=0 where ξ d = ρ d ρ R ρ d γ 1 βρ d β λ 4 ρ d λ 5 ρ d and ξ u = 1 ρ R φ π + ρ R ρ u ] ρ 2 u. β λ 4 ρ u λ 5 ρ u B.3 Solution to the Model Now we use the solution of two-step ahead expected output 26 and solve the model for all the endogenous variables Y t π t and R t. First solve 21 for E t π t+1 and E t π t+2 as E t π t+1 = β βγ π t β 1 γπ t 1 β 1 κ ϕ + 1 Y t η ] Y t 1 u t 28 and E t π t+2 =β βγ E t π t+1 β 1 γπ t β 1 κ ] ϕ + 1 = β βγ 2 β 1 γ π t β βγ γπ t 1 β 1 κ ϕ + 1 E t Y t+1 + β βγ κ η + β βγ κ E t Y t+1 Y t 1 β 1 ρ u + β βγ ] u t. η ] Y t E t u t+1 ϕ + 1 ] η + β 1 κ Y t 19
20 Plug the solution for E t Y t+2 E t π t+1 and E t π t+2 into 23 to obtain where Φ Y1 E t Y t+1 = Φ Y0 Y t + Φ π0 π t + Φ Y 1 Y t 1 + Φ π 1 π t 1 + Φ d0 d t + Φ u0 u t 29 Φ Y1 =1 λ 1 + λ 2 + λ 3 + η + ρ R + 1 ρ R φ Y + κβ 1 ϕ + 1 ] Φ Y0 =η + ρ R λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 + ρ R η + 1 ρ R φ π κβ 1 ϕ + 1 β βγ κ ϕ + 1 η κβ 1 ] Φ π0 = β 1 β βγ 2 γ β βγ 1 ρ R φ π + ρ R η Φ Y 1 = λ 1 λ 2 λ 3 ρ R η 1 ρ R φ π κβ 1 + β βγ κ η ρ R κβ 1 Φ π 1 = 1 ρ R φ π β 1 γ β 2 γ 1 + βγ + ρ R β 1 γ Φ d0 = ρ d ρ R 1 + ρ d γ 1 βρ d β λ 4 ρ d λ 5 ρ d ] + ρ R κβ 1 ϕ + 1 Φ u0 = 1 ρ R β 1 φ π β 1 ρ u + β βγ ] + ρ R β 1 1 ρ R φ π + ρ R ρ u ] ρ 2 u. β λ 4 ρ u λ 5 ρ u η Now eliminate E t Y t+1 and E t π t+1 from 20 using the solution for E t Y t+1 in 29 and E t π t+1 in 28 and eliminate R t using 22 to get where Ψ π0 π t = Ψ Y0 Y t + Ψ Y 1 Y t 1 + Ψ π 1 π t 1 + Ψ R 1 R t 1 + Ψ d0 d t + Ψ u0 u t + Ψ εr 0ε Rt 30 Ψ π0 = Φ 1 Y1 Φ π0 1 ρ R φ π β βγ ] Ψ Y0 = 1 Φ 1 Y1 Φ Y0 + η + Ψ Y 1 = η Φ 1 Y1 Φ Y 1 κβ 1 η Ψ π 1 = Φ 1 Y1 Φ π 1 + β 1 γ Ψ R 1 = ρ R Ψ d0 = Φ 1 Y1 Φ d0 1 Ψ u0 = Φ 1 Y1 Φ u0 + β 1 Ψ εr 0 =. 1 ρ R φ Y + κβ 1 ϕ + 1 From 30 we get another expression for E t π t+1. After substituting 29 for E t Y t+1 in this expression for E t π t+1 equate it with 28 to obtain ] Φ Y0 Y t = Φ π0 π t + Φ Y 1 Y t 1 + Φ π 1 π t 1 + Φ R 1 R t 1 + Φ d0 d t + Φ u0 u t + Φ εr 0ε Rt 20
21 where Φ Y0 = Ψ 1 π0 Ψ Y0 Φ 1 + κβ 1 ϕ + 1 ] Y1 Φ Y0 + Ψ Y 1 + Ψ R 1 1 ρ R φ Y Φ π0 = β βγ Ψ 1 π0 Ψ Y0 Φ 1 Y1 Φ π0 + Ψ π 1 + Ψ R 1 1 ρ R φ π η Φ Y 1 = κβ 1 Ψπ0 1 Ψ Y0Φ 1 Y1 Φ Y 1 Φ π 1 = β 1 γ Ψ 1 π0 Ψ Y0Φ 1 Y1 Φ π 1 Φ R 1 = Ψ 1 π0 Ψ R 1ρ R Φ d0 = Ψ 1 π0 Φ u0 = β 1 Ψ 1 π0 Φ εr 0 = Ψ 1 π0 Ψ R 1. Ψ Y0 Φ 1 Y1 Φ d0 + Ψ d0 ρ d Ψ Y0 Φ 1 Y1 Φ u0 + Ψ u0 ρ u Finally using 30 we can eliminate π t and solve for Y t as Y t = ΦY0 Φ 1 π0 Ψ 1 π0 Ψ Y0 Φπ0 Ψ 1 π0 Ψ Y 1 + Φ Y 1 Y t 1 + ΦY0 Φ 1 π0 Ψ 1 π0 Ψ Y0 Φπ0 Ψ 1 π0 Ψ π 1 + Φ π 1 π t 1 1 Φπ0 Ψ 1 + ΦY0 Φ π0 Ψ 1 π0 Ψ Y0 π0 Ψ R 1 + Φ R 1 R t 1 + ΦY0 Φ 1 π0 Ψ 1 π0 Ψ Y0 Φπ0 Ψ 1 π0 Ψ d0 + Φ d0 d t + ΦY0 Φ 1 π0 Ψ 1 π0 Ψ Y0 Φπ0 Ψ 1 π0 Ψ u0 + Φ u0 u t + ΦY0 Φ 1 π0 Ψ 1 π0 Ψ Y0 Φπ0 Ψ 1 π0 Ψ ε R 0 + Φ εr 0 ε Rt. The solution for π t can be obtained from 30. The solution for R t is simply determined by the Taylor rule 22. B.4 Grid Search on the Parameter Space We do an exhaustive grid search on the parameter space to figure out the discrepancy between the sufficient condition 13 14] and the Generalized Taylor Principle 12. The following sets of values for each parameter are selected: β { } α { } ϕ { } η { } γ { } φ Y { } ρ d { } ρ u { } 21
22 ρ R { }. For φ π we use the set of values φ π + { } where φ π = 1 1 γ 1 β φ Y κ ϕ + 1 is the boundary value of φ π for determinacy given values for the other parameters. With firm-specific labor the Phillips curve slope parameter is computed as κ = 1 αβ1 α α1 + ϕθ where θ = 8 is the steady state value of the elasticity of substitution between differentiated goods. For each parameter value we check 1 how many roots fz = 0 has inside the unit circle; 2 whether there exists ω 0 2π] that solves 17 and if yes whether 14 is met; and 3 whether the Generalized Taylor Principle 12 is satisfied. 22
Price Indexation, Habit Formation, and the Generalized Taylor Principle
Crawford School of Public Policy CAMA Centre for Applied Macroeconomic Analysis Price Indexation Habit Formation and the Generalized Taylor Principle CAMA Working Paper 52/2013 August 2013 Saroj Bhattarai
Testing for Indeterminacy: An Application to U.S. Monetary Policy. Technical Appendix
Testing for Indeterminacy: An Application to U.S. Monetary Policy Technical Appendix Thomas A. Lubik Department of Economics Johns Hopkins University Frank Schorfheide Department of Economics University
2 Composition. Invertible Mappings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely,
HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:
HOMEWORK 4 Problem a For the fast loading case, we want to derive the relationship between P zz and λ z. We know that the nominal stress is expressed as: P zz = ψ λ z where λ z = λ λ z. Therefore, applying
Section 8.3 Trigonometric Equations
99 Section 8. Trigonometric Equations Objective 1: Solve Equations Involving One Trigonometric Function. In this section and the next, we will exple how to solving equations involving trigonometric functions.
4.6 Autoregressive Moving Average Model ARMA(1,1)
84 CHAPTER 4. STATIONARY TS MODELS 4.6 Autoregressive Moving Average Model ARMA(,) This section is an introduction to a wide class of models ARMA(p,q) which we will consider in more detail later in this
CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS
CHAPTER 5 SOLVING EQUATIONS BY ITERATIVE METHODS EXERCISE 104 Page 8 1. Find the positive root of the equation x + 3x 5 = 0, correct to 3 significant figures, using the method of bisection. Let f(x) =
The Simply Typed Lambda Calculus
Type Inference Instead of writing type annotations, can we use an algorithm to infer what the type annotations should be? That depends on the type system. For simple type systems the answer is yes, and
derivation of the Laplacian from rectangular to spherical coordinates
derivation of the Laplacian from rectangular to spherical coordinates swapnizzle 03-03- :5:43 We begin by recognizing the familiar conversion from rectangular to spherical coordinates (note that φ is used
EE512: Error Control Coding
EE512: Error Control Coding Solution for Assignment on Finite Fields February 16, 2007 1. (a) Addition and Multiplication tables for GF (5) and GF (7) are shown in Tables 1 and 2. + 0 1 2 3 4 0 0 1 2 3
Example Sheet 3 Solutions
Example Sheet 3 Solutions. i Regular Sturm-Liouville. ii Singular Sturm-Liouville mixed boundary conditions. iii Not Sturm-Liouville ODE is not in Sturm-Liouville form. iv Regular Sturm-Liouville note
Chapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Figure A.2: MPC and MPCP Age Profiles (estimating ρ, ρ = 2, φ = 0.03)..
Supplemental Material (not for publication) Persistent vs. Permanent Income Shocks in the Buffer-Stock Model Jeppe Druedahl Thomas H. Jørgensen May, A Additional Figures and Tables Figure A.: Wealth and
22 .5 Real consumption.5 Real residential investment.5.5.5 965 975 985 995 25.5 965 975 985 995 25.5 Real house prices.5 Real fixed investment.5.5.5 965 975 985 995 25.5 965 975 985 995 25.3 Inflation
Homework 3 Solutions
Homework 3 Solutions Igor Yanovsky (Math 151A TA) Problem 1: Compute the absolute error and relative error in approximations of p by p. (Use calculator!) a) p π, p 22/7; b) p π, p 3.141. Solution: For
C.S. 430 Assignment 6, Sample Solutions
C.S. 430 Assignment 6, Sample Solutions Paul Liu November 15, 2007 Note that these are sample solutions only; in many cases there were many acceptable answers. 1 Reynolds Problem 10.1 1.1 Normal-order
Areas and Lengths in Polar Coordinates
Kiryl Tsishchanka Areas and Lengths in Polar Coordinates In this section we develop the formula for the area of a region whose boundary is given by a polar equation. We need to use the formula for the
Every set of first-order formulas is equivalent to an independent set
Every set of first-order formulas is equivalent to an independent set May 6, 2008 Abstract A set of first-order formulas, whatever the cardinality of the set of symbols, is equivalent to an independent
6.3 Forecasting ARMA processes
122 CHAPTER 6. ARMA MODELS 6.3 Forecasting ARMA processes The purpose of forecasting is to predict future values of a TS based on the data collected to the present. In this section we will discuss a linear
Notes on the Open Economy
Notes on the Open Econom Ben J. Heijdra Universit of Groningen April 24 Introduction In this note we stud the two-countr model of Table.4 in more detail. restated here for convenience. The model is Table.4.
Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)
Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts
Monetary Policy Design in the Basic New Keynesian Model
Monetary Policy Design in the Basic New Keynesian Model Jordi Galí CREI, UPF and Barcelona GSE June 216 Jordi Galí (CREI, UPF and Barcelona GSE) Monetary Policy Design June 216 1 / 12 The Basic New Keynesian
Section 7.6 Double and Half Angle Formulas
09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ)
Appendix S1 1. ( z) α βc. dβ β δ β
Appendix S1 1 Proof of Lemma 1. Taking first and second partial derivatives of the expected profit function, as expressed in Eq. (7), with respect to l: Π Π ( z, λ, l) l θ + s ( s + h ) g ( t) dt λ Ω(
1 1 1 2 1 2 2 1 43 123 5 122 3 1 312 1 1 122 1 1 1 1 6 1 7 1 6 1 7 1 3 4 2 312 43 4 3 3 1 1 4 1 1 52 122 54 124 8 1 3 1 1 1 1 1 152 1 1 1 1 1 1 152 1 5 1 152 152 1 1 3 9 1 159 9 13 4 5 1 122 1 4 122 5
Chapter 6: Systems of Linear Differential. be continuous functions on the interval
Chapter 6: Systems of Linear Differential Equations Let a (t), a 2 (t),..., a nn (t), b (t), b 2 (t),..., b n (t) be continuous functions on the interval I. The system of n first-order differential equations
Statistical Inference I Locally most powerful tests
Statistical Inference I Locally most powerful tests Shirsendu Mukherjee Department of Statistics, Asutosh College, Kolkata, India. shirsendu st@yahoo.co.in So far we have treated the testing of one-sided
forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with
Week 03: C lassification of S econd- Order L inear Equations In last week s lectures we have illustrated how to obtain the general solutions of first order PDEs using the method of characteristics. We
Approximation of distance between locations on earth given by latitude and longitude
Approximation of distance between locations on earth given by latitude and longitude Jan Behrens 2012-12-31 In this paper we shall provide a method to approximate distances between two points on earth
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit
Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit Ting Zhang Stanford May 11, 2001 Stanford, 5/11/2001 1 Outline Ordinal Classification Ordinal Addition Ordinal Multiplication Ordinal
3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β
3.4 SUM AND DIFFERENCE FORMULAS Page Theorem cos(αβ cos α cos β -sin α cos(α-β cos α cos β sin α NOTE: cos(αβ cos α cos β cos(α-β cos α -cos β Proof of cos(α-β cos α cos β sin α Let s use a unit circle
ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ. Ψηφιακή Οικονομία. Διάλεξη 7η: Consumer Behavior Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών
ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ Ψηφιακή Οικονομία Διάλεξη 7η: Consumer Behavior Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών Τέλος Ενότητας Χρηματοδότηση Το παρόν εκπαιδευτικό υλικό έχει αναπτυχθεί
Concrete Mathematics Exercises from 30 September 2016
Concrete Mathematics Exercises from 30 September 2016 Silvio Capobianco Exercise 1.7 Let H(n) = J(n + 1) J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2) J(2n+1) = (2J(n+1) 1) (2J(n)+1)
Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.
Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο The time integral of a force is referred to as impulse, is determined by and is obtained from: Newton s 2 nd Law of motion states that the action
Other Test Constructions: Likelihood Ratio & Bayes Tests
Other Test Constructions: Likelihood Ratio & Bayes Tests Side-Note: So far we have seen a few approaches for creating tests such as Neyman-Pearson Lemma ( most powerful tests of H 0 : θ = θ 0 vs H 1 :
Second Order RLC Filters
ECEN 60 Circuits/Electronics Spring 007-0-07 P. Mathys Second Order RLC Filters RLC Lowpass Filter A passive RLC lowpass filter (LPF) circuit is shown in the following schematic. R L C v O (t) Using phasor
Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
Inverse trigonometric functions & General Solution of Trigonometric Equations. 1. Sin ( ) = a) b) c) d) Ans b. Solution : Method 1. Ans a: 17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin
Homework 8 Model Solution Section
MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx
Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1
Eon : Fall 8 Suggested Solutions to Problem Set 8 Email questions or omments to Dan Fetter Problem. Let X be a salar with density f(x, θ) (θx + θ) [ x ] with θ. (a) Find the most powerful level α test
Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee
Appendi to On the stability of a compressible aisymmetric rotating flow in a pipe By Z. Rusak & J. H. Lee Journal of Fluid Mechanics, vol. 5 4, pp. 5 4 This material has not been copy-edited or typeset
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3
Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3 1 State vector space and the dual space Space of wavefunctions The space of wavefunctions is the set of all
ST5224: Advanced Statistical Theory II
ST5224: Advanced Statistical Theory II 2014/2015: Semester II Tutorial 7 1. Let X be a sample from a population P and consider testing hypotheses H 0 : P = P 0 versus H 1 : P = P 1, where P j is a known
Section 9.2 Polar Equations and Graphs
180 Section 9. Polar Equations and Graphs In this section, we will be graphing polar equations on a polar grid. In the first few examples, we will write the polar equation in rectangular form to help identify
Matrices and Determinants
Matrices and Determinants SUBJECTIVE PROBLEMS: Q 1. For what value of k do the following system of equations possess a non-trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z
Uniform Convergence of Fourier Series Michael Taylor
Uniform Convergence of Fourier Series Michael Taylor Given f L 1 T 1 ), we consider the partial sums of the Fourier series of f: N 1) S N fθ) = ˆfk)e ikθ. k= N A calculation gives the Dirichlet formula
Solutions to Exercise Sheet 5
Solutions to Eercise Sheet 5 jacques@ucsd.edu. Let X and Y be random variables with joint pdf f(, y) = 3y( + y) where and y. Determine each of the following probabilities. Solutions. a. P (X ). b. P (X
Partial Differential Equations in Biology The boundary element method. March 26, 2013
The boundary element method March 26, 203 Introduction and notation The problem: u = f in D R d u = ϕ in Γ D u n = g on Γ N, where D = Γ D Γ N, Γ D Γ N = (possibly, Γ D = [Neumann problem] or Γ N = [Dirichlet
Lecture 2. Soundness and completeness of propositional logic
Lecture 2 Soundness and completeness of propositional logic February 9, 2004 1 Overview Review of natural deduction. Soundness and completeness. Semantics of propositional formulas. Soundness proof. Completeness
Second Order Partial Differential Equations
Chapter 7 Second Order Partial Differential Equations 7.1 Introduction A second order linear PDE in two independent variables (x, y Ω can be written as A(x, y u x + B(x, y u xy + C(x, y u u u + D(x, y
10.7 Performance of Second-Order System (Unit Step Response)
Lecture Notes on Control Systems/D. Ghose/0 57 0.7 Performance of Second-Order System (Unit Step Response) Consider the second order system a ÿ + a ẏ + a 0 y = b 0 r So, Y (s) R(s) = b 0 a s + a s + a
Math221: HW# 1 solutions
Math: HW# solutions Andy Royston October, 5 7.5.7, 3 rd Ed. We have a n = b n = a = fxdx = xdx =, x cos nxdx = x sin nx n sin nxdx n = cos nx n = n n, x sin nxdx = x cos nx n + cos nxdx n cos n = + sin
Risk! " #$%&'() *!'+,'''## -. / # $
Risk! " #$%&'(!'+,'''## -. / 0! " # $ +/ #%&''&(+(( &'',$ #-&''&$ #(./0&'',$( ( (! #( &''/$ #$ 3 #4&'',$ #- &'',$ #5&''6(&''&7&'',$ / ( /8 9 :&' " 4; < # $ 3 " ( #$ = = #$ #$ ( 3 - > # $ 3 = = " 3 3, 6?3
.5 Real consumption.5 Real residential investment.5.5.5 965 975 985 995 25.5 965 975 985 995 25.5 Real house prices.5 Real fixed investment.5.5.5 965 975 985 995 25.5 965 975 985 995 25.3 Inflation rate.3
6.1. Dirac Equation. Hamiltonian. Dirac Eq.
6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2
Quadratic Expressions
Quadratic Expressions. The standard form of a quadratic equation is ax + bx + c = 0 where a, b, c R and a 0. The roots of ax + bx + c = 0 are b ± b a 4ac. 3. For the equation ax +bx+c = 0, sum of the roots
HW 3 Solutions 1. a) I use the auto.arima R function to search over models using AIC and decide on an ARMA(3,1)
HW 3 Solutions a) I use the autoarima R function to search over models using AIC and decide on an ARMA3,) b) I compare the ARMA3,) to ARMA,0) ARMA3,) does better in all three criteria c) The plot of the
Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is
Pg. 9. The perimeter is P = The area of a triangle is A = bh where b is the base, h is the height 0 h= btan 60 = b = b In our case b =, then the area is A = = 0. By Pythagorean theorem a + a = d a a =
Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.
Exercises 0 More exercises are available in Elementary Differential Equations. If you have a problem to solve any of them, feel free to come to office hour. Problem Find a fundamental matrix of the given
Math 6 SL Probability Distributions Practice Test Mark Scheme
Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry
PARTIAL NOTES for 6.1 Trigonometric Identities
PARTIAL NOTES for 6.1 Trigonometric Identities tanθ = sinθ cosθ cotθ = cosθ sinθ BASIC IDENTITIES cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ PYTHAGOREAN IDENTITIES sin θ + cos θ =1 tan θ +1= sec θ 1 + cot
ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007
Οδηγίες: Να απαντηθούν όλες οι ερωτήσεις. Αν κάπου κάνετε κάποιες υποθέσεις να αναφερθούν στη σχετική ερώτηση. Όλα τα αρχεία που αναφέρονται στα προβλήματα βρίσκονται στον ίδιο φάκελο με το εκτελέσιμο
These derivations are not part of the official forthcoming version of Vasilaky and Leonard
Target Input Model with Learning, Derivations Kathryn N Vasilaky These derivations are not part of the official forthcoming version of Vasilaky and Leonard 06 in Economic Development and Cultural Change.
Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics
Fourier Series MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018 Introduction Not all functions can be represented by Taylor series. f (k) (c) A Taylor series f (x) = (x c)
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in
Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in : tail in X, head in A nowhere-zero Γ-flow is a Γ-circulation such that
Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.
Chemistry 362 Dr Jean M Standard Problem Set 9 Solutions The ˆ L 2 operator is defined as Verify that the angular wavefunction Y θ,φ) Also verify that the eigenvalue is given by 2! 2 & L ˆ 2! 2 2 θ 2 +
Tridiagonal matrices. Gérard MEURANT. October, 2008
Tridiagonal matrices Gérard MEURANT October, 2008 1 Similarity 2 Cholesy factorizations 3 Eigenvalues 4 Inverse Similarity Let α 1 ω 1 β 1 α 2 ω 2 T =......... β 2 α 1 ω 1 β 1 α and β i ω i, i = 1,...,
2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.
EAMCET-. THEORY OF EQUATIONS PREVIOUS EAMCET Bits. Each of the roots of the equation x 6x + 6x 5= are increased by k so that the new transformed equation does not contain term. Then k =... - 4. - Sol.
D Alembert s Solution to the Wave Equation
D Alembert s Solution to the Wave Equation MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018 Objectives In this lesson we will learn: a change of variable technique
Managing Economic Fluctuations. Managing Macroeconomic Fluctuations 1
Managing Economic Fluctuations -Keynesian macro: - -term nominal interest rates. - - P. - - P. Managing Macroeconomic Fluctuations 1 Review: New Keynesian Model -run macroeconomics: - π = γ (Y Y P ) +
5. Choice under Uncertainty
5. Choice under Uncertainty Daisuke Oyama Microeconomics I May 23, 2018 Formulations von Neumann-Morgenstern (1944/1947) X: Set of prizes Π: Set of probability distributions on X : Preference relation
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C
DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C By Tom Irvine Email: tomirvine@aol.com August 6, 8 Introduction The obective is to derive a Miles equation which gives the overall response
Problem Set 3: Solutions
CMPSCI 69GG Applied Information Theory Fall 006 Problem Set 3: Solutions. [Cover and Thomas 7.] a Define the following notation, C I p xx; Y max X; Y C I p xx; Ỹ max I X; Ỹ We would like to show that C
k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +
Chapter 3. Fuzzy Arithmetic 3- Fuzzy arithmetic: ~Addition(+) and subtraction (-): Let A = [a and B = [b, b in R If x [a and y [b, b than x+y [a +b +b Symbolically,we write A(+)B = [a (+)[b, b = [a +b
1 String with massive end-points
1 String with massive end-points Πρόβλημα 5.11:Θεωρείστε μια χορδή μήκους, τάσης T, με δύο σημειακά σωματίδια στα άκρα της, το ένα μάζας m, και το άλλο μάζας m. α) Μελετώντας την κίνηση των άκρων βρείτε
Depth versus Rigidity in the Design of International Trade Agreements. Leslie Johns
Depth versus Rigidity in the Design of International Trade Agreements Leslie Johns Supplemental Appendix September 3, 202 Alternative Punishment Mechanisms The one-period utility functions of the home
Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013
Notes on Average Scattering imes and Hall Factors Jesse Maassen and Mar Lundstrom Purdue University November 5, 13 I. Introduction 1 II. Solution of the BE 1 III. Exercises: Woring out average scattering
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS
CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3
Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1
Conceptual Questions. State a Basic identity and then verify it. a) Identity: Solution: One identity is cscθ) = sinθ) Practice Exam b) Verification: Solution: Given the point of intersection x, y) of the
Fractional Colorings and Zykov Products of graphs
Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is
Module 5. February 14, h 0min
Module 5 Stationary Time Series Models Part 2 AR and ARMA Models and Their Properties Class notes for Statistics 451: Applied Time Series Iowa State University Copyright 2015 W. Q. Meeker. February 14,
Strain gauge and rosettes
Strain gauge and rosettes Introduction A strain gauge is a device which is used to measure strain (deformation) on an object subjected to forces. Strain can be measured using various types of devices classified
Μηχανική Μάθηση Hypothesis Testing
ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ Μηχανική Μάθηση Hypothesis Testing Γιώργος Μπορμπουδάκης Τμήμα Επιστήμης Υπολογιστών Procedure 1. Form the null (H 0 ) and alternative (H 1 ) hypothesis 2. Consider
TMA4115 Matematikk 3
TMA4115 Matematikk 3 Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet Trondheim Spring 2010 Lecture 12: Mathematics Marvellous Matrices Andrew Stacey Norges Teknisk-Naturvitenskapelige Universitet
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM
SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM Solutions to Question 1 a) The cumulative distribution function of T conditional on N n is Pr T t N n) Pr max X 1,..., X N ) t N n) Pr max
Mean-Variance Analysis
Mean-Variance Analysis Jan Schneider McCombs School of Business University of Texas at Austin Jan Schneider Mean-Variance Analysis Beta Representation of the Risk Premium risk premium E t [Rt t+τ ] R1
Bayesian statistics. DS GA 1002 Probability and Statistics for Data Science.
Bayesian statistics DS GA 1002 Probability and Statistics for Data Science http://www.cims.nyu.edu/~cfgranda/pages/dsga1002_fall17 Carlos Fernandez-Granda Frequentist vs Bayesian statistics In frequentist
Finite difference method for 2-D heat equation
Finite difference method for 2-D heat equation Praveen. C praveen@math.tifrbng.res.in Tata Institute of Fundamental Research Center for Applicable Mathematics Bangalore 560065 http://math.tifrbng.res.in/~praveen
Lecture 21: Properties and robustness of LSE
Lecture 21: Properties and robustness of LSE BLUE: Robustness of LSE against normality We now study properties of l τ β and σ 2 under assumption A2, i.e., without the normality assumption on ε. From Theorem
Solution Series 9. i=1 x i and i=1 x i.
Lecturer: Prof. Dr. Mete SONER Coordinator: Yilin WANG Solution Series 9 Q1. Let α, β >, the p.d.f. of a beta distribution with parameters α and β is { Γ(α+β) Γ(α)Γ(β) f(x α, β) xα 1 (1 x) β 1 for < x
CRASH COURSE IN PRECALCULUS
CRASH COURSE IN PRECALCULUS Shiah-Sen Wang The graphs are prepared by Chien-Lun Lai Based on : Precalculus: Mathematics for Calculus by J. Stuwart, L. Redin & S. Watson, 6th edition, 01, Brooks/Cole Chapter
SOLVING CUBICS AND QUARTICS BY RADICALS
SOLVING CUBICS AND QUARTICS BY RADICALS The purpose of this handout is to record the classical formulas expressing the roots of degree three and degree four polynomials in terms of radicals. We begin with
ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ. Ψηφιακή Οικονομία. Διάλεξη 6η: Basics of Industrial Organization Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών
ΕΛΛΗΝΙΚΗ ΔΗΜΟΚΡΑΤΙΑ ΠΑΝΕΠΙΣΤΗΜΙΟ ΚΡΗΤΗΣ Ψηφιακή Οικονομία Διάλεξη 6η: Basics of Industrial Organization Mαρίνα Μπιτσάκη Τμήμα Επιστήμης Υπολογιστών Course Outline Part II: Mathematical Tools Firms - Basics
( y) Partial Differential Equations
Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate
The challenges of non-stable predicates
The challenges of non-stable predicates Consider a non-stable predicate Φ encoding, say, a safety property. We want to determine whether Φ holds for our program. The challenges of non-stable predicates
ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω
0 1 2 3 4 5 6 ω ω + 1 ω + 2 ω + 3 ω + 4 ω2 ω2 + 1 ω2 + 2 ω2 + 3 ω3 ω3 + 1 ω3 + 2 ω4 ω4 + 1 ω5 ω 2 ω 2 + 1 ω 2 + 2 ω 2 + ω ω 2 + ω + 1 ω 2 + ω2 ω 2 2 ω 2 2 + 1 ω 2 2 + ω ω 2 3 ω 3 ω 3 + 1 ω 3 + ω ω 3 +
Instruction Execution Times
1 C Execution Times InThisAppendix... Introduction DL330 Execution Times DL330P Execution Times DL340 Execution Times C-2 Execution Times Introduction Data Registers This appendix contains several tables
5.4 The Poisson Distribution.
The worst thing you can do about a situation is nothing. Sr. O Shea Jackson 5.4 The Poisson Distribution. Description of the Poisson Distribution Discrete probability distribution. The random variable
SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018
Journal of rogressive Research in Mathematics(JRM) ISSN: 2395-028 SCITECH Volume 3, Issue 2 RESEARCH ORGANISATION ublished online: March 29, 208 Journal of rogressive Research in Mathematics www.scitecresearch.com/journals
Congruence Classes of Invertible Matrices of Order 3 over F 2
International Journal of Algebra, Vol. 8, 24, no. 5, 239-246 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ija.24.422 Congruence Classes of Invertible Matrices of Order 3 over F 2 Ligong An and