HOMEWORK#1. t E(x) = 1 λ = (b) Find the median lifetime of a randomly selected light bulb. Answer:

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1 HOMEWORK# 李亞晟 Eercise 2. The lifetime of light bulbs follows an eponential distribution with a hazard rate of. failures per hour of use (a) Find the mean lifetime of a randomly selected light bulb. Let X denote the lifetime of light bulbs,then the hazard rate h() =.. So the survival function S() = ep{- is f() = - d S() = ep{ } ; >. d du } = ep{ }, and the p.d.f. of X X has eponential distribution with λ= E() = λ =. (b) Find the median lifetime of a randomly selected light bulb. Set S(m) = ep{ m } =.5 m = log(.5) m = - log(.5) = 3.3 median lifetime is 3.3 hours (c) What is the probability a light bulb will still function after 2, hours of use? P{X>2} = S(=2) = ep{ 2 } = e 2 =.35 Eercise 2.2 The time in days to development of a tumor for rats eposed to a carcinogen follows a Weibull distribution with α=2 and λ=.. (a) What is the probability a rat will be tumor free at 3 days? 45 days? 6 days? Let X denote the time in days to development of a tumor for rats eposed to a carcinogen, and X has Weibull distribution with α=2 and λ=.. So the p.d.f. of X is f() = 2 ep { 2 }. P{X>t} = 2 t ep { 2 } d = ep { 2 } t = ep { t2 } p{x>3} = ep { 32 } =.47 ; p{x>45} = ep { 452 } =.32 p{x>6} = ep { 62 } =.27

2 (b) What is the mean time to tumor? Because X has Weibull distribution with α=2 and λ=.. E(X) = ( ) 2 Γ( + 2 ) = Γ( 2 ) 2 = (b) Find the hazard rate of the time to tumor appearance at 3 days, 45 days, and 6 days. Because S() = P{X>} = ep { 2 }, hazard rate is h() = - d d logs(). h() = h(3) = 3 =.6 ; h(45) = 45 =.9 ; h(3) = 6 = (d) Find the median time to tumor. Set S(m) =ep { m2 } =.5 m 2 = - log(.5) m= log(.5) tne median time to tumor is days Eercise 2.3 The time to death (in days) following a kidney transplant follows a log logistic distribution with α=.5 and λ=.. (a) Find the 5,, and 5 day survival probabilities for kidney transplantation in patients. Let X denote the time to death following a kidney transplant,because X has log logistic distribution with α=.5 and λ=.,the p.d.f of X is f() =.5.5 (+..5 ) 2 The survival function S() = P{X>} =.5 u.5 (+. u.5 ) 2 du = +. u.5 = P{X>5} = S(5) = =.22 P{X>} = =.9 ; P{X>5} = =.52 (b) Find the median time to death following a kidney transplant. Set S(m) = +. m.5 =.5 +. m.5 = 2 m = the median time to death is days.5 = 2.544

3 (c) Show that the hazard rate is initially increasing and, then, decreasing over time. Find the time at which the hazard rate changes from increasing to decreasing. Hazard rate h()= f() S() =.5.5 (+..5 ) 2 ( +..5 ) = h () = d d =.5.5 (.5..5 ) (+..5 ) 2 to solve h () =.5.5 (.5..5 ) = = = 5 (+..5 ) 2 = We can know if > 3.572,then h () will be negative;similarly,if < 3.572,then h () will be positive. h() is increasing if < < ; h() is decreasing if > And the change point is = (d) Find the mean time to death. f() =.5.5 E() =.5.5 (+..5 ) 2 (+..5 ) 2 d let u = du =.5.5 d = (+..5 ) 2 ( ) u <u<.5 E() = ( u ) du = ( ) u.5 du = u.5( u).5 du by beta integral E() = u.5.5 ( u) du the mean time to death is 52.2 days = Γ(.5.5 )Γ(2.5.5 ) Γ( ) = Eercise 2.4 A model for lifetimes, with a bathtub-shaped hazard rate, is the eponential power distribution with survival function S()=ep { ep [(λ) α ]} (a) If α=.5, show that the hazard rate has a bathtub shape and find the time at which the hazard rate changes from decreasing to increasing. h() = - d d logs() = - d d (- ep [(λ) 2]) = 2 (λ ) 2 ep [(λ) 2] h () = d d 2 (λ ) 2 ep [(λ) 2]= 4 λ ep [(λ) 2] + 4 λ ep [(λ) 2]

4 let h () = 4 λ ep [(λ) 2] = 4 λ ep [(λ) 2] = λ if > λ,then h () > h() is increasing ; if < λ,then h () < h() is decreasing the change point is = λ (b) If α= 2, show that the hazard rate of is monotone increasing. h() = - d logs() = - d (- ep d d [(λ)2 ]) = 2 λ 2 ep [(λ) 2 ] h () = d d 2 λ2 ep [(λ) 2 ] = 2 λ 2 ep[(λ) 2 ] + 4 λ 4 2 ep [(λ) 2 ] because 2 λ 2 ep[(λ) 2 ] > and 4 λ 4 2 ep [(λ) 2 ] > for all and λ, we can obtain that h () > fot all andλifα=2 h() is monotone increasing the hazard rate of is monotone increasing. Eercise 2.5 The time to death (in days) after an autologous bone marrow transplant, follows a log normal distribution with μ= 3.77 and σ = Find (a) the mean and median times to death; Let X denote the time to death after an autologous bone marrow transplant,because, X has log normal distribution, and let X = e Y where Y~N(μ=3.77, σ=2.84) the mean is E(X) E(X) = E(e Y ) = ep {μ + 2 σ2 } = ep { } = S() = P{X>} = P{e Y > } = P{Y>log} = P{ Y 3.77 > log Where Z~N(,), now denoteφ(t) = P{Z<t},so S() = -Φ( log 3.77 ) 2.84 Let S(m) = -Φ( logm 3.77 ) = Φ(logm 3.77 ) = logm m = ep{3.77} = the median time is days } = P{Z > log 3.77 } 2.84 =Φ - ( 2 )= (b) the probability a individual survives, 2, and 3 days following a transplant P{X>}=S()=-Φ( log 3.77 )=.247(by check the normal distribution table) 2.84 P{X>2}=-Φ( log )=.54 P{X>3}=-Φ( log )=

5 y (c) plot the hazard rate of the time to death and interpret the shape of this function. h()= - d d logs() = - d d where φ(. ) is p.d.f. of Z graph: log (-Φ(log 3.77)) = φ(log 3.77 ) 2.84 Φ( log 3.77 ) R-code: =seq(,5,.); y=ep(-((log()-3.77)/2.84)^2/2)/(sqrt(2*pi)*2.48**(-pnorm((log()-3.7 7)/2.84,,))); plot(,y,type="l",lim=c(,5),ylim=c(,.8)) h() is increasing,when > ,h() is decreasing. Eercise 2.8 The battery life of an internal pacemaker, in years, follows a Pareto distribution with θ= 4 and λ= 5. (a) What is the probability the battery will survive for at least years? Let X denote the battery life,because X has Pareto distribution,the p.d.f of X is given by f() = ; > 5 P{X > } = d = =.625

6 (b) What is the mean time to battery failure? E(X) = d = d 5 the mean time to battery failure is years. = = (c) If the battery is scheduled to be replaced at the time t, at which 99% of all batteries have yet to fail (that is, at t so that Pr (X > t) =.99), find t. P{X > t } = t d = t = 5 4 t -4 = = t 4 4 t = = (years) Eercise 2. A model used in the construction of life tables is a piecewise, constant hazard rate model. Here the time ais is divided into k intervals, [τi-, τi ), i =,..., k, with τ = and τk =. The hazard rate on the ith interval is a constant value, θi; that is θ < τ θ 2 τ < τ 2 h() = θ k τ k 2 < τ k { θ k τ k (a) Find the survival function for this model. For < τ S() = ep { h(u) du}=ep { h(u) du} = ep{ θ } For τ < τ 2 S() = ep { h(u) du} ep{ h(u) du} τ For τ k 2 < τ k =ep{ [θ τ + θ 2 ( τ )]} τ k 2 S() = ep { h(u) du} ep{ h(u) du} ep{ h(u) du} τ k 2 τ k 3 = ep { θ k ( τ k 2 )} ep{ θ k 2 (τ k 2 τ k 3 )} ep{ θ τ } ep{ θ } For < τ τ τ So S() = ep{ [θ τ + θ 2 ( τ )]} For τ < τ 2 ep{ [θ τ + + θ k ( τ k 2 )]} For τ k 2 < τ k { ep{ [θ τ + + θ k ( τ k )]} For τ k

7 (b) Find the mean residual-life function. Mean residual-life function e() = S(t) dt S() For < τ e() = ep{ θ t}dt ep{ θ } = τ ep{ θ t}dt+ S(t) dt τ ep{ θ } = ep{ θ t} τ ep{ θ } θ + S(t) dt τ ep{ θ } = ep{ τ θ }+ep{ θ } + θ ep{ θ } S(t) dt τ ep{ θ } For τ < τ 2 e() = ep{ [θ τ +θ 2 (t τ )]} dt ep{ [θ τ +θ 2 ( τ )]} = τ2 ep{ [θ τ +θ 2 (t τ )]}dt+ S(t) dt τ2 ep{ [θ τ +θ 2 ( τ )]} = ep{ [θ τ +θ 2 (t τ )]} ep{ [θ τ +θ 2 ( τ )]} θ 2 τ 2 + S(t) dt τ2 ep{ [θ τ +θ 2 ( τ )]} = ep{ [θ τ +θ 2 (τ 2 τ )]}+ep{ [θ τ +θ 2 ( τ )]}+θ 2 S(t) dt τ2 θ 2 ep{ [θ τ +θ 2 ( τ )]} For τ k 2 < τ k e() = ep{ [θ τ + +θ k (t τ k 2 )]} dt ep{ [θ τ + +θ k ( τ k 2 )]} = τ k ep{ [θ τ + +θ k (t τ k 2 )]} dt+ S(t) dt τ k ep{ [θ τ + +θ k ( τ k 2 )]} = = ep{ [θ τ + +θ k (t τ k 2 )]} ep{ [θ τ + +θ k ( τ k 2 )]} θ k τ k + S(t) dt τ k ep{ [θ τ + +θ k ( τ k 2 )]} ep{ [θ τ + +θ k (τ k τ k 2 )]} +ep{ [θ τ + +θ k ( τ k 2 )]} +θ k S(t) dt τ k θ k ep{ [θ τ + +θ k ( τ k 2 )]} { e() = ep{ τ θ }+ep{ θ } S(t) dt τ + θ ep{ θ } ep{ θ } ep{ [θ τ +θ 2 (τ 2 τ )]}+ep{ [θ τ +θ 2 ( τ )]}+θ 2 S(t) dt τ2 θ 2 ep{ [θ τ +θ 2 ( τ )]} ep{ [θ τ + +θ k (τ k τ k 2 )]} +ep{ [θ τ + +θ k ( τ k 2 )]} +θ k τ S(t) dt k θ k ep{ [θ τ + +θ k ( τ k 2 )]} ep{ [θ τ + +θ k ( τ k )]} θ k ep{ [θ τ + +θ k ( τ k )]} For < τ = θ k For τ k For τ < τ 2 For τ k 2 < τ k

8 (c) Find the median residual-life function. Set S(m) =.5,if τ i 2 m < τ i ;<i<k ep{ [θ τ + + θ k (m τ k 2 )]}=.5 ep{ [θ k (m τ k 2 )]} = ep{[θ τ + + θ k 2 (τ k 2 τ k 3 )]}.5 m = θ k {[θ τ + + θ k 2 (τ k 2 τ k 3 )] + log(.5)}+τ k 2 Eercise 2. In some applications, a third parameter, called a guarantee time, is included in the models discussed in this chapter. This parameter ϕ is the smallest time at which a failure could occur. The survival function of the three-parameter Weibull if < ϕ distribution is given by S() ={ ep{ λ( φ) α } if ϕ (a) Find the hazard rate and the density function of the three- parameter Weibull distribution h() = d logs() = { d αλ( φ) α because h() = f() S() f() = h() S() so f() = { αλ( φ) α ep{ λ( φ) α } if < ϕ if ϕ if < ϕ if ϕ (b) Suppose that the survival time X follows a three-parameter Weibull distribution with α =, λ =.75 and ϕ =. Find the mean and median lifetimes E(X) = ϕ αλ( φ) α ep{ λ( φ) α } d let u = - φ <u E(X) = (u + φ) αλu α ep{ λu α } du = αλu α ep{ λu α } du + φ αλu α ep{ λu α } du (by gamma integral) = λ α Γ ( + α ) + ϕ =.75 Γ( + ) + = Set S(m) =ep{.75(m ) } =.5 m = log(.5) m = the median lifetimes is

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