Probability and Random Processes (Part II) 1. If the variance σ x of d(n) = x(n) x(n 1) is one-tenth the variance σ x of a stationary zero-mean discrete-time signal x(n), then the normalized autocorrelation function R xx (k)/σ x at k = 1 is (a) 0.95 (b) 0.90 Soln. The variance σ X = E[(X μ X ) ] Where μ X (mean value) = 0 (c) 0.10 (d) 0.05 [GATE 00: Marks] σ d = E[{X(n) X(n 1)} ] σ d = E[X(n)] + E[X(n 1)] E[X(n)X(n 1)] σ X 10 = σ X + σ X R XX (1) σ X = 0σ X 0R XX (1) R XX σ X = 19 = 0. 95 0 Option (a). Let Y and Z be the random variables obtained by sampling X(t) at t = and t = 4 respectively. Let W = Y Z. The variance of W is (a) 13.36 (c).64 (b) 9.36 (d) 8.00 [GATE 003: Marks]
Soln. W = Y Z Given R XX(τ) = 4(e 0. τ + 1) Variance[W] = E[Y Z] σ W = E[Y ] + E[Z ] E[YZ] Y and Z are samples of X(t) at t = and t = 4 E[Y ] = E[X ()] = R XX(0) = 4[e. 0 + 1] = 8 E[Z ] = E[X (4)] = 4[e 0. 0 + 1] = 8 E[YZ] = R XX() = 4[e 0.(4 ) + 1] = 6. 68 σ W Option (c) = 8 + 8 6. 68 =. 64 3. The distribution function F X (x) of a random variable X is shown in the figure. The probability that X = 1 is F x (X) (a) Zero (b) 0.5 1.0 0.55 0.5-0 1 3 X (c) 0.55 (d) 0.30 [GATE 004: 1 Mark] Soln. The probability that X = 1 = F X (x = 1 + ) F X (x = 1 ) Option (d) P(x = 1) = 0. 55 0. 5 = 0. 30
4. If E denotes expectation, the variance of a random variable X is given by (a) E[X ] E [X] (c) E[X ] (b) E[X ] + E [X] (d) E [X] [GATE 007: 1 Mark] Soln. The variance of random variable X σ X = E[(X μ X ) ] Where μ X is the mean value = E[X] σ X = E[X ] + E[μ X ] μ X E[X] = E[X ] + μ X μ X μ X = E[X ] μ X = mean square value square of mean value Option (a) 5. If R(τ) is the auto-correlation function of a real, wide-sense stationary random process, then which of the following is NOT true? (a) R(τ) = R( τ) (b) R(τ) R(0) (c) R(τ) = R( τ) (d) The mean square value of the process is R(0) [GATE 007: 1 Mark] Soln. If all the statistical properties of a random process are independent of time, it is known as stationary process. The autocorrelation function is the measure of similarity of a function with it s delayed replica. 1 R(τ) = lim T T T f(t τ) f (t)dt T
1 for τ = 0, R(0) = lim T T T f(t) f (t)dt T 1 = lim T T T f(t) dt T R(0) is the average power P of the signal. R(τ) = R ( τ)exhibits conjugate symmety R(τ) = R( τ) for real function R(0) R(τ) for all τ R(τ) = R( τ) is not true (since it has even symmetry) Option (c) 6. If S(f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (a) S(0) S(f) (d) S(f)df = 0 (b) S(f) 0 (c) S( f) = S(f) [GATE 007: 1 Mark] Soln. Power spectral density is always positive S(f) 0 Option (b) 7. P X (x) = M exp( x ) + N exp( 3 x )is the probability density function for the real random variable X over the entire X axis M and N are both positive real numbers. The equation relating M and N is (a) M + N = 1 (c) M + N = 1 3 (b) M + 1 N = 1 (d) M + N = 3 3 [GATE 008: Marks]
Soln. P X (x) dx = 1 (M. e x + N. e 3x ) dx = 1 (M. e x + N. e 3x ) dx = 1 0 M. e x N. e 3x + 3 0 0 = 1 M + N 3 = 1 or, M + N 3 = 1 Option (a) 8. A white noise process X(t) with two-sided power spectral density 1 10 10 W/Hz is input to a filter whose magnitude squared response is shown below. x(t) 1 y(t) -10 KHz 10 KHz f The power of the output process y (t) is given by (a) 5 10 7 W (b) 1 10 6 W (c) 10 6 W (d) 1 10 5 W [GATE 009: 1 Mark]
Soln. Power spectral density of white noise at the input of a filter = G i (f) G i (f) = 1 10 10 (W/Hz) PSD at the output of a filter G 0 (f) = H(f) G i (f) = 1 ( 10 103 1) 10 10 = 10 6 W Option (b) Soln. 9. Consider two independent random variables X and Y with identical distributions. The variables X and Y take value 0,1 and with probabilities 1, 1 respectively. What is the conditional 4 and 1 4 probability (X + Y = X Y = 0)? (a) 0 (b) 1/16 P(X = 0) = P(Y = 0) = 1 (c) 1/6 (d) 1 [GATE 009: Marks] P(X = 1) = P(Y = 1) = 1 4 P(X = ) = P(Y = ) = 1 4 P(X Y = 0) = P(X = 0, Y = 0) + P(X = 1, Y = 1) +P(X =, Y = ) = 1 1 + 1 4 1 4 + 1 4 1 4 = 6 16 P(X + Y = ) = P(X = 1, Y = 1) = 1 4 1 4 = 1 16
P(X + Y = X Y=0 ) = 1 16 6 16 =1/6 Option (c) 10. X (t) is a stationary random process with autocorrelation function R X (τ) = exp( πτ ) this process is passed through the system below. The power spectral density of the output process Y(t) is X(t) + _ Σ Y(t) Soln. (a) (4π f + 1) exp( πf ) (b) (4π f 1) exp( πf ) (c) (4π f + 1) exp( πf) (d) (4π f 1) exp( πf) Y(f) = jπf X(f) X(f) [GATE 011: Marks] PSD S Y (f) = (jπf 1) S X (f) S X (f) = FT{R X (τ)} = FT(e πτ ) = e πf S Y (f) = (4π f + 1)e πf Option (a)
11. Two independent random variables X and Y are uniformly distributed in the interval [ 1,1]. The probability that max [X, Y] is less than 1/ is (a) 3/4 (c) 1/4 (b) 9/16 (d) /3 [GATE 01: 1 Mark] Soln. y (-1,1) 1 (1,1) 1/ -1 1/ 1 x (-1, -1) -1 (1, -1) 1 X 1 and 1 Y 1 The region in which maximum of [X, Y] is less than 1/ is shown as shaded region inside the rectangle. P [max(x, Y) < 1 Area of shaded region ] = Area of entire region = 3 3 = 9 4 4 = 9 16 Option (b)
1. A power spectral density of a real process X (t) for positive frequencies is shown below. The values of [E[X (t)] and E[X(t)] ] respectively are 6 (a) 6000/π, 0 (b) 6400/π, 0 0 9 10 11 (c) 6400/π, 0/(π ) (d) 6000/π, 0/(π ) [GATE 01: 1 Mark] Soln. The mean square value of a stationary process equals the total area under the graph of power spectral density E[X (t)] = S X (f)df = 1 π S X(ω) dω = π S X(ω) dω 0 = 1 [area under the triangle π + integration under delta function] = 1 π [ (1 1 6 103 ) + 400]
= 6400 π E[X(t)] is the absolute value of mean of signal X(t) which is also equal to value of X(ω) at ω = 0 From PSD S X (ω) ω=0 = 0 X(ω) = 0 X(ω) = 0 Option (b) 13. Let U and V be two independent zero mean Gaussian random variables of variances 1 and 1 respectively. The probability P(3V U) is 4 9 (a) 4/9 (c) /3 (b) 1/ (d) 5/9 [GATE 013: Marks] Soln. pdf of W P(3V U) = P(3V U 0) = P(W 0) W = 3V U
W is the Gaussian Variable with zero mean having pdf curve as shown below P(W 0) = 1 (area under the curve from 0 to ) Option (b) 14. Let X 1, X, and X 3 be independent and identically distributed random variables with the uniform distribution on [0,1]. The probability P{X 1 is the largest} is Soln. Probability P[X 1 ] = P[X ] = P[X 3 ] P 1 + P + P 3 = 1 P(X 1 ) + P(X ) + P(X 3 ) = 1 3P(X 1 ) = 1 [GATE 014: 1 Mark] P(X 1 ) = 1 3 15. Let X be a real-valued random variable with E[X] and E[X ] denoting the mean values of X and X, respectively. The relation which always holds (a) (E[X]) > E[X ] (b) E[X ] (E[X]) Soln. variance σ X = E[X ] m X X m X = mean square value square of mean value σ X = E[X ] [E(X)] (c) E[X ] = (E[X]) (d) E[X] > (E[X]) Variance is always positive so E[X ] [E(X) ] And can be zero Option (b) [GATE 014: Marks]
16. Consider a random process X(t) = sin(πt + φ), where the random phase ϕ is uniformly distributed in the interval[0,π]. The autocorrelation E[X(t 1 )X(t )] is (a) cos[π(t 1 + t )] (b) sin[π(t 1 t )] (c) sin[π(t 1 + t )] (d) cos[π(t 1 t )] [GATE 014: Marks] Soln. E[X(t 1 ) X(t )] = E[A sin(πt 1 + φ) A sin(πt + φ)] = A E[cos π(t 1 t ) cos π(t 1 + t + φ)] = A cos π(t 1 t ) E[cos π(t 1 + t + φ)] = 0 Option (d) Soln. 17. Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E[X] is [GATE 014: 1 Mark] E[X] = 1 + 3 + 5 + (n 1) 50 Where n = 50 = n 50 = 50 18. The input to a 1-bit quantizer is a random variable X with pdf f X (x) = e x for x 0 andf X (x) = 0 for x < 0. For outputs to be of equal probability, the quantizer threshold should be [GATE 014: Marks]
Soln. The input to a 1-bit quantizer is a random variable X with pdf f X (x) = e x for x 0 And f X (x) = 0 for x < 0 let Vthr be the quantizer threshold V thr e x dx = e x dx V thr V thr = e x 0 dx = e x V thr dx f X (x) = 0 for x < 0 e x V thr = e x 0 V thr ( e V thr + e 0 ) = (0 e V thr) e V thr = 1 V thr = ln ( 1 ) = ( 0. 693) V thr = 0. 693 = 0.346