EE 302: Probabilistic Methods in Electrical Engineering

Transcription

1 EE : Probabilistic Methods in Electrical Engineering Print Name: Solution (//6 --sk) Test II : Chapters.5 4 //98, : PM Write down your name on each paper. Read every question carefully and solve each problem in a legible and ordered manner. Make sure you write down all your answers without skipping details. However, you could use short cuts provided you have a valid argument and you clearly state it on paper. I don t give credit for wrong answers and partial credit will be given only when sufficient details have been provided.. Consider the following probability density function which is also shown in Figure : f X (x) kx x < k x < 4 k(6 x) 4 x 6 otherwise Compute the following ( points):. (a) Find the value of k which makes f X (x) a valid probability density function. ( points). (b) Find the cumulative distribution function F X (x). Use the space indicated for that in Figure. ( points). (a) f X (x)dx () 4 6 kxdx + kdx + k(6 x)dx () 4 k + 4k + k 8k () so k /8. (b) First we check for discontinuities at the ends of the intervals: k (4) k k (5) k k(6 4) (6) k(6 6). (7) Finding no discontinuities we need only integrate f X (x) to determine F X (x). F X (x) x kydy y [, ) k + x kdy y [, 4) 6k + x 4 k(6 y)dy y [4, 6] y > 6, (8)

2 EE : Probabilistic Methods in Electrical Engineering which yields F X (x) x < kx / x [, ) k(x ) x [, 4) kx / + 6kx k x [4, 6] x > 6. (9)

3 EE : Probabilistic Methods in Electrical Engineering. Assume that X and X are coded scores on two intelligence tests, and the probability density function of [X,X ] is given by f X,X (x,x ) { 6x x x, x otherwise Compute the following ( points): (a) Find the E(X x ). (5 points). (b) Find the variance of the score on test No. given the score on test No., σ X x E(X x ) µ X x. (5 points). (c) Find the correlation coefficient between the two coded scores. ( points). (a) We will need the conditional PDF f X X (x x ) and to find that we ll need the marginal f X (x ), so let s calculate that first. f X (x ) f X,X (x, x )dx 6x x dx 6x x x. () Thus the conditional PDF is and f X X (x x ) f X,X (x, x ) f X (x ) E[X X ] { x x [, ] and x [, ] otherwise () x f X X (x x )dx /. () (b) To find σ X X we will need E[X X ] so we calculate that first. Thus E[X X ] σ X X x f X X (x x )dx /. () E[X X ] (E[X X ]) / 4/9 8. (4) (c) We will need f X (x ) and f X (x ) in order to determine the correlation coefficient. We already determined f X (x ), and f X (x ) f X,X (x, x )dx 6x x dx x x x. (5) At this point, we providentially notice that the joint PDF is the product of the marginal PDF s, hence the random variables X and X are independent, hence their covariance is zero, hence the correlation coefficient is zero, thereby saving ourselves a lot of computation.

4 EE : Probabilistic Methods in Electrical Engineering 4. The Bernoulli distribution function is given as P X (x) { p x q x < p <, x or otherwise Compute the following ( points): (a) Find the moment generating function Ψ X (u). ( points). (b) Find the mean and variance using the results from part (a). ( points). (a) The moment generating function is φ X (s)e sx e s P X () + e s P X () (6) p ( p) + e s p ( p) (7) ( p) + pe s. (8) (b) E[X] d ds (( p) + pes ) pe p. (9) s Similarly E[X ] d ds (( p) + pes ) pe p () s so V ar[x] p p p( p).

5 EE : Probabilistic Methods in Electrical Engineering 5 4. Two discrete random variables X and X have joint probability distribution function as given in the following Table: x j... xi 4 P X (x j ) 4 4 P X (x i ) Px (x) Compute the following: ( points). (a) The marginal distributions P X (x i ) and P X (x j ). ( points). (b) Find the correlation coefficient ρ between X and X. ( points). (a) Summing columns we obtain P X (x ) 7/ x {,, } 8/ x / x 4 otherwise. Summing rows we obtain 8/ x 9/ x 6/ x P X (x ) 4/ x / x 4 otherwise. () () (b) I don t see any way of avoiding the calculations here. As a simple check, showing that P X,X (x, x ) P X ()P X ()) will show that the random

6 EE : Probabilistic Methods in Electrical Engineering 6 variables X and X are not independent. I found so E[X ] 7 () + 7 () + 8 () + 7 () + 48 (4) E[X ] 8 () + 9 () + 6 () + 4 () + 45 (4) E[X ] 7 () + 7 () + 8 () + 7 () + (4) 8 E[X ] 8 () + 9 () + 6 () + 4 () + (4) ( ( ) E[X X ] () + () + () ) ( ( ) () + () ) ( ) 4 + () ( () (4) (5) (6) ) + (7) 44, (8) Cov[X, X ] E[X X ] E[X ]E[X ] 44 48(45) () 84 9 V ar[x ] 8 ( ) V ar[x ] ( 45 ) 5 9 and thus the correlation coefficient is (9) () () ρ X,X (5). () We know that the correlation coefficient should be between and, which it is, so this answer is plausible.

7 EE : Probabilistic Methods in Electrical Engineering 7 5. Given the Gaussian probability density function f X (x) ( x µx ) e σx, σx >, < x < πσx and its associated standard Gaussian probability density function f Z (z) π e z, < z <, where the two random variables are related by Z X µx σ x. Answer the following questions: ( points). (a) Given two random variables X and X with respective mean and variance (µ,σ ) and (µ,σ ). Consider the case where µ µ and σ σ. Is the area under the Gaussian curve between (µ σ,µ +σ ) greater, equal, or less than the area under the curve between (µ σ,µ + σ )? Explain your answer. (4 points). The area under the curve between µ σ and µ +σ equals the area under the curve between µ σ and µ +σ. This is true whether or not the means are equal. To see this, note that ( ) ( ) µ +σ e x µ σ (µ + σ ) µ φ µ σ πσ and so does ( ) ( ) µ +σ e x µ σ (µ + σ ) µ φ µ σ πσ σ φ ( ) (µ σ ) µ φ() φ( ) () σ φ σ ( ) (µ σ ) µ φ() φ( ). (4) σ (b) Suppose you are a contestant at some TV show. You are given a pair of cards. The first card is shown to you and contains a value of f X (x), for some unknown value of x. The second card is not shown to you but you are told that it will contain a value X x. You are asked to select a value X x, then only you will know what value of X x the second card contains. What value of X x would you choose in order to determine the variance of the random variable X using the information from both cards? Explain how you would solve this. Refer to the above Gaussian density. (4 points). I don t understand the question, so don t worry if you don t either. (c) Using the Gaussian Table provided, compute the area under the curve between (µ σ, µ+σ), i.e., P X (µ σ X µ + σ) (4 points). No table having been provided, we ll have to skip this one too. How disappointing. :)

8 EE : Probabilistic Methods in Electrical Engineering 8 (d) Are any of the following four statements false? If so, briefly explain why (4 points). () median f X (x)dx.5. () µ x+σ x µ x f X (x)dx.68. () Coefficient of skewness is always γ s < for a Gaussian probability density. (4) The combined effect of Z X µx σ x is to cause a shift towards the origin, thus a change in the shape of the Gaussian distribution, but does not affect the scale. () is false. It would be true if we replaced median by mean or µ x. () is false. µx+σ x µ x σ x f X (x)dx.687. () is not covered in this course. (But in case you are interested, skewness is a measure of asymmetry, hence it is zero for symmetric distributions. Since the Gaussian distribution is symmetric, () is false.) (4) I would say false because shifting the curve so that it is centered at the origin does not change the shape. (e) Towards what function (well known among electrical engineers) does the Gaussian probability density converge to when σ X? Explain why. (4 points). From geometrical considerations, I d say the impulse function (because the area under the curve is one and as σ X decreases the curve becomes taller and narrower). However, to prove this would require mathematical theory well beyond the scope of this course.