ECE32 Exam 2 Version A April 21, 214 1 Name: Solution Score: /1 This exam is closed-book. You must show ALL of your work for full credit. Please read the questions carefully. Please check your answers carefully. Calculators may NOT be used. Please leave fractions as fractions, but simplify them, etc. I do not want the decimal equivalents. Cell phones and other electronic communication devices must be turned off and stowed under your desk. Please do not write on the backs of the exam or additional pages. The instructor will grade only one side of each page. Extra paper is available from the instructor. Please write your name on every page that you would like graded. 1 2 3 4 5 6 7 8 9 1 11
ECE32 Exam 2 Version A April 21, 214 2 1. (1 points) The PDF of the random variable Y is { 2y < y 1, f Y (y) (a) Find the CDF of Y. The CDF is obtained by integrating the PDF, y 2rdr r 2 y. For y the integrand is zero. Thus for y [, 1] r 2 y r2 y y2. For y > 1, the integrand is again zero so r 2 y r2 1 1 The CDF is thus y <, F Y (y) y 2 y < 1, 1 y 1. (b) Find P[1/3 < Y 2/3]. Using the CDF obtained in the previous part, P[1/3 < Y 2/3] F Y (2/3) F Y (1/3) 4/9 1/9 3/9 1/3. (c) Find E[Y ]. The expected value is E[Y ] (d) Find Var[Y ]. yf Y (y)dy y(2y)dy 2y 3 /3 1 2/3. To compute the variance we need E[Y 2 ], which is Thus E[Y ] yf Y (y)dy y 2 (2y)dy 2y 4 /4 1 1/2. Var[Y ] E[Y 2 ] (E[Y ]) 2 1/2 4/9 1/18.
ECE32 Exam 2 Version A April 21, 214 3 2. (12 points) The joint PMF of the random variables X and Y is 1/3 (x, y) (1, 1), 1/6 (x, y) {(2, 1), (2, 2)}, P X,Y (x, y) 1/9 (x, y) {(3, 1), (3, 2), (3, 3)}, (a) Find the marginal PMF P X (x). The marginal is obtained by summing (for discrete random variables) or integrating (for continuous random variables) over all values taken by the other variable(s), so P X (1) y S Y P X,Y (1, y) 1/3, P X (2) y S Y P X,Y (2, y) 2(1/6) 1/3, P X (3) y S Y P X,Y (3, y) 3(1/9) 1/3, so the marginal PMF is 1/3 x 1, 1/3 x 2, P X (x) 1/3 x 3, (b) Find the conditional PMF P Y X (y 2). The conditional PMF is defined on the range of values of y for which x 2 to be P Y X (y 2) P X,Y (x, y). P X (2) Thus, for each of the two values taken by y when x 2 we obtain 1/6 1/2 so 1/3 1/2 y 1, P Y X (y 2) 1/2 y 2, (c) Find E[Y X 2]. The set of values that y takes when X 2 is just {1, 2}, so E[Y X 2] yp Y X (y 2) 1(1/2) + 2(1/2) 3/2. y {1,2}
ECE32 Exam 2 Version A April 21, 214 4 3. (6 points) The joint PMF of the random variables X and Y is 1/3 (x, y) (1, 1), 1/6 (x, y) {(2, 1), (2, 2)}, P X,Y (x, y) 1/9 (x, y) {(3, 1), (3, 2), (3, 3)}, Let B {(1, 1), (2, 2), (3, 3)}. (a) Find P[B]. The probability of the event is the sum of the probabilities of the outcomes that comprise the event, so P [B] 1/3 + 1/6 + 1/9 11/18. (b) Find the conditional PMF P X,Y B (x, y). For (x, y) B, so P X,Y B (x, y) P X,Y B (x, y) P (X,Y )(x, y) P (B) (1/3)/(11/18) (x, y) (1, 1), (1/6)/(11/18) (x, y) (2, 2), (1/9)/(11/18) (x, y) (3, 3), otherwise, 6/11 (x, y) (1, 1), 3/11 (x, y) (2, 2), 2/11 (x, y) (3, 3),
ECE32 Exam 2 Version A April 21, 214 5 4. (1) The random vector X has PDF { 1e 5x 1 2x 2 x f X (x) 1, x 2, (a) Find the marginal PDF f x1 (x 1 ). The marginal is obtained by integrating the PDF over x 2, to obtain on S X1, Thus f x1 (x 1 ) f x1 (x 1 ) f X (x)dx 2 1e 5x 1 2x 2 dx 2 5e 5x 1 2e 2x 2 dx 2 5e 5x 1 e 2x 2 5e 5x 1 ( ( 1)). { 5e 5x 1 x 1 To verify that this is a valid PDF, we first note that it is i. non-negative, as required, and ii. integrates over x 1 to one, as shown below: f x1 (x 1 )dx 1 5e 5x 1 dx 1 e 5x 1 ( 1) 1. as required. (b) Are X 1 and X 2 independent? (Explanation required.) By symmetry, we see that f x2 (x 2 ) { 2e 2x 2 x 2 Thus f X (x) f x1 (x 1 )f x2 (x 2 ) so X 1 and X 2 are independent. (c) Are X 1 and X 2 uncorrelated? (Explanation required.) Random variables that are independent are uncorrelated. is not necessarily true.) (The reverse
ECE32 Exam 2 Version A April 21, 214 6 5. (1 points) The joint CDF of random variables X and Y is x <, y <, F X,Y (x, y) xy x < 1, y < 1, 1 x 1, y 1. Let Z [X, Y ] T. (a) Find E [Z]. By definition E [Z] E [[ X Y ]] [ E[X] E[Y ] so we will need the marginals f X (x) and f Y (y). These are so f X (x) and by symmetry, Then E[X] and by symmetry, ], xydy xy 2 /2 1 x/2 x [, 1], f X (x) f Y (y) xf X (x)dx { x/2 x [, 1], otherwise, { y/2 y [, 1], x 2 /2 dx x 3 /6 1 1/6, Thus E[Y ] yf Y (y)dy E[Z] y 2 /2 dy y 3 /6 1 1/6, [ 1/6 1/6 ].
ECE32 Exam 2 Version A April 21, 214 7 (b) Find the vector covariance matrix C Z. The vector covariance of the random vector Z is C Z E[(Z E[Z])(Z E[Z]) T ] [[ ] X E[X] [ ] ] E X E[X] Y E[Y ] Y E[Y ] [[ ]] X E 2 E[X] 2 XY XY Y 2 E[Y ] 2 [ ] E[X 2 ] E[XY ] E[XY ] E[Y 2. ] We need the expected values of the squares: and by symmetry, E[X 2 ] E[Y 2 ] x 2 (x/2)dx x 4 /8 1 1/8, y 2 (y/2)dy y 4 /8 1 1/8, and, as calculated in the next problem, E[XY ] 1/4, so [ ] E[X C Z 2 ] E[X] 2 E[XY ] E[X]E[Y ] E[XY ] E[X]E[Y ] E[Y 2 ] E[Y 2 ] [ ] 1/8 (1/6) 2 1/4 (1/6) 2 1/4 (1/6) 2 1/8 (1/6) 2 [ ] 7/72 2/9. 2/9 7/72
ECE32 Exam 2 Version A April 21, 214 8 6. (1 points) The joint CDF of random variables X and Y is x <, y <, F X,Y (x, y) xy x < 1, y < 1, 1 x > 1, y > 1. (a) Find the joint PDF f X,Y (x, y). On [, 1] [, 1], so f X,Y (x, y) 2 x y F X,Y (x, y) 2 x y xy x x 1 { 1 x 1, y 1, f X,Y (x, y) Because X and Y are continuous, P [(X, Y ) (x, y)] for any specific (x, y) lr 2, it would be valid to define the joint PDF to be one on (, 1) (, 1) or [, 1) (, 1) or [, 1) [, 1) etc. instead. (b) Find E[XY ]. E[XY ] x xdx xyf X,Y (x, y)dydx xydxdy ydydx ydy x 2 /2 1 y2 /2 1 1/4.
ECE32 Exam 2 Version A April 21, 214 9 (c) Find the vector cross-covariance matrix C X,Y. The vector cross-covariance of a pair of random vectors X and Y is C XY E [ (X E [X])(Y E [Y]) ] T. Of course here our random vectors are of length one. Thus the cross covariance is a scalar. In the previous problem we found that E[X] E[Y ] 1/6. Above we found that Thus E[XY ] 1/4. C XY E[(X E[X])(Y E[Y ])] E[XY XE[Y ] E[X]Y + E[X]E[Y ]] E[XY ] 2E[X]E[Y ] + E[X]E[Y ] E[XY ] E[X]E[Y ] 1/4 (1/6) 2 2/9.
ECE32 Exam 2 Version A April 21, 214 1 7. (6 points) Find the moment generating function of the random variable X where 1/3 x, 1/2 x 1, P X (x) 1/6 x 2, φ X (s) E [ e sx] x S X e sx f X (x) e s() /3 + e s(1) /2 + e s(2) /6 1/3 + e s /2 + e 2s /6 8. (6 points) Find the moment generating function of the random variable X where { 1/(b a) a < x < b, f X (x) φ X (s) E [ e sx] b e sx a b a dx e sx b s(b a) esb e sa s(b a). a
ECE32 Exam 2 Version A April 21, 214 11 9. (1 points) Let W X 1 + X 2 + + X n, where n is a constant and the X i are iid Poisson(β) random variables with MGF φ X (s) e β(es 1). (a) Find the moment generating function of W. Because the X i are independent φ W (s) n φ Xi (s). i1 Because they are identically distributed Poisson(β), φ Xi (s) φ X (s) i [n]. Thus φ W (s) (φ X (s)) n e nβ(es 1). (b) Find the PMF of W. The moment generating function of W is that of a Poisson(nβ) random variable so { (nβ) w e nβ w, 1, 2,... P W (s) w!
ECE32 Exam 2 Version A April 21, 214 12 1. (1 points) Let W X 1 + X 2 + + X n, where N is a Bernoulli(p) random variable and the X i are iid Poisson(β) random variables with MGF φ X (s) e β(es 1). Find the moment generating function of W. First we need the moment generating function of N. E[e sn ] n S N e sn P N (n) e s(1) p + e s() (1 p) pe s + 1 p Now the moment generating function of W is The natural log is so φ W (s) φ N (ln φ X (s)). ln e β(es 1) β(e s 1), φ W (s) pe β(es 1) + 1 p p ( e β(es 1) 1 ) + 1
ECE32 Exam 2 Version A April 21, 214 13 11. (1 points) The central limit theorem approximation to the CDF of a random variable W n X 1 + X 2 + + X n that is the sum of n iid random variables X i, each with E[X i ] µ X and Var[X] σ 2 X is (a) What are a, b, and c? ( ) a b F Wn (w) Φ. c The central limit theorem approximation is ( ) w nµ X F Wn (w) Φ.. nσ 2 X (b) What does iid stand for? independent identically distributed Note that order is important. Identically modifies distributed not independent.