ECE 302 Division 2 Exam 2 Solutions, 11/4/2009.

Transcription

1 NAME: ECE 32 Division 2 Exam 2 Solutions, /4/29. You will be required to show your student ID during the exam. This is a closed-book exam. A formula sheet is provided. No calculators are allowed. Total number of points: 5. This exam counts for 3% of your final grade. You have hour and 45 minutes to complete FIVE problems. Be sure to fully and clearly explain all your answers. There will not be any discussion of grades. All re-grade requests must be submitted in writing, as stated in the course information handout. Problem Points Score TOTAL 5

2 Some random variables and their distributions: Random variable PMF or PDF Mean Variance Bernoulli p for k = ; p for k =. p p( p) Discrete uniform n, k = k +,k + 2,...,k + n k + n+ 2 Geometric ( p) k p, k =,2,3,... p Binomial n k n 2 2 p 2 p ( p) n k p k, k =,,...,n pn np( p) Continuous uniform b a, a x b Exponential λe λx, x λ λ 2 Normal (Gaussian) b+a 2 2πσ e (x µ)2 2σ 2 µ σ 2 (b a) 2 2 Lognormal Cauchy Rayleigh Pareto where x (ln x µ) 2 2πσ e 2σ 2, x > e µ+σ2 /2 (e σ2 )e 2µ+σ2 [ πγ + ( x x γ x 2 σ 2xe 2σ 2, x a ca x a+, x c, ) 2 ] undefined undefined ( n k ) = n! (n k)!k!. σ π 2 4 π 2 σ2 ac a, a > ac 2 (a ) 2 (a 2), a > 2 2

3 Problem (5 points). The CDF of a continuous random variable X is:, x < F X (x) = x/2, x 2, x > 2 a. (2 points). Find the PDF of X. Solution. Differentiating the CDF, we get:, x < f X (x) = /2, < x < 2, x > 2 Thus, X is a continuous random variable, uniformly distributed between and 2. Note that the derivative of the CDF does not exist at x = and at x = 2; however, we would not deduct any points for including x = and/or x = 2 in the above expression. b. (2 points). Find E[X]. Solution. E[X] = ( + 2)/2 =. c. (2 points). Find var(x). Solution. var(x) = (2 ) 2 /2 = /3. d. (3 points). Find E[X 3 ]. Solution. E[X 3 ] = 2 x 3 /2dx = x4 8 2 = 6/8 = 2. e. (3 points). Find the conditional PDF of X given that < X < /2. Solution. Given that < X < /2, the conditional PDF of X is zero outside of the region < x < /2. Inside the region, it is equal to the PDF of X divided by the probability of < X < /2. For < x < /2, the PDF is /2 as found in Part a. P( < X < /2) = /2 /2dx = /4. Hence, we get:, x < f X <X</2 (x) = 2, < x < /2, x > /2 An alternative way of arriving at the same answer is simply to notice that, since the PDF of X is uniform over x 2, the conditional PDF of X given < X < /2 must be uniform over < x < /2. f. (3 points). Find the PDF of the random variable Y = X. Solution. First, find the CDF: F Y (y) = P(Y y) = P(X y) = P(X y + ) = F X (y + ) 3

4 Then differentiate to find the PDF: f Y (y) = d dy F X(y + ) = f X (y + ) =, y < /2, < y <, y > As a quick check, note that, since X is between and 2 with probability, Y = X must be between and with probability. An alternative method is to note that by subtracting from a random variable, we move its PDF to the left by. Hence, the PDF of Y is uniform from to. Fully and clearly substantiate all your answers. 4

5 Problem 2 (5 points). Consider the following joint PDF of two continuous random variables, X and Y : {, x and y f XY (x,y) =, otherwise. a. (3 points) Find the marginal PDF f X (x). Solution. To get the marginal density of X, integrate the joint density: f X (x) = = So X is uniformly distributed on [,]. f XY (x,y)dy (u(x) u(x ))(u(y) u(y ))dy = (u(x) u(x )) = (u(x) u(x )) = u(x) u(x ). (u(y) u(y ))dy b. (3 points) Specify all values y for which the conditional PDF f X Y (x y) exists. For all these values, find f X Y (x y). Solution. The conditional PDF f X Y (x y) exists for all values of y for which f Y (y) >. The marginal distribution of Y is uniform over [,]. This can be verified either by integrating the joint density over x, or by noticing that the joint density is symmetric (in the sense that swapping X and Y does not change the joint density) and therefore Y must have exactly the same marginal density as the marginal density of X, which was found in Part a. Thus, f X Y (x y) exists for y. For these values of y, dy f X Y (x y) = f X,Y (x,y) f Y (y) = u(x) u(x ), i.e., the conditional distribution of X given any value Y = y [, ] is uniform between and. c. (3 points) Are X and Y independent? Fully substantiate your answer. No partial credit will be given for the correct answer without a full and correct justification. Solution. Since f X (x)f Y (y) = f XY (x,y), X and Y are independent. d. (2 points) Find the probability P(X > 3/4). Solution. This probability is the integral from 3/4 to of the uniform density found in Part a, which is /4. e. (2 points) Find the probability P(X < Y ). Solution. Since the joint density of X and Y is symmetric, P(X < Y ) = P(X > Y ). But P(X < Y ) + P(X > Y ) =, and hence P(X < Y ) = /2. 5

6 f. (2 points) Find the conditional probability P(X > 3/4 Y < /2). Solution. Since the two random variables are independent, the conditional probability of X > 3/4 given Y < /2 is the same as the unconditional probability of X > 3/4, which was found to be /4 in Part d. Fully and clearly substantiate all your answers. 6

7 Problem 3 ( points). Y is a random variable with mean and variance 2. Z is a random variable with mean and variance 8. The correlation coefficient of Y and Z is ρ(y,z) =.5. a. (2 points). Find E[Y + Z]. Solution. E[Y + Z] = E[Y ] + E[Z] =. b. (2 points). Find E[Z 2 + ]. Solution. E[Z 2 + ] = E[Z 2 ] + = var(z) + (E[Z]) 2 + = = 9. c. (3 points). Find E[Y Z]. Solution. cov(y,z) = ρ(y,z)σ Y σ Z = = 3. Therefore, d. (3 points). Find var(y + Z). E[Y Z] = cov(y,z) + E[Y ]E[Z] = cov(y,z) = 3. Solution. var(y + Z) = var(y ) + var(z) + 2cov(Y,Z) = = 26. Fully and clearly substantiate all your answers. 7

8 Problem 4 (5 points). X is a Gaussian random variable with mean. a. (2 points). Find the probability of the event {X is an integer}. Solution. P(X = n) = n n f X(x)dx = for any n, and therefore P(X is an integer) = n= P(X = n) =. b. (3 points). List the following probabilities in the increasing order, from smallest to largest: P( X ), P( X ), P( X 2), P( X 2). Fully substantiate your answer. No partial credit will be given for the correct answer without a full and correct proof. Solution. Since a Gaussian PDF with zero mean is an even function, P( X ) = f X (x)dx = f X ( x)dx = Since a Gaussian PDF with zero mean is monotonically decreasing for x, f X (x)dx = P( X ). () P( X ) > P( X 2). (2) Since the event { X 2} is the union of the events { X } and { X 2} each of which has nonzero probability, we have: Finally, P( X ) < P( X 2). P( X 2) = P( X ) + P( X 2) Eq. (2) < 2P( X ) Eq. () = P( X ) + P( X ) = P( X ). Putting everything together, we have the following answer: P( X 2) < P( X ) < P( X 2) < P( X ) Fully and clearly substantiate all your answers. 8

9 d(r 2 ) d(r 2 ) = d(r ) + ( x) x x d(r 2 ) = d(r ) ( x) x x d(r ) Figure : For those who finished Problems, 2, 3, and 4 quickly, it may be beneficial to go back and carefully check your solutions before proceeding to Problem 5. Problem 5 (5 points). Let c be a fixed point in a plane, and let C be a circle of length centered at c. In other words, C is the set of all points in the plane whose distance from c is /(2π). (Note that C does NOT include its interior, i.e., the points whose distance from c is smaller than /(2π).) Points R and R 2 are each uniformly distributed over the circle, and are independent. (The fact that R and R 2 are uniformly distributed over a circle of unit length means that, for any arc A whose length is p, the following identities hold: P(R A) = p and P(R 2 A) = p.) Point r is a non-random, fixed point on the circle. We let R R 2 be the arc with endpoints R and R 2, obtained by traversing the circle clockwise from R to R 2. We let R rr 2 be the arc with endpoints R and R 2 that contains the point r: { R R rr 2 = R 2, if r R R 2 R 2 R, if r R R 2 a. (2 points). The random variable L is defined as the length of the arc R R 2. Find E[L]. Solution. Let L be the length of the arc R 2 R. Due to symmetry between R and R 2, E[L ] = E[L]. But the sum of the lengths of the two arcs is equal to the length of the circle: L + L =. Hence, we have E[L] + E[L ] =, and E[L] = /2. b. (3 points). The random variable L is defined as the length of the arc R rr 2. Find E[L ]. Solution. For any point x C, let d(x) be the length of the arc rx, i.e., the distance between r and x along the circle in the clockwise direction. Since R and R 2 are independent uniform points on the circle, d(r ) and d(r 2 ) are independent random variables, each uniformly distributed 9

10 between and. Then the length of the arc between R and R 2 that does not contain r, is d(r ) d(r 2 ). The length of the arc that contains r is L = d(r ) d(r 2 ). Since both d(r ) and d(r 2 ) are between and with probability, L is also between and with probability. To get its probability density function, we first find its CDF on the interval [, ], i.e., P(L x): P(L x) = P( d(r ) d(r 2 ) x) = P( d(r ) d(r 2 ) x) = P( d(r ) d(r 2 ) x) = P(d(R ) d(r 2 ) x) + P(d(R 2 ) d(r ) x) The joint distribution of d(r ) and d(r 2 ) is uniform over the square [,] [,] shown in Fig.. Since the square has unit area, the probability of any set is the area of its intersection with the square. The event d(r ) d(r 2 ) x corresponds to the triangle in the lower-right corner of the square in Fig.. Therefore, its probability is the area of the triangle, which is x 2 /2. Similarly, the probability of the event d(r 2 ) d(r ) x is the area of the upper-left triangle, which is also x 2 /2. We therefore have: P(L x) = x 2. Differentiating, we get the PDF for L on the interval [,]: 2x. The expectation is therefore equal to: E[L ] = x 2xdx = 2x3 3 It may seem counterintuitive that the answers to Parts a and b are different. The reason that E[L ] > /2 is that the arc connecting R and R 2 and covering an arbitrary point r is likely to be larger than the arc connecting R and R 2 that does not cover r. = 2 3. Fully and clearly substantiate all your answers.