Assignment 6 Solutions Iii

ECE 316 – Probability Theory and Random Processes Chapter 6 Solutions (Part 3) Xinxin Fan

Problems 42. The joint density function of X and Y is given by f (x, y) = xe−x(y+1) x > 0, y > 0 (a) Find the conditional density of X, given Y = y, and that of Y , given X = x. (b) Find the density function of Z = XY . Solution. (a) Using the definition of the conditional PDF, we have fX|Y (x|y) =

xe−x(y+1) R∞ = (y + 1)2 xe−x(y+1) , x > 0 −x(y+1) dx xe 0

fY |X (y|x) =

xe−x(y+1) R∞ = xe−xy , −x(y+1) dy xe 0

y>0

(b) We first compute the CDF as follows: Z

∞ Z a/x

P {XY < a} =

Z −x(y+1)

xe 0

∞

dydx =

0

(1 − e−a )e−x dx = 1 − e−a .

0

Differentiation yields fXY (a) = e−a , a > 0. 43. The joint density of X and Y is f (x, y) = c(x2 − y 2 )e−x

0 ≤ x < ∞, −x ≤ y ≤ x

Find the conditional distribution of Y , given X = x. Solution. We first find the conditional PDF as follows: (x2 − y 2 )e−x 3 = 3 (x2 − y 2 ), −x < y < x. 2 2 −x 4x −x (x − y )e dy

fY |X (y|x) = R x

With the conditional PDF, we can compute the conditional CDF from the definition: µ ¶ Z y 3 3 y 3 2x3 FY |X (y|x) = 3 (x2 − y 2 )dy = 3 x2 y − , −x < y < x. + 4x −x 4x 3 3 1

Theoretical Exercises 14. Suppose that X and Y are independent geometric random variables with the same parameter p. (a) Without any computations, what do you think is the value of P {X = i | X + Y = n}? (b) Verify your conjecture in part (a). Solution. (a) Imagine that we continually flip a coin having probability p of coming up heads. Let X be the time that the first head occurs and Y be the number of times that we continue flipping the coin until the second head occurs. Given the condition that X + Y = n, we know that the second head occurs on the nth flip. Note that X and Y are independent. Therefore, the first head can occur at any step of the first n − 1 flips and the value of 1 . P {X = i | X + Y = n} should be n−1 (b) We can verify our conjecture by the following computations: P {X = i | X + Y = n} = P {X = i, Y = n − i}/P {X + Y = n} p(1 − p)i−1 p(1 − p)n−i−1 1 ¡n−1¢ = = . 2 n−2 n−1 1 p (1 − p) 18. Let U denote a random variable uniformly distributed over (0, 1). Compute the conditional distribution of U given that (a) U > a; (b) U < a; where 0 < a < 1. Solution. We can compute the conditional CDF as follows: P {U > s|U > a} = P {U > s}/P {U > a} 1−s = , a<s<1 1−a P {U < s|U < a} = P {U < s}/P {U < a} = s/a, 0 < s < a. Hence, U |U > a is uniform on (a, 1), whereas U |U < a is uniform over (0, a).

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