{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 5 Solution

# Homework 5 Solution - EE 351K PROBABILITY RANDOM PROCESSES...

This preview shows pages 1–3. Sign up to view the full content.

EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi [email protected] Homework 5 Solution Problem 1 Let Y be a random variable with probability density function (PDF) f Y ( y ) = 1 + y, - 1 y 0 , y, 0 < y 1 , 0 , otherwise . Find (a) P ( | Y | < 0 . 5) . (b) P ( Y > 0 | Y < 0 . 5) . (c) E [ Y ] . Sol : The pdf of Y is as shown below: So based on this, you have two ways of calculating: one is using the structure of graph, the other is using a standard way, i.e. integral. (a) By inspection from the figure, P ( | Y | < 0 . 5) = 0 . 5 . Or you can get it by P ( | Y | < 0 . 5) = P ( - 0 . 5 < Y < 0 . 5) = 0 - 0 . 5 (1 + y ) dy + 0 . 5 0 ydy = 0 . 5 . (b) By inspection from the figure or integral, P ( Y > 0 | Y < 0 . 5) = P (0 < Y < 1 2 ) P ( Y < 1 2 ) = 1 5 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(c) We have that E [ Y ] = 0 - 1 y (1 + y ) dy + 1 0 y 2 dy = [ y 2 2 + y 3 3 ] 0 y = - 1 + [ y 3 3 ] 1 y =0 = 1 6 . Problem 2 The random variable X has the PDF f X ( x ) = { cx - 2 , 1 x 2 0 , otherwise (a) Determine the value of c . (b) Find the CDF of X . (c) Let A be the event { X > 1 . 5 } . Calculate P ( A ) and the conditional PDF of X given that event A has occurred.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}