9.Two discrete random variables X and Y have joint probability mass function(pmf)f(x) =k x = 1,2,n(n+1)otherwise{=01,2,...,n; y = 1,2, x.·(d)Use the fact that E(Y) = Ex (Ey\x(Y|X)), where Ex( ) and Ey|x ( ) are theexpected values with respect to X and with respect to Y given X, respectively,n(n+3)to show that E(Y)4

To find the expected value of Y, E(Y), using the law of total expectation, we can break it down as…

Two discrete random variables X and Y have joint probability mass function (pmf) ‚n;_y=1,2, x. f(x) = بر k x = 1, 2, n(n+1) 0 otherwise (d) Use the fact that E(Y) = Ex (Ey|x (YX)), where Ex() and Exx() are the expected values with respect to X and with respect to Y given X, respectively, n(n+3) to show that E(Y) 4 =

Two discrete random variables X and Y have joint probability mass function (pmf) ‚n;_y=1,2, x. f(x) = بر k x = 1, 2, n(n+1) 0 otherwise (d) Use the fact that E(Y) = Ex (Ey|x (YX)), where Ex() and Exx() are the expected values with respect to X and with respect to Y given X, respectively, n(n+3) to show that E(Y) 4 =

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

9.
Two discrete random variables X and Y have joint probability mass function
(pmf)
f(x) =
k x = 1,2,
n(n+1)
otherwise
{
=
0
1,2,...,n; y = 1,2, x.
·
(d)
Use the fact that E(Y) = Ex (Ey\x(Y|X)), where Ex( ) and Ey|x ( ) are the
expected values with respect to X and with respect to Y given X, respectively,
n(n+3)
to show that E(Y)
4

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I don't understand how 4(n+1)/(n-1) matches n(n+3)/4.

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