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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
Transcribed Image Text: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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Follow-up Question
Hi!
I don't understand how 4(n+1)/(n-1) matches n(n+3)/4.
They are not equal as far as I can see or understand. Is there an error in the solution or am I missing something?
Brgds
MB
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