for the first part, could u tell me why the lower bound is 0 rather than x? We can see from the question that 0<x<y<1, if u set the lower bound=0, x might be bigger than y Q 4.2. Let (X, Y) be a random variable with the follow...Elyla):=== (2) argelJyflylaladyRyzdyy=9= 2(1-x2)21-12X[2]1(la)y=ox (13-03)] ?ņwhy is 0 ?

Expressing E(Y∣X) in terms of X:The conditional expectation of Y given X is defined as:E(Y∣X)=…

Answered: for the first part, could u tell me why…

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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for the first part, could u tell me why the lower bound is 0 rather than x? We can see from the question that 0<x<y<1, if u set the lower bound=0, x might be bigger than y

Q 4.2. Let (X, Y) be a random variable with the follow...
Elyla):
=
=
= (2) argel
Jyflylalady
Ryzdy
y=9
= 2
(1-x2)
2
1-12X
[2]
1
(la)
y=o
x (13-03)
] ?
ņ
why is 0 ?

Expert Solution

Step 1: Solving E(Y|X)

Expressing $E (Y ∣ X)$ in terms of $Error converting from MathML to accessible text.$ :

The conditional expectation of $Y$ given $X$ is defined as:

$integral subscript negative infinity end subscript superscript infinity$

Where $f (y ∣ X)$ is the conditional probability density function (PDF) of $Y$ given $X$ .

In this case, the joint PDF is given by:

$f subscript X comma Y end subscript left parenthesis x comma y right parenthesis space equals space 15 space x squared y$

To find the conditional PDF $f (y ∣ X)$ , we need to find the marginal PDF of $X$ . To do this, integrate the joint PDF over the possible values of $Y$ :

$f subscript X left parenthesis x right parenthesis space equals space integral subscript x superscript 1 15 space x squared y space d y$

$open vertical bar y squared over 2 close vertical bar subscript x superscript 1$

Now, the conditional PDF $Error converting from MathML to accessible text.$ is given by:

$f (y ∣ X) = f ^{X, Y} ( x , y ) / f(x)$

$= 2y / (1 - x 2) ; x < y < 1$

Now, the conditional expectation $E (Y ∣ X)$ is:

$integral subscript x superscript 1 y space fraction numerator 2 y over denominator 1 minus x squared end fraction space d y$

$fraction numerator 2 over denominator 3 left parenthesis 1 minus x squared right parenthesis end fraction space open vertical bar y cubed close vertical bar subscript x superscript 1$

$= 2(1 - x 3) / 3(1 - x 2)$

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