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5:23PM Thu 14 Nov ? \( 64 \% \) Take quiz Exit Question 5 1 pts Suppose that \( X \) has density \[ f_{X}(x)=\frac{x^{2}}{Z} \text { for } 0<x<2 \text { (and } f_{X}(x)=0 \text { otherwise). } \] Here \( \boldsymbol{Z} \) is a constant that needs to be determined. Also suppose that the conditional density of \( Y \) given \( X=x \) (where \( x>0 \) ) is given by \[ f_{Y \mid X}(y \mid x)=\frac{2 y}{x^{2}} \quad \text { for } \quad 0<y<x \quad \text { (and } f_{Y \mid X}(y \mid x)=0 \text { otherwise). } \] Using the Law of Total Expectation (Partition Theorem for Expectation), or otherwise, determine \( \mathbb{E}(Y) \), the expected value of \( Y \). \( \square \)

          5:23PM Thu 14 Nov
? \( 64 \% \)
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Question 5
1 pts

Suppose that \( X \) has density
\[
f_{X}(x)=\frac{x^{2}}{Z} \text { for } 0<x<2 \text { (and } f_{X}(x)=0 \text { otherwise). }
\]

Here \( \boldsymbol{Z} \) is a constant that needs to be determined.
Also suppose that the conditional density of \( Y \) given \( X=x \) (where \( x>0 \) ) is given by
\[
f_{Y \mid X}(y \mid x)=\frac{2 y}{x^{2}} \quad \text { for } \quad 0<y<x \quad \text { (and } f_{Y \mid X}(y \mid x)=0 \text { otherwise). }
\]

Using the Law of Total Expectation (Partition Theorem for Expectation), or otherwise, determine \( \mathbb{E}(Y) \), the expected value of \( Y \).
\( \square \)

5:23PM Thu 14 Nov
? 64 %
Take quiz
Exit

Question 5
1 pts

Suppose that X has density

fX(x)=(x^2)/(Z) for 0<x<2 (and fX(x)=0 otherwise).

Here Z is a constant that needs to be determined.
Also suppose that the conditional density of Y given X=x (where x>0 ) is given by

fY | X(y | x)=(2 y)/(x^2) for 0<y<x (and fY | X(y | x)=0 otherwise).

Using the Law of Total Expectation (Partition Theorem for Expectation), or otherwise, determine 𝔼(Y), the expected value of Y.
□

Added by Ruben B.

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