GATE DS&AI 2024 | Question: 49

Question

Consider a joint probability density function of two random variables $X$ and $Y$
\[
f_{X, Y}(x, y)=\left\{\begin{array}{rll}2 x y, & 0<x<2, & 0<y<x \\ 0, & \text { otherwise } & \end{array}\right.
\]
Then, $E[Y \mid X=1.5]$ is $\_\_\_\_\_\_\_\_\_$

Answer 1

Here, we have given the joint probability density function as:

$f(x,y) = 2xy \;\;$ if $0<y<x<2$

and 0, otherwise.

We need to find the conditional expectation $\mathbb{E}[Y|X=x]$ where $x=1.5$

If $u(Y)$ is a function of $Y$ then the conditional expectation of $u(Y)$ given that $X=x,$ if it exists, is given by,

  

$\mathbb{E}[u(Y)|x] = \int_{-\infty}^{\infty}u(Y)f_{Y|X}(y|x) \ dy$

Where, conditional pdf $f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)}$

Here, the marginal pdf of $X$ is given by,

$f_X(x)=\int_{0}^{x}f_{X,Y}(x,y) \ dy=\int_{0}^{x}2xy \ dy=x^3$

So,

$f_{Y|X}(y|x) =\frac{2xy}{x^3}=\frac{2y}{x^2}$ for $0<y<x$ and zero elsewhere.

Now, here $u(Y)=Y$ 

So,

$\mathbb{E}[Y|x] = \int_{0}^{x}yf_{Y|X}(y|x) \ dy$

$\mathbb{E}[Y|x] = \int_{0}^{x}y\frac{2y}{x^2} \ dy$

$\mathbb{E}[Y|x] = \frac{2}{x^2}\int_{0}^{x}y^2 \ dy$

$\mathbb{E}[Y|x] =\frac{2x}{3}$

Therefore,

$\mathbb{E}[Y|1.5] =\frac{2\times1.5}{3}=1$

$\mathbb{E}[Y|1.5] =1$

Comments

marked it as wrong because it is not a valid joint pdf.

credit @redshellspy

https://gateoverflow.in/423666/gate-da-2024-answer-key-challenge