Question

The random variables X and Y are jointly continuous, with a joint PDF of the form fX,Y(x,y)={cxy,0,if 0≤x≤y≤1,otherwise, where c is a normalizing constant. For x∈[0,0.5] , the conditional PDF fX|Y(x|0.5) is of the form axb . Find a and b .

Answer 1

Answer:

The value of a and b are 8 and 1 respectively

Step-by-step explanation:

$f_{x,y}(x,y)=\left\{\begin{matrix}cxy ,\text{if} 0 \leq x \leq y \leq 1\\ 0 , \text{otherwise}\end{matrix}\right.$

$\int_{0}^{1}\int_{x}^{1}cxy \text{dy dx}=1\\\\c\int_{0}^{1}x\left ( \frac{y^2}{2}\right )_{x}^{1} \text{ dx}=1\\\\\frac{c}{2}\int_{0}^{1}x\left ( 1-y^2\right )\text{dx}=1\\\\\frac{c}{2}\int_{0}^{1}(x-x^3)\text{dx}=1\\\\\frac{c}{2}\left [ \frac{x^2}{2}-\frac{x^5}{4} \right ]_{0}^1=1\\\\\frac{c}{2}\left [ \frac{1}{2}-\frac{1}{4} \right ]=1\\\\\frac{c}{2}\left ( \frac{1}{4} \right )=1\\\\c=8$

Conditional pdf of $f_{x,y}(x|0.5)$

$f_{x,y}(x|0.5)=\frac{f_{xy}(x,y=0.5)}{f_{y}y}$

Normalizing pdf of 1

$f_{y}y=\int_{0}^{y} 8yx \text{dx}$

$f_{y}y=8y[\frac{x^2}{2}]_{0}^{y}\\f_{y}y=4y(y^2)\\f_{y}y=4y^3\\f_{x|y}(x|0.5)=\frac{8x(0.5)}{4(0.5)^3}=\frac{2x}{0.25}=8x$

We are given that PDF $f_{x|y}(x|0.5)$ is of the form $ax^b$

So, on comparing 8x with $ax^b$

So, a = 8 , b = 1

So, the value of a and b are 8 and 1 respectively