Question

(1 point) If the joint density function of X and Y is f(x,y)=c(x2−y2)e−2x, with 0≤x<∞ and −x≤y≤x, find each of the following. (a) The conditional probability density of X, given Y=y>0. Conditional density fX|Y(x,y)= 4(x^2-y^2)e^(-2x)/(1-2y^2) (Enter your answer as a function of x, with y as a parameter.) (b) The conditional probability distribution of Y, given X=x. Conditional distribution FY|X(y|x)= 3/4(x^2-y^2) (for −x≤y≤x). (Enter your answer as a function of y, with x as a parameter.)

Answer 1

Before you do anything, you have to find $c$ such that $f_{X,Y}(x,y)$ is a proper joint density function. Doing the math, you'll find that $c=2$.

Now, determine the marginal densities:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_{-x}^x2(x^2-y^2)e^{-2x}\,\mathrm dy$

$\implies f_X(x)=\dfrac83x^3e^{-2x}$

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^\infty2(x^2-y^2)e^{-2x}\,\mathrm dx$

$\implies f_Y(y)=\dfrac12-y^2$

a. Then the density of $X$ conditioned on $Y=y$ is

$f_{X\mid Y}(x\mid Y=y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\dfrac{4(x^2-y^2)e^{-2x}}{1-2y^2}$

b. The density of $Y$ conditioned on $X=x$ is

$f_{Y\mid X}(y\mid X=x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\dfrac{3(x^2-y^2)}{4x^3}$

and so the distribution of $Y$ conditioned on $X=x$ is

$F_{Y\mid X}(y\mid X=x)=\displaystyle\int_{-\infty}^uf_{Y\mid X}(y\mid X=x)\,\mathrm du$

$F_{Y\mid X}(y\mid X=x)=\begin{cases}0&\text{for }yx\end{cases}$