(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.)

Before you do anything, you have to find [tex]c[/tex] such that [tex]f_{X,Y}(x,y)[/tex] is a proper joint density function. Doing the math, you'll find that [tex]c=2[/tex].Now, determine the marginal densities:[tex]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[/tex][tex]\implies f_X(x)=\dfrac83x^3e^{-2x}[/tex][tex]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[/tex][tex]\implies f_Y(y)=\dfrac12-y^2[/tex]a. Then the density of [tex]X[/tex] conditioned on [tex]Y=y[/tex] is[tex]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}[/tex]b. The density of [tex]Y[/tex] conditioned on [tex]X=x[/tex] is[tex]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}[/tex]and so the distribution of [tex]Y[/tex] conditioned on [tex]X=x[/tex] is[tex]F_{Y\mid X}(y\mid X=x)=\displaystyle\int_{-\infty}^uf_{Y\mid X}(y\mid X=x)\,\mathrm du[/tex][tex]F_{Y\mid X}(y\mid X=x)=\begin{cases}0&\text{for }y<-x\\\frac{2x^3+3x^2y-y^3}{4x^3}&\text{for }-x\le y\le x\\1&\text{for }y>x\end{cases}[/tex]