Question #190372

The joint pdf of random variables ‘X’ and ‘Y’ is

 

Find

a. The marginal pdfs, fX(x) and fY(y).

b. The conditional pdfs, fX/Y(x/y) and fY/X(y/x)

c. The E(X/Y=1)

d. Are ‘X’ and ‘Y’ statistically independent?



1
Expert's answer
2021-05-07T14:31:03-0400

Since joint pdf is not given in the question hence we can assume that

Joint pdf of variables X and Y is

fX,Y(x,y)=6xyf_{X,Y}(x,y) = 6xy when 0x1,0yx0 \le x \le 1, 0\le y \le \sqrt{x}

Otherwise


a.) fX(x)=fXY(x,y)dyf_{X}(x) = \int_{-\infty}^{\infty} f_{XY}(x,y)dy

=0x6xydy= \int_{0}^{\sqrt{x}}6xy dy

=3x2= 3x^2

Thus, fX(x)=3x20x1f_{X}(x) = 3x^2 \hspace{3mm}0 \le x\le 1

0 Otherwise


To find fY(y)for0y1f_{Y}(y) \hspace{2mm}for \hspace{2mm} 0 \le y \le 1 , we can write,


fY(y)=fXY(x,y)dxf_{Y}(y) = \int_{-\infty}^{\infty} f_{XY}(x,y)dx


=y216xydx= \int_{y^2}^{1} 6xydx


=3y(1y4)= 3y(1-y^4)


fY(y)=3y(1y4)f_{Y}(y ) = 3y(1-y^4) 0y10 \le y \le 1

=0= 0 Otherwise



b.) The conditional pdf,


fYX(yx)=fX,Y(x,y)fX(x)f_{Y|X}(y|x) = \dfrac{f_{X,Y}(x,y)}{f_X(x)}


=6xy3x2=2yx= \dfrac{6xy}{3x^2} = \dfrac{2y}{x}


fXY(xy)=fX,Y(x,y)fY(x)f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(x)}

=6xy3y(1y4)=2x(1y4)= \dfrac{6xy}{3y(1-y^4)} = \dfrac{2x}{(1-y^4)}


c.) The E(X/Y=y)E(X/Y = y)


=xfXY(xy)dx= \int_{-\infty}^{\infty} xf_{X|Y}(x|y)dx


=y21x2x1y4dx= \int_{y^2}^{1} x \dfrac{2x}{1-y^4}dx


=2(1y6)3(1y4)= \dfrac{2(1-y^6)}{3(1-y^4)}


d.) Yes X and Y are statistically independent.



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