Question

Let X and Y be jointly distributed with p.d.f:
f XY (x,y)={1/4(1+xy),|x|<1,|y|<1; 0 otherwise
Show that X and Y are not independent by X^2 and Y^2 are independent.

Answer 1

The intervals for x and y is (-1,1). X and Y are independent if $E(XY)=E(X)×E(Y)$

$E(X)=\iint(\frac{X}{4}(1+XY))dx dy X,Y\isin(-1,1)$

$=\intop(\frac{X^2}{8}+\frac{X^3Y}{12})dy X,Y\isin(-1,1)$

$=\frac{X^2Y}{8}+\frac{X^3Y^2}{24}. X,Y\isin(-1,1)$

$=\frac{Y}{8}+\frac{Y^2}{24}-\frac{Y}{8}+\frac{Y^2}{24}, X\isin(-1,1)$

$=\frac{1}{24}+\frac{1}{24}-\frac{1}{24}-\frac{1}{24}=0$

Since both X and Y have same domain and are at same position on the distribution, $E(X)=E(Y) =0$

$E(XY)=\iint(\frac{XY}{4}(1+XY))dx dy, X,Y\isin(-1,1)$

$=\intop(\frac{X^2Y}{8}+\frac{X^3Y^2}{12})dy ,X,Y\isin(-1,1)$

$=\frac{X^2Y^2}{16}+\frac{X^3Y^3}{36}X,Y\isin(-1,1)$

$=\frac{Y^2}{16}+\frac{Y^3}{36}-\frac{Y^2}{16}+\frac{Y^3}{36} , Y\isin(-1,1)$

$=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{1}{9}$

Since E(XY) is not equal to the product of E(X) and E(Y), X and Y are not independent.

$E(X^2)=\iint(\frac{X^2}{4}(1+XY))dx dy X,Y\isin(-1,1)$

$=\intop(\frac{X^3}{12}+\frac{X^4Y}{16})dy X,Y\isin(-1,1)$

$=\frac{X^3Y}{12}+\frac{X^4Y^2}{32}X,Y\isin(-1,1)$

$=\frac{Y}{12}+\frac{Y^2}{32}+\frac{Y}{12}-\frac{Y^2}{32},Y\isin(-1,1)$

$=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{3}$

Similarly, since both X and Y have same domain and are at same position on the distribution, $E(X^2)=E(Y^2)=\frac{1}{3}$

$E(X^2Y^2)=\iint(\frac{X^2Y^2}{4}(1+XY))dx dy, X,Y\isin(-1,1)$

$=\intop(\frac{X^3Y^2}{12}+\frac{X^4Y^3}{16})dy ,X,Y\isin(-1,1)$

$=\frac{X^3Y^3}{36}+\frac{X^4Y^4}{64} ,X,Y\isin(-1,1)$

$=\frac{Y^3}{36}+\frac{Y^4}{64}+\frac{Y^3}{36}-\frac{Y^4}{64} ,Y\isin(-1,1)$

$=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{1}{9}$

$E(X^2)×E(Y^2)=\frac{1}{3}×\frac{1}{3}=\frac{1}{9}$

Since the product of $E(X^2)$ and $E(Y^2)$ is equal to $E(X^2Y^2),$ $X^2$ and $Y^2$ are independent.