Answer to Question #103368 in Statistics and Probability for Biraj Chhetri

Question

Let X and Y be jointly distributed with p.d.f:


fxy(x,y)=1/4(1+xy) ; |x|>1 |y|>1
=0. ; Otherwise
Show that X and Y are not independent but X2 and Y2 are independent

Answer 1

Let X and Y be jointly distributed with p.d.f:  

$f_{X,Y}(x,y)= \begin{cases} {1 \over 4}(1+xy) & |x|<1, |y|<1 \\ 0 &\text{Otherwise } \end{cases}$

Check

$\displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dydx=$

$=\displaystyle\int_{-1}^{1}\displaystyle\int_{-1}^{1} {1 \over 4}(1+xy)dydx=$

$=\displaystyle\int_{-1}^{1}{1 \over 4}\bigg[(y+{xy^2 \over 2})\bigg]\begin{matrix} 1 \\ -1 \end{matrix}dx=$

$=\displaystyle\int_{-1}^{1}{1 \over 4}(1+{x(1)^2 \over 2}-(-1+{x(-1)^2 \over 2}))dx=$

$=\displaystyle\int_{-1}^{1}{1 \over 2}dx={1 \over 2}\big[x\big]\begin{matrix} 1 \\ -1 \end{matrix}$

Marginal distribution of X

$g(x)=\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dy=$

$=\displaystyle\int_{-1}^{1}{1 \over4}(1+xy)dy={1 \over 4}\bigg[(y+{xy^2 \over 2})\bigg]\begin{matrix} 1 \\ -1 \end{matrix}=$

$={1 \over 4}(1+{x(1)^2 \over 2}-(-1+{x(-1)^2 \over 2}))={1 \over 2}$

Marginal distribution of Y

$h(y)=\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dx=$

$=\displaystyle\int_{-1}^{1}{1 \over4}(1+xy)dx={1 \over 4}\bigg[(x+{x^2y \over 2})\bigg]\begin{matrix} 1 \\ -1 \end{matrix}=$

$={1 \over 4}(1+{y(1)^2 \over 2}-(-1+{y(-1)^2 \over 2}))={1 \over 2}$

$f_{X}(x,y)= \begin{cases} {1 \over 2} & |x|<1, |y|<1 \\ 0 &\text{Otherwise } \end{cases}$ $f_{Y}(x,y)= \begin{cases} {1 \over 2} & |x|<1, |y|<1 \\ 0 &\text{Otherwise } \end{cases}$

The support set is a rectangle, so we need to check if it is true that $f_{XY}(x,y)=f_{X}(x)f_{Y}(y),$ for all $(x, y).$ We easily find counterexamples:

$f_{X,Y}({1 \over 2},{1 \over 2})={1 \over 4}(1+{1 \over 2}({1 \over 2}))={5 \over 16}$

$f_{X}({1 \over 2})f_{Y}({1 \over 2})={1 \over 2}({1 \over 2})={1 \over 4}$

$f_{X,Y}({1 \over 2},{1 \over 2})\not=f_{X}({1 \over 2})f_{Y}({1 \over 2})$

Therefore, X and Y are not independent.

$U=X^2, V=Y^2, 0\leq u<1, \leq v<1$

If $-1<x<0 \ and -1<y<0$

$X=-\sqrt{U}, Y=-\sqrt{V}$

The Jacobian

$J(u, v)=\begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix}=$ $=\begin{vmatrix} -1/(2\sqrt{u}) & 0 \\ 0 & -1/(2\sqrt{v}) \end{vmatrix}=1/(4\sqrt{uv})$

If $0\leq x<1 \ and\ 0\leq y<1$

$X=\sqrt{U}, Y=\sqrt{V}$

$J(u, v)=\begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix}=$ $=\begin{vmatrix} 1/(2\sqrt{u}) & 0 \\ 0 & 1/(2\sqrt{v}) \end{vmatrix}=1/(4\sqrt{uv})$

${1 \over 4}(1+\sqrt{uv})({1 \over 4\sqrt{uv}})= {1 \over 16}(1+{1 \over \sqrt{uv}})$

If $-1<x<0 \ and\ 0\leq y<1$ or $\ 0\leq x<1 \ and\ 0\leq y<1$

$X=-\sqrt{U}, Y=\sqrt{V}$

$J(u, v)=\begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix}=$ $=\begin{vmatrix} -1/(2\sqrt{u}) & 0 \\ 0 & 1/(2\sqrt{v}) \end{vmatrix}=-1/(4\sqrt{uv})$

If $0\leq x<1 \ and\ -1< y<0$

$X=\sqrt{U}, Y=-\sqrt{V}$

$J(u, v)=\begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix}=$ $=\begin{vmatrix} 1/(2\sqrt{u}) & 0 \\ 0 & -1/(2\sqrt{v}) \end{vmatrix}=-1/(4\sqrt{uv})$

${1 \over 4}(1-\sqrt{uv})(-{1 \over 4\sqrt{uv}})= {1 \over 16}(1-{1 \over \sqrt{uv}})$

If $-1<x<0 \ and -1<y<0$

$f_{U,V}(u,v)= \begin{cases} {1 \over 16}(1+{1 \over \sqrt{uv}}) & 0<u<1,0<v<1 \\ 0 &\text{Otherwise } \end{cases}$ $q(u)=\displaystyle\int_{-\infin}^{\infin}f_{U,V}(u,v)dv=$ $=\displaystyle\int_{0}^{1} {1 \over 16}(1+{1 \over \sqrt{uv}})dv={1 \over 16}(1+{1 \over \sqrt{u}})$

$w(v)=\displaystyle\int_{-\infin}^{\infin}f_{U,V}(u,v)du=$ $=\displaystyle\int_{0}^{1} {1 \over 16}(1+{1 \over \sqrt{uv}})du={1 \over 16}(1+{1 \over \sqrt{v}})$

$f_{U}(v)f_{V}(u)={1 \over 16}(1+{1 \over \sqrt{u}})({1 \over 16})(1+{1 \over \sqrt{v}})\not=$ $\not= {1 \over 16}(1+{1 \over \sqrt{uv}})=f_{U,V}(u,v)$

Therefore, X² and Y² are not independent.