1
$\begingroup$

I have the following problem:

Given two independent standard normal random variables, call them $X$ and $Y$, how can I show that $Z = X^2 + Y^2$ and $W=\frac{X}{Y}$ are also independent?

I know that because $X$ and $Y$ are standard normal I can write their distributions as $$f_X(x) = \frac{1}{2\pi}e^\frac{-(x)^2}{2}$$ $$f_Y(y) = \frac{1}{2\pi}e^\frac{-(y)^2}{2}$$ Then because they are independent I can write their joint probability distribution as $$f_{X,Y}(x,y) = \frac{1}{2\pi}e^\frac{-(x^+y^2)}{2}$$

Now I know that to show $Z$ and $W$ are independent, I need to show that their joint probability distribution is equal to the product of the marginal distribution functions. But I'm not sure how to find $f_{Z,W}(z,w)$, $f_Z(z)$ and $f_W(w)$.

Any help is much appreciated.

$\endgroup$

marked as duplicate by StubbornAtom, Community Oct 21 '18 at 22:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

The joint distribution for $(X^2,Y^2)$ is $$\frac{x^{-1/2}e^{-x/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)\times \frac{y^{-1/2}e^{-y/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)=\frac{(xy)^{-1/2}e^{-(x+y)/2}}{2(\Gamma(1/2))^2}1_{(0,+\infty)}(x)1_{(0,+\infty)}(y)$$ Since $X^2=Z-\frac{Z}{W^2+1}=\frac{ZW^2}{W^2+1},Y^2=\frac{Z}{W^2+1}$, the Jacobian matrix is $$ \left|\begin{array}{cc} \frac{W^2}{W^2+1}& \frac{1}{W^2+1}\\ \frac{2ZW}{(W^2+1)^2} & -\frac{2ZW}{(W^2+1)^2} \end{array}\right| =\frac{2ZW}{(W^2+1)^2} $$ Therefore, the joint pdf of $(Z,W)$ is $$\frac{(\frac{ZW^2}{W^2+1}\frac{Z}{W^2+1})^{-1/2}e^{-(\frac{ZW^2}{W^2+1}+\frac{Z}{W^2+1})/2}}{2(\Gamma(1/2))^2}\times \frac{2ZW}{(W^2+1)^2}1_{(0,+\infty)}(z)=\frac{1}{w^2+1}\frac{e^{-z/2}}{\pi}1_{(0,+\infty)}(z).$$ Hence, the pdf of $W$ and $Z$ are $\frac{1}{\pi (1+w^2)}$ and $\frac{1}{2}e^{-z/2}1_{(0,+\infty)}(z)$. For now, one can see the independence of $W$ and $Z$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.