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I have two random variables, $X,Y\sim$ Geo$(p)$. Define two more random variables, $U=\min{(X,Y)}$ and $V=X-Y$. I am tasked with showing that $U$ and $V$ are independent.

This means I need the joint distribution $f_{U,V}(u,v)$. So

$$f_{U,V}(u,v)=P(U=u,V=v)=P(\min{(X,Y)}=u, X-Y=v)$$

This is confusing me. So i thougth I look at the support of $U$ and $V$. Since $X,Y$ is geometric, then the minimum of $X$ and $Y$ is also over the same support of $X$ and $Y$ which would just be $\mathbb{N}$. But the support of $V$ would then be $\mathbb{Z}$ since $V$ can take on negative values ($x=3, y=4$, for example). So do I need to consider cases of $V$ then and compare the distributions of all three?

For example, if $V>0$, then $X>Y$ and $\min{(X,Y)}=Y$. If $V<0$, then $Y>X$ and $\min{(X,Y)}=X$. For the first case,

$$V>0\Rightarrow P(\min{(X,Y)}=u, X-Y=v)=P(Y=u, X=v+y)=P(X=v+u,Y=u)$$

So since $X,Y$ are independent,

$$f_{U,V}(u,v)=P(X=v+u,Y=u+v)=P(X=u+v)P(Y=u)=p(1-p)^{u+v}p(1-p)^u$$

So $f_{U,V}(u,v)=p^2(1-p)^{2u+v}$. Is this correct? And is this the correct approach to now consider $V<0$ and then $V=0$?

AveryJessup
  • 956

1 Answers1

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$U$ and $V$ are independent if and only if $f_{U,V}(u,v)$ can be written as $f(u)f(v)$.

Now our variables are $$ X = \left\{ \begin{array}{cl} U&V\geq 0\\ U+V& V<0\end{array}\right.\\ Y = \left\{ \begin{array}{cl} U-V&V\geq 0\\ U& V<0\end{array}\right. $$ Then using our knowledge of $f_X(x)$ and $f_Y(y)$ we can fill in $f_{U,V}(u,v)$ as $$ f_{U,V}(u,v) = \left\{ \begin{array}{cl} (1-p)^{u-1}p(1-p)^{u-v-1}p &v\geq 0\\ (1-p)^{u+v-1}p(1-p)^{u-1}p & v<0\end{array}\right. $$ and these expressions can be written as $$ f_{U,V}(u,v) = (1-p)^{2u+|v|-2}p^2 = \left[(1-p)^{2u-1}p \right] \left[(1-p)^{|v|-1}p \right]=f(u)f(v) $$

Mark Fischler
  • 42,507