Let $X, Y \sim G (p)$ be independent Geometric random variables ($p \in (0, 1)$). Show that $P (X = Y) = p / (2−p)$.

I'm not sure how to approach this problem - any help would be appreciated.


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Sometimes "geometric distribution" means a distribution supported on $\{0,1,2,3,\ldots\}$ and sometimes it means a distribution supported on $\{1,2,3,\ldots\}$. Assuming the latter, you have \begin{align} & \Pr(X=Y) \\[10pt] = {} & \Pr(X=Y=1 ) + \Pr(X=Y=2 ) + \Pr(X=Y=3 ) + \Pr(X=Y=4 ) + \cdots \\[10pt] = {} & p^2 + p^2 ( 1- p)^2 + p^2 (1-p)^4 + p^2 ( 1-p)^6 + \cdots. \end{align}

Remember that $$ a+ar+ar^2+ar^3+\cdots = \frac a {1-r}. $$ In this case $a=p^2$ and $r = (1-p)^2$.

So the sum comes to $\dfrac p {2-p}$.

In the case of the support being $\{0,1,2,3,\ldots\}$ you'd start the sum with $\Pr(X=Y=0)$, but the rest is the same.

  • $\begingroup$ what is a $+p^2$ in between?.. $\endgroup$ – Upstart Mar 20 '16 at 20:29
  • $\begingroup$ I believe that's just a typo, and there should be no + sign between p^2 and (1-p)^4 $\endgroup$ – Michael Meng Mar 20 '16 at 20:51

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