-1
$\begingroup$

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.

$\endgroup$

closed as off-topic by heropup, colormegone, Leucippus, John B, Shailesh Mar 21 '16 at 0:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, colormegone, Leucippus, John B, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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