Conditional pmf and mean for geometric rv

Question

I have a geometric r.v. p(1-p)^(n-1), n = 1, 2, 3, 4, ...

I was able to figure out the conditional mean conditioned on X > a. My answer to this is E[X] + a. I am pretty sure that is correct but if not can anyone let me know where I went wrong.

I am not able to find the conditional pmf for the same condition. Any pointers that can help? Thanks

The answer is simplest if $a$ is a non-negative integer. Use the memorylessness of the geometric, or prove it. In the second part, you will be essentially proving memorylessness. Again, it is simplest if $a$ is a non-negative integer. — André Nicolas, Mar 08 '15 at 23:08

score 1 · Accepted Answer · answered Mar 08 '15 at 22:50

1

Write $X \sim \operatorname{Geo}(p)$ as your RV. Bayes's theorem: $$ P(X=r \mid X>a ) = \frac{P(X>a \mid X=r)P(X=r)}{P(X>a)}. $$ The first term in the numerator is just $1$ when $r>a$. Can you see how to do the rest?

answered Mar 08 '15 at 22:50

Chappers

67,606

Thank you. So would the pmf be p(1-p)^(n-a-1) for n > a, and 0 otherwise? – user100503 Mar 08 '15 at 22:54
Yes, that looks right. As a simple sanity check, the sum of that from $a+1$ to $\infty$ is $1$. Of course, now you can use that to compute $E[X\mid X>a]$ directly. – Chappers Mar 08 '15 at 23:06
Thank you very much. I appreciate your help. – user100503 Mar 08 '15 at 23:24

Conditional pmf and mean for geometric rv

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