Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a geometric random variable and let $A$ denote the event $\{X>3\}$ find the conditional probability mass function of $X$ with respect to the event $A$ and then compute $E[X\mid A]$.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

We have that $$ P(X=k)=p(1-p)^k,\quad k=0,1,2,\ldots $$ and hence $P(X>3)=(1-p)^4$. Now if $k=0,1,2,3$, then of course $P(X=k\mid X>3)=0$ and if $k=4,5,\ldots$, then $$ P(X=k\mid X>3)=\frac{P(X=k,X>3)}{P(X>3)}=\frac{P(X=k)}{P(X>3)}=p(1-p)^{k-4}. $$ The conditional expectation is thus given by $$ E[X\mid X>3]=\sum_{k=4}^\infty kP(X=k\mid X>3)=\sum_{k=4}^\infty kp(1-p)^{k-4}=\sum_{k=0}^\infty (k+4)p(1-p)^k=4+E[X]. $$

share|improve this answer
    
There is something wrong here. Conditioned on $X>3$, the conditional expectation of $X$ is $3+E[X]$, not $E[X]$. If $p = 0.5$ so that $E[X]=2$, why should the average value of $X$ be $2$ when we are given that $X > 3$? –  Dilip Sarwate Dec 14 '12 at 3:33
    
@DilipSarwate: Yes, you're absolutely right. –  Stefan Hansen Dec 14 '12 at 3:40
    
I edited my answer. Though I'm getting $4+E[X]$, but I think it depends on whether you define the geometric distribution on $\{0,1,2,\ldots\}$ or on $\{1,2,\ldots\}$. –  Stefan Hansen Dec 14 '12 at 3:45
    
Thank you! This really helped, hopefully I remember it on my final tomorrow! –  Olivia Irving Dec 14 '12 at 4:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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