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.

I know that the expected value of a geometrically distributed random variable is $\frac1p$ but how do we get there. This is what I got so far: $$\sum_{x=1}^\infty xP(X=x)$$ where X is the number of failures until first success. Since it's geometric we have:$$\begin{align} \sum_{x=1}^\infty xp(1-p)^{x-1}\\ \frac{p}{1-p} \sum_{x=1}^\infty x(1-p)^x\\ .... \end{align}$$ How do we sum that?

share|improve this question
4  
Differentiate a well-known series. –  M.B. Nov 12 '12 at 20:51
add comment

2 Answers

up vote 4 down vote accepted

Set $r=1-p$ and recall geometric series formula $$ \sum\limits_{x=1}^\infty r^x=\frac{r}{1-r} $$ Then $$ \sum\limits_{x=1}^\infty x r^x= r\sum\limits_{x=1}^\infty x r^{x-1}= r\sum\limits_{x=1}^\infty \frac{d}{dr} r^x= r \frac{d}{dr}\sum\limits_{x=1}^\infty r^x= r\frac{d}{dr}\frac{r}{1-r} $$ Is the rest clear?

share|improve this answer
    
then wouldn't we have $\sum_{x=1}^\infty xr^x$? How do we take care of the extra x? –  TheHopefulActuary Nov 12 '12 at 20:53
1  
Oh sorry, i'll fix it –  Norbert Nov 12 '12 at 20:56
add comment

An experiment has probability of success $p\gt 0$, and probability of failure $1-p$. We repeat the experiment until the first success. Let $X$ be the total number of trials. We want $E(X)$.

Do the experiment once. So we have used $1$ trial. If this trial results in success (probability: $p$), then the expected number of further trials is $0$. If the first trial results in failure (probability: $1-p$), our experiment has been wasted, and the expected number of trials remains at $E(X)$. Thus $$E(X)=1+(p)(0)+(1-p)E(X).$$ If $E(X)$ exists, we can solve for $E(X)$ and obtain $E(X)=\dfrac{1}{p}$.

share|improve this answer
    
I like the insight you made, but can you tell me why 1 isn't multiplied by $p$? since $E[X]= \displaystyle\sum_{k=0}^{\infty} x(Pr(X=k))$? –  adam Dec 18 '13 at 4:33
    
We could do it your way. With probability $p$ we get $X=1$ (giving a term $p$, or with probability $1-p$, $E(X)=1+E(X)$. That gives $E(X)=p+(1-p)(1+E(X))$, giving exactly the same thing for $E(X)$ as was obtained in (I think) a marginally simpler way. –  André Nicolas Dec 18 '13 at 6:48
    
let me try to see if I follow with what you are doing. Are you evaluating the summation by parts? meaning $\displaystyle\sum_{k=1}^1 1(Pr(X=1))?$ and than representing the other half of the sum as E[X]? –  adam Dec 18 '13 at 7:38
    
The argument uses conditional expectation, conditioning on the first outcome. In this example, it can also be viewed as a summation trick or technique. –  André Nicolas Dec 18 '13 at 14:50
    
so its the probability of success p + the probability of failure 1-p times the expected value if we fail the 1st trial 1 + E [X]. if I got the logic down. –  adam Dec 18 '13 at 18:49
show 2 more comments

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.