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I watched $\text{Statistics} \space 110$ from Harvard University through YouTube.

From lecture 9, the expected value of the geometric distribution is: $$\sum\limits_{k=0}^{\infty} kpq^k=p\sum\limits_{k=1}^{\infty}kq^k=\frac{pq}{p^2}=\frac{q}{p}$$ where $X$ = number of failures before the 1st success

But I cannot understand the derivation below intuitively - what is the meaning of $0$, $p$, $q$ and $(1+c)$: $$ c=E(X)=0\times p+(1+c)\times q =q+cq=\frac{q}{p} $$

Starlight
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sunmi yoon
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1 Answers1

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Suppose we are tossing repeatedly a coin that has probability $p$ of landing heads. Then the probability of a tail is $1-p$, which we call $q$. Let $X$ be the number of tails before the first head. We want to find $E(X)$.

Let $c=E(X)$. We condition on the result of the first toss. If we get a head on that toss, then $X=0$, and therefore $E(X)=0$. More formally, the conditional expectation of $X$, given that we got a head on the first toss, is $0$.

If we got a tail on the first toss, then we already have $1$ tail. And the conditional expectation of the number of additional tails is $c$, because the process has no memory. Thus the conditional expectation of $X$, given we got a tail, is $1+c$. It follows that $$c=(p)(0)+(q)(1+c).$$

André Nicolas
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  • The probability is $q$, the conditional expectation is $1+c$. I put the probabilities first in each product. The thing you quote put them second. Makes no difference. (I assume you deleted the comment because you understood. I will delete this comment in a few minutes.) – André Nicolas May 19 '15 at 06:45
  • I have further question related to translate that equation using conditional probability. What is the better that asking 'new' question or just adding comment? – sunmi yoon May 19 '15 at 06:58
  • I am using conditional expectation, not conditional probability. Please see Wikipedia, the Law of Total Expectation. About new versus additional, if it is really different I would suggest a new question. For one thing, it is quite late here. I will be sleeping, and a comment or addendum to your post would be largely invisible to the rest of the MSE community. – André Nicolas May 19 '15 at 07:05