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My first related question is this link. enter link description here

Expected value of geometric distribution is $$c=E(X)$$ $$c=0\times p+(1+c)\times q =q+cq$$ $$c=\frac{q}{p}$$

when $X= number \ of \ failures \ before \ the \ 1st \ success$

I just learned about conditioning through $Statistics \ 110$, lecture in Harvard.

So I want to proof that equation using conditioning.

$$P(X)=P(X\mid A_1)P(A_1)+P(X\mid A_2)P(A_2)$$

So, $$E(X)=\sum\limits_{x=1}^{\infty} xP(X=x)=\sum\limits_{x=1}^{\infty}xP(X=x\mid A_1)P(A_1)+\sum\limits_{x=1}^{\infty}xP(X=x\mid A_2)P(A_2)$$ $$when \ P(A_1)=P(1st \ success)=p, \ P(A_2)=P(1st \ failure)=q$$ I know $$\sum\limits_{x=1}^{\infty}xP(X=x\mid A_1)P(A_1)=0$$ because $P(X=x\mid A_1)=0 \ ,\ for \ any \ positive \ integer \ x $. And now we have $q\sum\limits_{x=1}^{\infty}xP(X=x\mid A_2)$ only.

But I have no idea that covert $\sum\limits_{x=1}^{\infty}xP(X=x\mid A_2)$ into $(1+c)$ algebraicly.

I want to know that this process has any error and some idea to complete this proof.

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sunmi yoon
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    Actually, you want to prove that $$E(X)=E(X\mid A_1)P(A_1)+E(X\mid A_2)P(A_2)$$ and this follows directly from the definitions and from the fact that $(A_1,A_2)$ is a partition of $\Omega$. – Did May 19 '15 at 07:56
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    As @Did remarked: your $P$ should be an $E$ (or $\mathbb E$ if you like). If the first toss is a succes, then how many failures do you "expect"? Exactly $0$ right? In mathematical notation: $\mathbb E(X\mid A_1)=0$. If the first toss is not a success then one failure is there and you will start over again the same process. In mathematical notation: $\mathbb E(X\mid A_2)=1+\mathbb EX$. – drhab May 19 '15 at 08:05
  • @drhab Hmmm... I seem to have missed that the question was actually to understand why $E(X\mid A_1)$ and $E(X\mid A_2)$ are what they are. Sorry about that. – Did May 19 '15 at 08:12
  • @drhab why $E(X \mid A_2)=1+E(X)$ ? Memoriless means 'following next steps are completely independent from the previous step? – sunmi yoon May 20 '15 at 11:10
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    $\mathsf E(X\mid X>0)$ is the expected count of failures before the first success given that the first trial was a failure. That will be the first trial plus the expected count of failures after that first trial and before the first success. That is: $1+\mathsf E(X)$. – Graham Kemp May 20 '15 at 11:23
  • @sunmiyoon Yes, and I expressed that by saying that after the first failure "you will start over again the same process" (unaffected by the first toss). This in probabilistic sense. Consequently the expected failures is $\mathbb EX$ again. But. one failure is allready there so we come to $1+\mathbb EX$ expected failures in total under condition $A_2$. – drhab May 20 '15 at 12:55

1 Answers1

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$X$ is the count of Bernoulli trials before the first success.

$A_1$ and $A_2$ are the events of a success and failure (respectively) on the first trial.   These events are mutually exclusive and exhaustive; they partition the outcome space.

$\begin{align} \mathsf E(X) & = \sum_{x=0}^\infty x\mathsf P(X=x) & \text{by definition of expectation} \\[1ex] & = \sum_{x=0}^\infty x\,\big(\mathsf P(X=x\cap A_1)+\mathsf P(X=x\cap A_2)\big) & \text{Law of Total Probability} \\[1ex] & = \sum_{x=0}^\infty x\,\big(\mathsf P(X=x\mid A_1)\mathsf P(A_1)+\mathsf P(X=x\mid A_2)\mathsf P(A_2)\big) & \text{by Conditional Probability} \\[1ex] & = \mathsf P(A_1)\sum_{x=0}^\infty x\,\mathsf P(X=x\mid A_1)+\mathsf P(A_2)\sum_{x=0}^\infty x\, \mathsf P(X=x\mid A_2) & \text{Rearranging} \\[1ex] & = \mathsf P(A_1)\,\mathsf E(X\mid A_1)+\mathsf P(A_2)\,\mathsf E(X\mid A_2) & \text{by definition of expectation} \end{align}$

This result is called the Law of Iterated Expectation.

Notice that $A_1$ is the event that $X=0$ , and that $A_2$ is the event that $X>0$.

$\begin{align} \mathsf E(X) & = \mathsf P(X=0)\,\mathsf E(X\mid X=0) + \mathsf P(X>0)\,\mathsf E(X\mid X>0) & \text{Law of Iterated Expectation} \\[1ex] & = p\,\mathsf E(X\mid X=0) + (1-p)\,\mathsf E(X\mid X>0) & \mathsf P(X=0)=p \\[1ex] & = p\cdot 0 + (1-p)\,\mathsf E(X\mid X>0) & \mathsf E(X\mid X=0)=0 \\[1ex] & = p\cdot 0 + (1-p)\,(1+\mathsf E(X)) & \mathsf E(X\mid X>0)=1+\mathsf E(X) \\[0ex] && \text{because the distribution is memoriless.} \\[2ex] \therefore \mathsf E(X) & = \frac{1-p}p \end{align}$

why $\mathsf E(X\mid A_2)=1+\mathsf E(X)$ ? Memoriless means 'following next steps are completely independent from the previous step'?

Yes, and that means the expected number of failures after the $n^\text{th}$ trial and before the next success is the same as the expected number of failures after the $0^\text{th}$ trial and before the first success, for any $n$.

$\mathsf E(X\mid X>0)$ is the expected count of failures before the first success given that the first trial was a failure.   That will be the first trial plus the expected count of failures after that first trial and before the first success.   That is: $1+\mathsf E(X)$.

Graham Kemp
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  • This answer is related to my question, but not totally cover. Because I want to know some algebric work to convert above things (if that method exists). But Law of lterated Expectation is very useful to understand with another perspective. Thank you. – sunmi yoon May 19 '15 at 13:34
  • Thank you so much for your additional explanation. It is really helpful. – sunmi yoon May 20 '15 at 12:35