Expected value of a geometric distribution with first step analysis.

Question

I am trying to understand the "story proof" found in this lecture.

I am a bit confused as how the expected value of a random variable differs from the the random variable itself when considering indicator functions.

Say there is a geometric distribution. $X$ counts the number of the failures before the first success, and $E(X)$ is the expected number of failures.

Now I want to compute the expected value given $p$, the probability of success, and $q$ otherwise.
Let $c=E(X)$.

Now I do first-step analysis, $$c=0\cdot p+(1+c)q.$$

In this step, I don't understand the coefficient of $q$. In $(1+c)$, $1$ makes sense but why $c$?
When computing the expected value: it is the $kp^kq^k$. So the coefficient $k$ is the value of the random variable. But in my example it is the expected value $c$ at the next step which confuses me.

Answer 1

Let $c=E(X)$. We use a conditional expectation argument, conditioning on the result of the first trial.

Either (i) we have a success on the first trial, or (ii) we have a failure on the first trial.

Case (i) has probability $p$.

Case (ii) has probability $q=1-p$. In that case, we have had a failure, and the expected number of additional failures before the first success is $c$. So the conditional expectation of $X$, given that there was a failure on the first trial, is $1+c$.

By the Law of Total Expectation, we therefore have $$E(X)=c=(p)(0)+(q)(1+c).$$ Finally, solve this linear equation for $c$. We get $c(1-q)=q$, and therefore $c=\frac{q}{1-q}=\frac{q}{p}=\frac{1}{p}-1$.

Answer 2

Another approach is that when $X$ counts the number of failures before the first success (in a sequence of iid Bernoulli trials of success rate $p$), then:

$$\begin{align}\mathsf E(X) =&~ \sum_{k=0}^\infty k~\mathsf P(X=k) \\[1ex] =&~ \sum_{k=0}^\infty k(1-p)^{k} p \\[1ex] =&~ 0(1-p)^0p+\sum_{k=1}^\infty k(1-p)^{k}p\\[1ex] =&~ 0\,p + \sum_{h=0}^\infty (h+1)(1-p)^{h+1}p \\[1ex] =&~ 0\,p + (1-p)~\sum_{h=0}^\infty (h+1)(1-p)^hp \\[1ex] =&~ 0\,p+(1-p)\mathsf E(X+1) \\[1ex] =&~ 0\,p+(1-p)\,(1+\mathsf E(X))\\[2ex]\therefore~\mathsf E(X) =&~ \tfrac {1-p}p \end{align}$$

That is a bit of a slog but should give some insight as to why (and how) partitioning on the result of the first trail works.   If the first trial is a success then we are done and the expected count is 0 (on that condition).   If the first trial is a failure then the expected count (on that condition) is one more than the expected count after that failure before the first success.   This little bit of reasoning saves some effort and we immediately jump to the end by using the Law of Total Expectation.

Let $X_1$ be the count of successes in the first trial; which is $1$ or $0$ ; with probabilities $p$ and $1-p$ respectively.

$$\begin{align}\mathsf E(X) =&~ \mathsf E(\mathsf E(X\mid X_1)) \\[1ex] =&~ \mathsf P(X_1=1)~\mathsf E(X\mid X_1=1)+\mathsf P(X_1=0)~\mathsf E(X\mid X_1=0) \\[1ex] =&~ p\cdot 0 + (1-p)~(1+\mathsf E(X))\end{align}$$

Answer 3

This answer is similar to others, but here the first-step analysis $(^*)$ is expressed entirely in terms of random variables before taking expectations, and then without using conditional expectation.

Let the sequence of i.i.d. Bernoulli trials be $(Y_1,Y_2,Y_3,...)$, with $P(Y_1=\text{success})=p\ $, and define $$X_i = \text{ number of failures before the first success in }(Y_i,Y_{i+1},Y_{i+2},...).$$

Note that $(X_1, X_2, X_3,...)$ are identically distributed, and that $X_i$ is independent of $Y_j$ for any $j<i$; therefore, using indicator functions $[...]$, $$X_1 = 0\cdot[Y_1=\text{success}] + (1 + X_2)\cdot[Y_1=\text{failure}],\tag{*}$$ so $$\begin{align}\mathbb{E}(X_1) &= \mathbb{E}(1 + X_2)\cdot\mathbb{E}[Y_1=\text{failure}]\\ &= (1 + \mathbb{E}(X_2))\cdot P(Y_1=\text{failure})\\ &= (1 + \mathbb{E}(X_1))\cdot (1-p)\\ \\ \therefore\ \mathbb{E}(X_1) &= \frac{1-p}{p}. \end{align} $$