# Calculating the expectation of a geometric distribution

Can someone please explain to me how we got from step 3 to step 4? I am extremely confused? Some background information for those curious... This is in regards to a sequence of Bernoulli trials, each with a probability $p$ of success and a probability $q = 1 - p$ of failure. Let variable $X$ denote the number of trials needed to obtain a success. Then $X$ has values in the range $\{0,1,2,\ldots\}$ and for $k \ge 1$,

$$E[X] = \sum\limits_{k=1}^\infty kq^{k-1}p$$ $$= \frac{p}{q} \sum\limits_{k=1}^\infty kq^{k}$$ $$= \frac{p}{q} \frac{q}{(1-q)^2}$$ $$= \frac{p}{q} \frac{q}{p^2}$$ $$= \frac{p}{q} \frac{q}{p^2}$$ $$= \frac{1}{p}$$

In geometric distributions, there are only two possible outcomes. Since either $p$ or $q$ happens, $p=1-q\implies(1-q)^2=p^2$.

• I'm an idiot..... -____- thanks
– ak1
Sep 29, 2015 at 2:57
• I have some question regrading this derivation. First, What is k here? Second, I know E[X]= E[I{A}], I define indicator random variable. We can say that I{A} is the indicator random variable of geometric distribution. Now, I{A} depend upon probability of success and failure. So I{A} = 1.pr{Success} + 0. pr{failure}. However, from the equation I cannot understand where 1.pr{Success} + 0. pr{failure} is used. Mar 6, 2022 at 6:22
• Also from where the 2nd step appear in the equation? Mar 6, 2022 at 6:34

Let me try. You have $$\frac{1}{1-q} = \sum_{n=0}^\infty q^n$$

Then, $$\frac{1}{(1-q)^2} = \sum_{n=1}^\infty nq^{n-1}$$

So, $$\frac{q}{(1-q)^2} = \sum_{n=1}^\infty nq^n$$

That step is certainly not explained in what you have posted. Possibly someone had in mind writing $$\sum_{k=1}^\infty kq^k = q \sum_{k=1}^\infty kq^{k-1} = q \frac d {dq} \sum_{k=1}^\infty q^k$$ and then summing the geometric series and then differentiating. One thing problematic about that is that, although the sum of the derivatives is equal to the derivative of the sum when there are only finitely many terms, the question of when it works with infinitely many terms is more involved than that.

Alternatively, one might say \begin{align} & \sum_{k=1}^\infty kq^k = \sum_{k=1}^\infty \sum_{j=1}^k q^k = \sum_{\begin{smallmatrix} j,k\,:\,1\le j\le k \end{smallmatrix}} q^k = \sum_{j=1}^\infty \sum_{k=j}^\infty q^k \overset{\star}= \sum_{j=1}^\infty \frac{q^j}{1-q} \\[10pt] = {} & \frac 1 {1-q} \sum_{j=1}^\infty q^j \overset{\star}= \frac 1 {1-q} \cdot \frac q {1-q} = \frac q {(1-q)^2} \end{align}

Both of the steps labeled $\text{“}\star\text{''}$ are just the summing of a geometric series.