Expected value of ${X \choose 2}$ where $X$ is geometrically distributed

Question

I'm trying to find $E[{X \choose 2}]$ where $X$ is geometrically distributed with parameter $p$, but I'm not sure how to finish my solution.

My first thought was to use the standard definition of expected value:

$$E[g(X)]=\sum_x g(x)P(X=x)$$

Since $P(X=x)=p(1-p)^{x-1}$, this gives me:

$$E[{X \choose 2}]=\sum_x {x \choose 2}p(1-p)^{x-1}$$

I couldn't see a clear way to simplify this sum, so I figured I would try another method instead, using the law of total expectation. If I let $A_1$ be the event $X=1$ and $A_2$ be the event $X>1$, then I get this:

$$E[{X \choose 2}]=E[{X \choose 2}|A_1]P(A_1)+E[{X \choose 2}|A_2]P(A_2)$$

This is what I know:

$$E[{X \choose 2}|A_1]=0$$ $$P(A_1)=p$$ $$P(A_2)=1-p$$

That just leaves $E[{X \choose 2}|A_2]$ for what I'm unsure about. My guess is that it equals $E[{X+1 \choose 2}]$, but if that is the case, then I wouldn't know how to solve the result:

$$E[{X \choose 2}]=0p+E[{X+1 \choose 2}](1-p)=E[{X+1 \choose 2}](1-p)$$

So this is where I'm now stuck and I would appreciate some guidance on how to proceed. Thanks!

Answer 1

Let $x' = x-1$. Then

$\sum\limits_{x=1}^{\infty}\dfrac{x(x-1)}{2}p(1-p)^{x-1}$

$= \sum\limits_{x = 2}^{\infty}\dfrac{x(x-1)}{2}p(1-p)^{x-1}$

$=\sum\limits_{x'=1}^{\infty}\dfrac{(x'+1)x'}{2}p(1-p)^{x'}$

$= (1-p)\sum\limits_{x'=1}^{\infty}\dfrac{x'^2 + x'}{2}p(1-p)^{x'-1}$

$= (1-p)\sum\limits_{x'=1}^{\infty}(\dfrac{x'^2-x'}{2} + x')p(1-p)^{x'-1}$

$= (1-p)\sum\limits_{x'=1}^{\infty}\dfrac{x'^2-x'}{2}p(1-p)^{x'-1} + (1-p)\sum\limits_{x'=1}^\infty x'p(1-p)^{x'-1}$

$= (1-p)\sum\limits_{x'=1}^{\infty}\dfrac{x'(x'-1)}{2}p(1-p)^{x'-1} + (1-p)\sum\limits_{x'=1}^\infty x'p(1-p)^{x'-1}$

Thus, $E[\binom{X}{2}] = (1-p)E[\binom{X}{2}] + (1-p)E[X]$

Note that by definition, $\binom{n}{k} = 0$ for $n<k$, so you can use $x(x-1)/2$ instead of $\binom{x}{2}$.

Answer 2

More generally, let $k \ge 0$ and consider the function $$f_k(z) = \sum_{n=k}^\infty \binom{n}{k} z^n.$$ Clearly, $f_0(z) = (1-z)^{-1}$. We also recall that $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k},$$ a direct consequence of Pascal's triangle (and is easily proven). Therefore $$\begin{align*} f_k(z) &= z \sum_{n=k}^\infty \binom{n-1}{k-1} z^{n-1} + \sum_{n=k}^\infty \binom{n-1}{k} z^n \\ &= z f_{k-1}(z) + \left( \binom{k-1}{k} z^k + \sum_{n=k+1}^\infty \binom{n-1}{k} z^n \right) \\ &= zf_{k-1}(z) + \sum_{n=k}^\infty \binom{n}{k}z^{n+1} \\ &= zf_{k-1}(z) + zf_k(z). \end{align*}$$ It follows that $$f_k(z) = \frac{z}{1-z} f_{k-1}(z).$$ This recurrence, along with the initial condition for $f_0(z)$, gives us the explicit formula $$f_k(z) = \frac{z^k}{(1-z)^{k+1}}.$$ Now the desired sum to be evaluated is $$\operatorname{E}\left[\binom{X}{2}\right] = \sum_{x=2}^\infty \binom{x}{2} p(1-p)^{x-1},$$ which is simply $$\frac{pf_2(1-p)}{1-p} = \frac{p(1-p)^2}{(1-p)(1-(1-p))^3} = \frac{1-p}{p^2}.$$ Moreover, it is now trivial to compute $$\operatorname{E}\left[\binom{X}{k}\right]$$ for any nonnegative integer $k$ and such geometrically distributed $X$.