# PMF of Discrete Random Variables

Prove that $$\frac{1} {(1-p)^2} = \sum_{x=2}^\infty x^2p^{x-2} \, .$$

The professor suggested that we start by taking the derivative of $$\frac{1} {(1-p)^2} = \sum_{x=1}^\infty xp^{x-1} \, .$$

So I have tried that, and then broken the derivative up into $$\sum_{x=2}^\infty x^2p^{x-2} \, - \sum_{x=2}^\infty xp^{x-2} \, .$$ by taking apart x(x-1)

but I can't get any further and I am extremely confused.

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1
You should consider adding some thoughts about your current approach to the problem. Without knowing how you're trying to do this, we don't have any idea what background you have or how best to help you. – Muphrid Feb 4 at 5:53

## 2 Answers

From $$\frac{1}{(1-p)^2} = \sum_{x \geq 1} x \cdot p^{x-1} \tag{1}$$ you obtain by differentiating (with respect to $p$):

$$\begin{align} \frac{2}{(1-p)^3} &= \sum_{x \geq 2} x \cdot (x-1) \cdot p^{x-2} = \sum_{x \geq 2} x^2 \cdot p^{x-2} - \underbrace{\sum_{x \geq 2} x \cdot p^{x-2}}_{\frac{1}{p} \cdot \sum_{x \geq 2} x \cdot p^{x-1}} \\ &= \sum_{x \geq 2} x^2 \cdot p^{x-2} - \frac{1}{p} \cdot \left( \sum_{x \geq 1} x \cdot p^{x-1} -1 \right) \\ &\stackrel{(1)}{=} \sum_{x \geq 2} x^2 \cdot p^{x-2} - \frac{1}{p} \cdot \left( \frac{1}{(1-p)^2} -1 \right) \end{align}$$

Hence

$$\sum_{x \geq 2} x^2 \cdot p^{x-2} = \frac{p^2-3p+4}{(1-p)^3}$$

(There has to be a mistake or typo in your claim, because if the equality $$\sum_{x \geq 2} x^2 \cdot p^{x-2} =\frac{1}{(1-p)^2}$$ holds, this would imply $$0 = \frac{1}{(1-p)^2} - \frac{1}{(1-p)^2} = \sum_{x \geq 2} x^2 \cdot p^{x-2} - \sum_{x \geq 1} x \cdot p^{x-1} = \sum_{x \geq 0} \underbrace{((x+2)^2-(x+1))}_{\geq 0} \cdot \underbrace{p^x}_{\geq 0}$$ by using (1). And this can only hold iff $$(x+2)^2-(x+1) = 0$$ for all $x \in \mathbb{N}$. And this is clearly not fulfilled.)

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For $\,|p|<1\,$ it is well known that

$$f(p):=\frac{1}{1-p}=\sum_{x=0}^\infty p^x\Longrightarrow f'(p)=\frac{1}{(1-p)^2}=\sum_{x=1}^\infty xp^{k-1}$$

So in fact your first equality's right hand must begin from $\,x=1\,$ and with $\,x\,$ but not squared...check this.

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That's exactly what Kathleen wrote in the second equation. (But as you can see in my answer, I would also say that there is some mistake in the claim.) – saz Feb 7 at 13:41
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