PMF of Discrete Random Variables

Question

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.

Answer 1

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.)

Answer 2

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.