Find $\mathbb{E}[X(X-1)]$ for the geometric distribution without using derivation

Question

I'm trying to find the $\mathbb{E}[X(X+1)]$ for the geometric distribution.

Everywhere I've looked explains how to do it using the derivative but I have not been taught that method. Is there another way to do this? I know that $\mathbb{E}[X(X+1)] = E[X^2]+\mathbb{E}[X]$, but I am unable to calculate $E[X^2]$. Any tips would be greatly appreciated. Thank you in advance!

yes I do, but after I find $E(X^2)$ I'm expected to find the variance so unfortunately I can't do it backwards like that — user229257, Dec 04 '20 at 02:07
there are two variants of the geometric distribution; what is the smallest value $X$ can take in your case? 0 or 1? — LinAlg, Dec 04 '20 at 02:09
1 in this case with the probability distribution = $p(1-p)^[x-1]$ — user229257, Dec 04 '20 at 02:12
see https://math.stackexchange.com/questions/3398487/proving-variance-of-geometric-distribution — LinAlg, Dec 04 '20 at 02:19

score 0 · Accepted Answer · 2020-12-04T02:22:50.457

0

Note that $$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$ Now, we know that $$\mathbb{V}ar[X]=\mathbb{E}[X^{2}]-[\mathbb{E}[X]]^{2}$$ Using that, we can see that $$\mathbb{E}[X(X+1)]=\mathbb{V}ar[X]+[\mathbb{E}[X]]^{2}+\mathbb{E}[X]$$ So, since that $X\sim \textbf{Geometric}(p)$ so, $$\mathbb{E}[X]=\frac{1}{p} \quad \text{and} \quad \mathbb{V}ar[X]=\frac{1-p}{p^{2}}$$

Other approach is by definition, we can see that $$\mathbb{E}[X(X+1)]=\mathbb{E}[X^{2}+X]=\mathbb{E}[X^{2}]+\mathbb{E}[X]$$ Now, we know that $\mathbb{E}X=\frac{1}{p}$ and we can see that $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]$$ with $x=1,2,\ldots$ and $\mathbb{P}[X=x]=p(1-p)^{x-1}$. So, you need to find $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}$$ Using that $0<p\leq 1$, so we can find that $$\mathbb{E}[X^{2}]=\sum_{x}x^{2}\mathbb{P}[X=x]=\sum_{x=1}^{\infty}x^{2}p(1-p)^{x-1}=\frac{2-p}{p^{2}}$$

edited Dec 04 '20 at 02:22

answered Dec 04 '20 at 02:07

is there another way to do this if I don't know the variance? I'm expected to calculate it and then use it to find variance – user229257 Dec 04 '20 at 02:09
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also, isn't Var(X) = $E(X^2) - (E(X))^2$ ? – user229257 Dec 04 '20 at 02:10
You can say $$E(X(X+1))=\sum_{k=1}^{\infty}k(k+1)P(X=k)$$ and then find the sum using calculus – Matthew H. Dec 04 '20 at 02:11
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@user229257 Yes, I have a typo. I corrected. – Dec 04 '20 at 02:14
I've tried that but get stuck when it comes to finding the $ E(X^2)$ part – user229257 Dec 04 '20 at 02:14
@АлександрПальма any tips on how to find$ \sum_{x=1}^{\infty} x^2 p(1-p)^[x-1]$? That's where I'm running into problems – user229257 Dec 04 '20 at 02:17
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@user229257 Try differentiating the geometric series twice to compute the following sum for $|x|<1$: $$\sum_{k=0}^{\infty} k^2 x^k$$ – projectilemotion Dec 04 '20 at 02:41
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@user229257 just use the hint of projectilemotion or LinAlg https://math.stackexchange.com/a/3398648/798113 – Dec 04 '20 at 03:16

Find $\mathbb{E}[X(X-1)]$ for the geometric distribution without using derivation

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