# Prove that if X∼ Geometric (p) then, E(X)=q/p, Var(X)=q/p^2, m_X(t)=p(1-qe^t).

Prove that if $X\sim$ Geometric (p) then, $E\left(X\right)=\frac{q}{p}\phantom{\rule{1em}{0ex}}\mathrm{Var}\left(X\right)=\frac{q}{{p}^{2}}\phantom{\rule{1em}{0ex}}{m}_{X}\left(t\right)=p\left(1-q{e}^{t}\right)$
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Belen Solomon
Step 1
For the variance, it suffices to compute $E\left[{X}^{2}\right]$. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for $\frac{1}{\left(1-x{\right)}^{3}}$.
Here is a direct derivation:
$\begin{array}{rl}E\left[{X}^{2}\right]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}\\ \left(1-q\right)E\left[{X}^{2}\right]=E\left[{X}^{2}\right]-qE\left[{X}^{2}\right]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{m\ge 0}{m}^{2}p{q}^{m+1}\\ & =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{k\ge 1}\left(k-1{\right)}^{2}p{q}^{k}\\ & =\sum _{k\ge 1}\left(2k-1\right)p{q}^{k}\\ \\ & =2\sum _{k\ge 1}kp{q}^{k}-\sum _{k\ge 1}p{q}^{k}\\ & =2E\left[X\right]-\left(1-p\right)=2\frac{q}{p}-q=\frac{q\left(2-p\right)}{p}.\end{array}$
Given the above expression for $\left(1-q\right)E\left[{X}^{2}\right]$, you can then do a few more steps compute the variance of X.
Step 2
For the MGF, ${m}_{X}\left(t\right)=E\left[{e}^{tX}\right]=\sum _{k\ge 0}{e}^{tk}p{q}^{k}=p\sum _{k\ge 0}\left({e}^{t}q{\right)}^{k}$ which is a geometric series that you should already know how to compute. By the way, there is a typo in your expression for the MGF.
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Austin Rangel
Step 1
I would do the MGF calculation first, then use the result to compute the expectation and variance. That's probably not the intent of the problem, but it is mathematically valid.
${M}_{X}\left(t\right)=\mathrm{E}\left[{e}^{tX}\right]=\sum _{x=0}^{\mathrm{\infty }}{e}^{tx}\left(1-p{\right)}^{x}p.$
Step 2
Then, $\mathrm{E}\left[X\right]={M}_{X}^{\prime }\left(0\right),\phantom{\rule{1em}{0ex}}\mathrm{E}\left[{X}^{2}\right]={M}_{X}^{″}\left(0\right),$, and $\mathrm{Var}\left[X\right]=\mathrm{E}\left[{X}^{2}\right]-\mathrm{E}\left[X{\right]}^{2}.$