Jensen Mclean

2022-09-03

Prove that if $X\sim $ Geometric (p) then, $E(X)=\frac{q}{p}\phantom{\rule{1em}{0ex}}\mathrm{Var}(X)=\frac{q}{{p}^{2}}\phantom{\rule{1em}{0ex}}{m}_{X}(t)=p(1-q{e}^{t})$

Belen Solomon

Beginner2022-09-04Added 5 answers

Step 1

For the variance, it suffices to compute $E[{X}^{2}]$. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for $\frac{1}{(1-x{)}^{3}}$.

Here is a direct derivation:

$\begin{array}{rl}E[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}\\ (1-q)E[{X}^{2}]=E[{X}^{2}]-qE[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{m\ge 0}{m}^{2}p{q}^{m+1}\\ & =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{k\ge 1}(k-1{)}^{2}p{q}^{k}\\ & =\sum _{k\ge 1}(2k-1)p{q}^{k}\\ \\ & =2\sum _{k\ge 1}kp{q}^{k}-\sum _{k\ge 1}p{q}^{k}\\ & =2E[X]-(1-p)=2\frac{q}{p}-q=\frac{q(2-p)}{p}.\end{array}$

Given the above expression for $(1-q)E[{X}^{2}]$, you can then do a few more steps compute the variance of X.

Step 2

For the MGF, ${m}_{X}(t)=E[{e}^{tX}]=\sum _{k\ge 0}{e}^{tk}p{q}^{k}=p\sum _{k\ge 0}({e}^{t}q{)}^{k}$ 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.

For the variance, it suffices to compute $E[{X}^{2}]$. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for $\frac{1}{(1-x{)}^{3}}$.

Here is a direct derivation:

$\begin{array}{rl}E[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}\\ (1-q)E[{X}^{2}]=E[{X}^{2}]-qE[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{m\ge 0}{m}^{2}p{q}^{m+1}\\ & =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{k\ge 1}(k-1{)}^{2}p{q}^{k}\\ & =\sum _{k\ge 1}(2k-1)p{q}^{k}\\ \\ & =2\sum _{k\ge 1}kp{q}^{k}-\sum _{k\ge 1}p{q}^{k}\\ & =2E[X]-(1-p)=2\frac{q}{p}-q=\frac{q(2-p)}{p}.\end{array}$

Given the above expression for $(1-q)E[{X}^{2}]$, you can then do a few more steps compute the variance of X.

Step 2

For the MGF, ${m}_{X}(t)=E[{e}^{tX}]=\sum _{k\ge 0}{e}^{tk}p{q}^{k}=p\sum _{k\ge 0}({e}^{t}q{)}^{k}$ 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.

Austin Rangel

Beginner2022-09-05Added 2 answers

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.

${M}_{X}(t)=\mathrm{E}[{e}^{tX}]=\sum _{x=0}^{\mathrm{\infty}}{e}^{tx}(1-p{)}^{x}p.$

Step 2

Then, $\mathrm{E}[X]={M}_{X}^{\prime}(0),\phantom{\rule{1em}{0ex}}\mathrm{E}[{X}^{2}]={M}_{X}^{\u2033}(0),$, and $\mathrm{Var}[X]=\mathrm{E}[{X}^{2}]-\mathrm{E}[X{]}^{2}.$

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.

${M}_{X}(t)=\mathrm{E}[{e}^{tX}]=\sum _{x=0}^{\mathrm{\infty}}{e}^{tx}(1-p{)}^{x}p.$

Step 2

Then, $\mathrm{E}[X]={M}_{X}^{\prime}(0),\phantom{\rule{1em}{0ex}}\mathrm{E}[{X}^{2}]={M}_{X}^{\u2033}(0),$, and $\mathrm{Var}[X]=\mathrm{E}[{X}^{2}]-\mathrm{E}[X{]}^{2}.$

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