91Ó°ÊÓ

First equation $e^{tX}=1+tX+\frac{t^2{X}^2}{2!}+\frac{t^3{X}^3}{3!}+…+\frac{t^n{X}^n}{n!}+…$

Second equation$M\left(t\right)=E\left[e^{tX}\right]=1+tE\left[X\right]+\frac{t^{2}E\left[X^2\right]}{2!}+…+\frac{t^{n}E\left[X^n\right]}{n!}+…$

We need to find that the expectation value of $e^{tx}$

So,

$M\left(t\right)=∫\left(e^{tx}\right)f\left(x\right)dx$

$M\left(t\right)=E\left[e^{\mathrm{Ä\pm }X}\right]$

$⇒E\left[e^{uX}\right]=E\left[1+tX+\frac{t^2{X}^2}{2!}+\frac{t^3{X}^3}{3!}+…+\frac{t^n{X}^n}{n!}+…\right]$

$⇒E\left[e^{tX}\right]=E\left[1\right]+E\left[tX\right]+E\left[\frac{t^2{X}^2}{2!}\right]+E\left[\frac{t^3{X}^3}{3!}\right]+…+E\left[\frac{t^n{X}^n}{n!}\right]+…$

$E\left[e^{\mathrm{ι}X}\right]=1+tâ‹\dots E\left[X\right]+\frac{t^2}{2!}E\left[X^2\right]+\frac{t^3}{3!}E\left[X^3\right]+…+\frac{t^n}{n!}E\left[X^n\right]+…$

The expectation value of$e^{tX}$is$E\left[e^{\mathrm{′}X}\right]=1+tâ‹\dots E\left[X\right]+\frac{t^2}{2!}E\left[X^2\right]+\frac{t^3}{3!}E\left[X^3\right]+…+\frac{t^n}{n!}E\left[X^n\right]+…$

First equation $e^{tX}=1+tX+\frac{t^2{X}^2}{2!}+\frac{t^3{X}^3}{3!}+…+\frac{t^n{X}^n}{n!}+…$

Second equation${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=E\left[X^n\right]$

We need to show that ${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=E\left[X^n\right]$

${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=\frac{d^n}{d{t}^n}{\left(1+tâ‹\dots E\left[X\right]+\frac{t^2}{2!}E\left[X^2\right]+\frac{t^3}{3!}E\left[X^3\right]+…+\frac{t^n}{n!}E\left[X^n\right]+…\right)}_{t=0}$

${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=\frac{d^n}{d{t}^n}(1{)}_{t=0}+\frac{d^n}{d{t}^n}(tâ‹\dots E\left[X\right]{)}_{t=0}+\frac{d^n}{d{t}^n}{\left(\frac{t^2}{2!}E\left[X^2\right]\right)}_{t=0}+\frac{d^n}{d{t}^n}{\left(\frac{t^3}{3!}E\left[X^3\right]\right)}_{t=0}+…$

$+\frac{d^n}{d{t}^n}{\left(\frac{t^n}{n!}E\left[X^n\right]\right)}_{t=0}+…$

The terms with $t^i$for $i<n$vanish because the $n^{\text{th}}$derivative is 0 . On the other hand, terms with $t^i$for $i>n$vanish because the derivative is evaluated for $t=0$. Hence, only the $n^{th}$term remains.

${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=\frac{n!}{n!}E\left[X^n\right]$

${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=E\left[X^n\right]$

It has been shown that in the solution that is${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=E\left[X^n\right]$