Question

Starting with

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

show that

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

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

Short answer

The answer is

1. The expectation value of $e^{tX}$islocalid="1647527908191" $E\left[e^{tX}\right]=1+t\cdot E\left[X\right]+\frac{t^2}{2!}E\left[X^2\right]+\frac{t^3}{3!}E\left[X^3\right]+\dots +\frac{t^n}{n!}E\left[X^n\right]+\dots$.
2. It has been shown that${\left.\frac{d^n}{d{t}^n}M\left(t\right)\right|}_{t=0}=E\left[X^n\right]$.

Step-by-step solution

Given information (Part a)

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

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

Solution (Part a)

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

So,

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

$M\left(t\right)=E\left[e^{ıX}\right]$

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

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

$E\left[e^{\iota X}\right]=1+t\cdot E\left[X\right]+\frac{t^2}{2!}E\left[X^2\right]+\frac{t^3}{3!}E\left[X^3\right]+\dots +\frac{t^n}{n!}E\left[X^n\right]+\dots$

Final answer (Part a)

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

Given information (Part b)

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

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

Solution (Part b)

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

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

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

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

Solution (Part b)

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.

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

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

Final answer (Part b)

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