$X\sim \text{Exp}(\lambda)$ use the moment generating function ($m_X(t)$) to find $E(X)$ and $E(X^2)$

Question

Q1) Let $X\sim\text{Exp}(\lambda)$. Find $m_X(t)$.

My attempt: $$m_X(t) = E[\text{e}^{tX}] = \int_{0}^{\infty}\, \text{e}^{tx} \lambda e^{-\lambda x}\,\text{d}x = \int_{0}^{\infty}\,e^{-\lambda x + tx} \lambda\,\text{d}x = \lambda \int_{0}^{\infty} \,\text{e}^{x(-\lambda+t)}\,\text{d}x\,.$$ Let $u = x(-\lambda+t)$, then $\text{d}u = (-\lambda +t)\,\text{d}x$, giving $\frac{1}{-\lambda +t}\,\text{d}u = \text{d}x$

If $t < \lambda$, then $$\frac{\lambda}{-\lambda +t} \int_{0(-\lambda +t) = 0}^{-\infty} e^{u}du = \frac{\lambda}{-\lambda +t} \left[e^{u} \right]_{0}^{-\infty} = \frac{\lambda}{\lambda - t}\,.$$

Q2) Use the moment generating function $m_X(t)$ to find $E(X)$ and $E(X^2)$.

My attempt: $$m_X(t) = \frac{\lambda}{\lambda - t} = \lambda(\lambda - t)^{-1}\,.$$

How?

Answer 1

Write $$m_X(t)=\mathbb{E}\big[\exp(tX)\big]=\mathbb{E}\left[\sum_{n=0}^\infty\,\frac{1}{n!}\,t^n\,X^n\right]=\sum_{n=0}^\infty\,\frac{t^n}{n!}\,\mathbb{E}\left[X^n\right]\,.$$ That is, $$\mathbb{E}\left[X^n\right]=m_X^{(n)}(0)\text{ for each }n=0,1,2,\ldots\,,$$ where $m_X^{(n)}$ denotes the $n$-th derivative of $m_X$.

On the other hand, $$m_X(t)=\dfrac{\lambda}{\lambda-t}=\left(1-\dfrac{t}{\lambda}\right)^{-1}=\sum_{n=0}^\infty\,\left(\frac{t}{\lambda}\right)^n$$ for $t\in\mathbb{C}$ with $|t|<\lambda$. You should be able to find $\mathbb{E}\left[X^n\right]$ for every $n\in\mathbb{Z}_{\geq 0}$ now.

We have $\mathbb{E}\left[X^n\right]=\dfrac{n!}{\lambda^n}$.

Answer 2

And you can calculate the first and second order derivatives to find$E(X),E(X^2)$

$E(X)=m'(0)=\frac{\lambda}{(\lambda-t)^2}_{|t=0}=\frac{1}{\lambda}$

$E(X^2)=m''(0)=\frac{2\lambda}{(\lambda-t)^3}_{|t=0}=\frac{2}{\lambda^2}.$

Gernerally,$m^{(k)}(0)=E(X^k)$

Answer 3

Suppose $X \sim Gamma(\alpha, \lambda)$

Now consider that an exponential distribution is a special case of a gamma distribution when $\alpha =1$

$$M(t) = \int_{0}^{\infty} e^{tx} \frac{\lambda^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} dx $$ $$ = \frac{\lambda^{\alpha}}{\Gamma(\alpha)} \int_{0}^{\infty} x^{\alpha-1} e^{x(t-\lambda)}$$

This converges for $ t < \lambda $

$$ M(t)= \frac{\lambda^{\alpha}}{\Gamma(\alpha)}\left( \frac{\Gamma(\alpha)}{(\lambda-t)^{\alpha}}\right) = \left( \frac{\lambda}{\lambda - t} \right)^{\alpha}$$

differentiating we get

$$ M^{'}(0) = E(X) = \frac{\alpha}{\lambda}$$

$$ M^{''}(0) = E(X^{2}) = \frac{\alpha(\alpha +1)}{\lambda^{2}}$$

inserting $\alpha =1$ $$ E(X) = \frac{1}{\lambda}$$ $$ E(X^{2}) = \frac{1(1+1)}{\lambda^{2} } = \frac{2}{\lambda^{2}}$$