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?
Best Answer
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.