[Math] Moment generating function of an exponential

moment-generating-functionsprobabilityprobability distributionsstatistics

I was usually given two terms in an exponential distribution which I combined and performed an integral to find the moment generating function.

What should I do here?

Let $X$~$Exponential(\lambda)$ for some $\lambda > 0$.

(a) Compute the moment generating function of $X$.

(b) Compute $E(X^n)$, $n\geq 1$.

(c) Let $Y=\lambda X$. Compute the moment generating function of $Y$.

Best Answer

(a) The moment generating function of $X \sim Exp(\lambda)$ is $M_X(t) = E[e^{t X}] = \int_{x \geq 0} \lambda e^{- \lambda x} e^{t x} dx = \frac{\lambda}{\lambda-t}$ for $t<\lambda$.

(b) To get the moments, note that $\frac{\lambda}{\lambda-t} = \lambda \frac{1}{1-\frac{t}{\lambda}} = \lambda \sum_{i=0}^\infty \frac{t^i}{\lambda^i}$ in a neighborhood of zero ($\sum_{i=0}^\infty x^i = \frac{1}{1-x}$). Compare this to $M_X(t) = \sum_{i=0}^\infty \frac{t^n E[X^i]}{i!}$ to read off the moments.

As for (c), note that $M_Y(t) = E[e^{t Y}] = E[e^{t \lambda X}] = E[e^{(t \lambda) X}] = M_X(t \lambda) = \frac{\lambda}{\lambda-t \lambda} = \frac{1}{1-t}$. Match this to the form of $M_X(t)$ for a particular value of $\lambda$ to see what distribution it is.