Solution: MGF Computation — Exponential and Gamma

Exercise: MGF Computation: Exponential and Gamma

Part 1

M_X(s) = \int_0^\infty e^{sx}\cdot \lambda e^{-\lambda x}\,dx = \lambda\int_0^\infty e^{-(\lambda - s)x}\,dx.

This converges iff $\lambda - s > 0$ , i.e. $s < \lambda$ . For such $s$ :

M_X(s) = \lambda\cdot \frac{1}{\lambda - s} = \frac{\lambda}{\lambda - s}, \qquad s < \lambda.

Part 2

$M_X'(s) = \lambda/(\lambda - s)^2$ . At $s = 0$ : $M_X'(0) = 1/\lambda = \mathbb{E}[X]$ . ✓

$M_X''(s) = 2\lambda/(\lambda - s)^3$ . At $s = 0$ : $M_X''(0) = 2/\lambda^2 = \mathbb{E}[X^2]$ .

So $\text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = 2/\lambda^2 - 1/\lambda^2 = 1/\lambda^2$ . ✓

Part 3

By independence and the product-of-MGFs rule:

M_Y(s) = \prod_{i=1}^n M_{Y_i}(s) = \prod_{i=1}^n \frac{\lambda}{\lambda - s} = \left(\frac{\lambda}{\lambda - s}\right)^n, \qquad s < \lambda. \quad \checkmark

Part 4

From the table of canonical MGFs, $(\lambda/(\lambda - s))^n$ is the MGF of a gamma distribution with shape $n$ and rate $\lambda$ — i.e. $Y \sim \text{Gamma}(n, \lambda)$ .

By the uniqueness theorem, this identification is rigorous: the sum of $n$ i.i.d. exponentials is gamma with integer shape $n$ .

Takeaways

Exponential MGF is a rational function $\lambda/(\lambda - s)$ , finite only for $s < \lambda$ . Limited domain is typical of exponential-tailed distributions.
MGF differentiation reads moments. First derivative at $0$ gives $\mathbb{E}[X]$ ; second derivative gives $\mathbb{E}[X^2]$ ; variance follows.
Gamma $=$ sum of i.i.d. exponentials (when shape is integer). This is the classical proof — three lines with MGFs. It also explains why gamma is the waiting time until the $n$ -th Poisson arrival.
Product-of-MGFs under independence is the engine. In nearly every "distribution of a sum" question, the MGF (or characteristic function) approach is faster and more rigorous than direct convolution.