Is this the correct way to use the Moment Generating Function?

Question

Let Sk = X₁+X₂+· · ·+X_k such that X_i for i ∈ {1, 2, . . . , k} are independently and identically exponentially distributed random variables with rate λ, i.e. X_i ∼ Exp(λ). Prove with adequate reasoning that Sk ∼ Gamma(x; k, λ)

My Solution: I used the moment generating function M_z(t) = $(\frac{1}{(1-(\frac{t}{λ})})^{k}$ to find E($X$) and E($X^2$) which allowed me to find $Var(X)$. The results I got were $E(X) = \frac{k}{λ}$ and $Var(X) = \frac{k}{λ^2}$. Then I set $k = 1$ and I got: $E(X) = \frac{1}{λ}$ and $Var(X) = \frac{1}{λ^2}$ which I believe are the correct values for mean and variance for the exponential distribution. Is this the correct solution?

Answer 1

Not really, you should start with the moment generating function of $X_i$, where $X_i \sim Exp (\lambda)$, denote it by $M_{X_i}(t)$. Now define $S_k = \sum_{i=1}^k X_i $, thus $M_{S_K}(t) = (M_X(t))^k$. Next, you should identify $M_{S_K}(t)$ as the moment generating function of $Gamma(k, \lambda)$, and then you can conclude that $S_k \sim Gamma(k, \lambda)$.

Answer 2

The mean and variance is not enough to characterize a distribution, but the MGF is.

Recall the MGF of a sum of independent RVs is the product of their MGFs. So the MGF of $S_k$ is

$$\prod_{i=1}^k M_{X_i}(t)=\left(\frac{1}{1-t/\lambda}\right)^k,t<\lambda.$$

Looks like you already found that; now just recognize this is the MGF of a gamma random variable with the stated parameters and you are done.