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

My Solution: I used the moment generating function Mz(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?

Uckfield 2
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    You are making some confusions : you already know that the $X_i$ are exponentially distributed. You are interested in the distribution of the sum $S_k$. Furthermore, in general, you can not deduce the distribution of a random variable just by knowing its expectation and variance, so that reasoning is not correct either. – Stratos supports the strike Feb 17 '22 at 18:06

2 Answers2

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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 $S_k \sim Gamma(k, \lambda)$.

V. Vancak
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  • Hi, thank you for your help. Does that mean this question wants me to 'show that' the moment generating function of Mz(t) is equal to $(\frac{1}{(1-(\frac{t}{λ})})^{k}$. If so, does the question essentially want me to find the moment generating function of the exponential distribution and then put it to the power k? – Uckfield 2 Feb 17 '22 at 19:39
  • It depends on what you've learned. But, you can indeed do what you suggested. Find the moment generating function (MGF) of the exponential distribution, then use the property (or prove it) that the MGF function of sum of $k$ i.i.d random variables is the original MGF in the power of $k$. Then just use the bijection between the MGFs and the density functions. – V. Vancak Feb 17 '22 at 19:49
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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.

Golden_Ratio
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