What is the distribution of the average of IID exponential RVs? (have to use MGFs)

Question

Just a quick question as I cannot figure this out. Here is the problem:

Q: What is the distribution of the following? ($Y_1 + Y_2 + Y_3 +\cdots+ Y_n$) $ /n$ if each $Y$ is independent and identically exponentially distributed (basically what is the distribution of the average)?

I am supposed to use the moment generating function to figure this out and here are my thoughts:

Lets call $z=(Y_1 + Y_2 + Y_3 +\cdots+ Y_n) / n $

So we want $M_z(t)= E (e^{tz}) $.

We also know that if we simply wanted the MGF of $z=(Y_1 + Y_2 + Y_3 +\cdots+ Y_n)$ we would have simply multiply the MGF of each RV (due to IID) and get $ (\lambda / \lambda - t)^n $ as the MGF.

How can we use this fact to get the MGF of the average RV $Z$ however? Help would be greatly appreciated and please show very detailed steps as I am DEFINITELY not mathematically/statistically inclined. Thank you so much!

Answer 1

1) The moment generating function of a sum of independent random variables is the product of the individual moment generating functions.

2) If $W=aV$, where $a$ is a constant, then the moment generating functions $M_V(t)$ and $M_W(t)$ are related by the equation $$M_W(t)=M_V(at).$$

If your exponentials $Y_i$ have parameter $\lambda$, then each has moment generating function $$M_{Y_i}(t)=\frac{1}{1-\frac{t}{\lambda}}.$$ Thus by applying 1), we find that $Y_1+Y_2+\cdots+Y_n$ has moment generating function $$\frac{1}{\left(1-\frac{t}{\lambda}\right)^n}.$$ Thus by 2), with $a=\frac{1}{n}$, the random variable $Z$ (I prefer upper case for rv) has moment generating function $$M_Z(t)=\frac{1}{\left(1-\frac{t}{n\lambda }\right)^n}.$$

To identify the distribution of $Z$, it is very helpful to have a "dictionary" of moment generating functions for various important families of distributions. If you have such a dictionary, you will recognize that $Z$ has gamma distribution. You can read off the parameters of the distribution of $Z$ from the shape of the mgf of the general gamma.

Answer 2

Say you have a random variable $Y=aX$, where X is a random variable and $a$ is a scalar. Then $$M_Y(t)=M_{aX}(t)=E[e^{t(aX)}]=E[e^{(ta)X}]=E[e^{(at)X}]=M_X(at)$$ Since you're taking an average, then your new random variable is the sum of your $Y_i$'s, each multiplied by the scalar $\frac1{n}$. By properties of the moment generating function, $$M_Z(t)=M_{(\frac1{n}\sum{Y_i})}(t)=\prod_{k=1}^n{M_{Y_k}\left(\frac{t}{n}\right)}=\left(\frac{\lambda}{\lambda-(t/n)}\right)^n$$