Question: > 
> Suppose that $X\_1,X\_2,\dotsc,X\_n$ is a random sample of size $n = 5$ of independent random variables from an exponential distribution with parameter $\beta= 4$.
> 
> 
> Suppose $Y= X\_1 + X\_2 + X\_3 + X\_4 + X\_5$. Use moment generating functions to find the
> distribution of $Y$.
> 
> 
> 

I'm just not sure where to start. I think the moment generating function for an exponential distribution is $\frac{1}{1-\beta t}$, which would mean each $X\_n$ would have a moment generating function of $\frac{1}{1-4t}$. I'm not sure how I would use this information to find the distribution of $Y$.

Answer: 
Use the fact that if $M\_X(t)$ and $M\_Y(t)$ are the MGFs of independent random variables $X$ and $Y$, then the MGF of their sum $X+Y$ is given by $$M\_{X+Y}(t) = \operatorname{E}[e^{tX + tY}] \overset{\text{ind}}{=} \operatorname{E}[e^{tX}]\operatorname{E}[e^{tY}] = M\_X(t) M\_Y(t);$$ that is to say, the MGF of a sum of independent random variables equals the product of the MGFs of each variable. From this, you can deduce that the sum $Y$ of $n$ IID exponential random variables $X\_1, \ldots, X\_n$ with common mean parameter $\beta$ has MGF $$M\_Y(t) = (1 - \beta t)^{-n}.$$ What type of random variable corresponds to such an MGF?
The moment generating function of the sum of independent identically distributed random variables is given by the product of moment generating functions:
$$m\_Y(t)= \prod\_{i=1}^5 m\_{X\_i}=(1-4t)^{-5}$$What distribution has this moment generating function?