Writing the pdf for a Gamma Distribution

Question

Let $X_1, \dots, X_5$ be 5 independent variables from the exponential distribution with the mean $2$.

a) The pdf of $T=X_1 + \dots + X_5$

b) $P(T > 5)$

c) $E(T/5)$

Having a little trouble understanding this questions. From the "$T=X_1 + \dots + X_5$" I can see this is of Gamma Distribution. So $G(n, \lambda)$ = $G(5,1/2)$.

The pdf for Gamma is:

$$ f(x) = \frac{\lambda ^{\alpha }}{\Gamma \left(\alpha \right)}x^{\alpha -1}e^{-\lambda x} $$

Now this is where I get confused. If I were to plug in the values:

$$\frac{\left(\frac{1}{2}\right)^5}{\Gamma \left(5\right)}x^{5-1}e^{-\frac{1}{2}x}, x > 0$$

Feels like im doing something wrong.

Answer 1

You can use moment generating functions to show that the sum of five iid exponential random variables has the gamma distribution you show. (In your specific PDF, you have forgotten to set one of the $\alpha$'s to 5.)

The distribution has $E(T) = \alpha/\lambda = 10$ and $V(T) = \alpha/\lambda^2 = 20.$ See the Wikipedia article on 'gamma distributions", under rate parameterization, if the details are not in your text.

Using statistical software, you can evaluate $P(T > 5).$ In R the computation is:

 1 - pgamma(5, 5, 1/2)
 ## 0.891178

The random variable $Y/5$ also has a gamma distribution, but you do not need to know that to find $E(\frac{1}{5}Y).$

Note: Suppose you are standing 4th in line waiting for a clerk to finish serving one customer and then the three people in line ahead of you. If the clerk's service time is exponential with rate is $\lambda = 1/2$ per minute (average service time 2 min.), then $P(T > 5)$ is the probability it will be more than five minutes before you are finished being served. (That's a total of five exponential waiting times before you're done.)

Answer 2

You are on the right track. To help you along, here are a number of things you should note:

1. $\Gamma (5)=4!$
2. $\alpha =5$, which you did not plug in everywhere
3. For, $P(T>5)$, compute the integral of the pdf with limits of integration between $5$ and $\infty $, which will yield the same result as $1-P(T\leq 5)=1-\int_{0}^{5}\frac{x^4e^{\frac{-x}{2}}}{768}dx=\frac{4169}{384e^\frac{5}{2}}\approx .89$.