Q. 7.51 - A First Course in Probability

Question

Use Table $7.2$ to determine the distribution of $\sum _{i=1}^n{X}_i$ when $X_1,\dots ,X_n$are independent and identically distributed exponential random variables, each having mean$1/\lambda$.

Step-by-Step Solution

Answer

The identically distributed exponential random variables value are$\sum X_i~Gamma(n,\lambda )$.

Step 1: Given Information

Determine the distribution of $\sum _{i=1}^n{X}_i$,when $X_1,\dots ,X_n$are independent.  

Step 2: Explanation

We have that $X_1,\dots ,X_n$ have distribution $Expo\left(\lambda \right)$ and that they are independent.

So, they have all common MGF

$M\left(t\right)=\frac{\lambda }{\lambda -t}$

Define $Y=\sum _{i=1}^n{X}_i$we have that,

$M_Y\left(t\right)=E\left(e^{tY}\right)=E\left(e^{t\sum _{i=1}^n{X}_i}\right)$$=\prod _{i=1}^{n}E\left(e^{t{X}_i}\right)=M(t{)}^n$.

Step 3: Explanation

Where the third and fourth equality hold since variables$X_i$ are independent and equally distributed.

So, we have that,

$M_Y\left(t\right)={\left(\frac{\lambda }{\lambda -t}\right)}^n$

Now, using the calculated MGF of $Y$, we can recognize that $Y$ has Gamma distribution with parameters $n$ and $\lambda$.

Step 4: Final answer

We can recognize that $Y$ has Gamma distribution with parameters $n$and$\lambda$ value are $\sum X_i~Gamma(n,\lambda )$.