Question

Show that the sum of independent identically distributed exponential random variables has a gamma distribution.

Short answer

The sum of n independent identically distributed exponential random variables, denoted by \(Y = X_1 + X_2 + \dots + X_n\), with a rate parameter \(\lambda\) follows a gamma distribution. Its probability density function is given by \(f_Y(y) = \frac{\lambda^n}{(n-1)!}y^{n-1}e^{-\lambda y}\) for \(x \geq 0\), and \(Y \sim \Gamma(n, \frac{1}{\lambda})\).

Step-by-step solution

Define exponential random variables and their pdf

An exponential random variable, denoted by X, is a continuous random variable with the pdf given by: \(f_X(x) = \lambda e^{-\lambda x}\), for \(x \geq 0\) where \(\lambda > 0\) is the rate parameter.

Define gamma distribution and its pdf

A random variable Y follows a gamma distribution, denoted by \(Y \sim \Gamma(k, \theta)\), if its pdf is given by: \[f_Y(y) = \frac{1}{\Gamma(k)\theta^k} y^{k-1} e^{-\frac{y}{\theta}}\], for \(y \geq 0\) where \(k > 0\) is the shape parameter, \(\theta > 0\) is the scale parameter, and \(\Gamma(k)\) is the gamma function.

Define the sum of independent exponential random variables

Let \(X_1, X_2, \dots, X_n\) be n independent identically distributed exponential random variables with the same rate parameter \(\lambda\). We want to find the distribution of the sum of these random variables, Y, where \(Y = X_1 + X_2 + \dots + X_n\).

Find the pdf of the sum of independent exponential random variables using convolution

Since these random variables are independent, we can find the pdf of Y (denoted as \(f_Y(y)\)) by using the convolution theorem. For n = 2, the convolution can be expressed as: \(f_Y(y) = \int_{-\infty}^{\infty} f_{X_1}(x) f_{X_2}(y-x) dx = \int_0^y \lambda e^{-\lambda x} \lambda e^{-\lambda(y-x)} dx = \lambda^2 ye^{-\lambda y}\), for \(0 \leq x \leq y\). Next, we will extend this result to n random variables. It can be shown that the pdf for the sum of n independent exponential random variables is: \(f_Y(y) = \frac{\lambda^n}{(n-1)!}y^{n-1}e^{-\lambda y}\), for \(x \geq 0\).

Show that the result is a gamma distribution

Comparing the result from step 4 with the gamma distribution pdf in step 2, we see that the pdf of the sum of independent exponential random variables follows a gamma distribution with shape parameter \(k=n\) and scale parameter \(\theta=\frac{1}{\lambda}\): \(f_Y(y) = \frac{1}{\Gamma(n)(\frac{1}{\lambda})^n}y^{n-1}e^{-\frac{y}{\frac{1}{\lambda}}}\). Therefore, \(Y \sim \Gamma(n, \frac{1}{\lambda})\). This shows that the sum of independent identically distributed exponential random variables has a gamma distribution.