Joint distribution of sums of exponential random variables
math
Author
Stefan Eng
Published
May 17, 2020
This is a problem from Ross’s Stochastic Processes [1]. Let \[
\begin{aligned}
S_1 &= X_1\\
S_2 &= X_1 + X_2\\
S_3 &= X_1 + X_2 + X_3
\end{aligned}
\] where \(X_1, X_2, X_3\) are i.i.d exponential random variables with rate \(\lambda\). Find the joint distribution of \(S_1, S_2, S_3\).
Let \(f\) be the PDF of each \(X_1, X_2, X_3\) (since identically distributed). Since \(X_1, X_2, X_3\) are independent the joint PDF is \[
f(x,y,z) = f(x) f(y) f(z) = \lambda^3 e^{-\lambda x} e^{-\lambda y} e^{-\lambda z}
\] Then we can find the joint CDF of \(S_1, S_2, S_3\)\[
\begin{aligned}
P(S_1 \leq t_1, S_2 \leq t_2, S_3 \leq t_3) &= \int_{0}^{t_1} \int_{0}^{t_2 - x} \int_{0}^{t_3 - x - y} f(x,y,z) ~dz~dy~dx\\
&= \int_{0}^{t_1} \int_{0}^{t_2 - x} \int_{0}^{t_3 - x - y} \lambda e^{-\lambda z} \lambda e^{-\lambda y} \lambda e^{-\lambda x}~dz~dy~dx\\
&= \int_{0}^{t_1} \lambda e^{-\lambda x} \int_{0}^{t_2 - x} (1 - e^{-\lambda (t_3 - x - y)}) \lambda e^{-\lambda y}~dy~dx\\
&= \int_{0}^{t_1} \lambda e^{-\lambda x} \left[\int_{0}^{t_2 - x} \lambda e^{-\lambda y}~dy - \int_{0}^{t_2 - x} e^{-\lambda (t_3 - x)}~dy\right]~dx\\
&= \int_{0}^{t_1} \lambda e^{-\lambda x} \left[(1 - e^{-\lambda (t_2 - x)}) - (t_2 - x) e^{-\lambda (t_3 - x)}\right]~dx\\
&= \int_{0}^{t_1} \lambda e^{-\lambda x} ~dx - \int_{0}^{t_1} \lambda e^{-\lambda t_2} ~dx - \int_{0}^{t_1} (t_2 - x) e^{-\lambda t_3}~dx\\
&= 1 - e^{-\lambda t_1} - \lambda t_1 e^{-\lambda t_2} - t_1 t_2 e^{-\lambda t_3} + \frac{1}{2} t_1^2 e^{-\lambda t_3}
\end{aligned}
\]
Simulation
We can confirm these results with a simple simulation in R.