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.

# Computed joint CDF
expect_joint <- function(t1, t2, t3, lambda = 1) {
  1 - exp(- lambda * t1) - lambda * t1 * exp(-lambda * t2) - 
    t1 * t2 * exp(-lambda * t3) + 1/2 * t1^2 *exp(- lambda * t3)
}

# Simulate 1000 realizations of S1, S2, S3
sim <- function(t1, t2, t3, rate = 1, n = 1000) {
  s1 <- rexp(n, rate)
  s2 <- s1 + rexp(n, rate)
  s3 <- s2 + rexp(n, rate)
  
  mean(s1 <= t1 & s2 <= t2 & s3 <= t3)
}

# P(S1 <= 1, S2 <= 2, S3 <= 3) with lambda = 1
t1 <- 1
t2 <- 2
t3 <- 3
rate <- 1
# Replicate the simulation 1000 times
sim_res <- replicate(1000, sim(t1, t2, t3, rate))

cat("Simulated mean:", round(mean(sim_res), 3))
Simulated mean: 0.422
cat("Expected joint distribution:", round(expect_joint(t1, t2, t3, rate), 3))
Expected joint distribution: 0.422

```

References

[1]
Ross, S.M. et al. 1996. Stochastic processes. Wiley New York.