Find $ \mathbb P \{ 2S_1 + 3S_2 \gt 2 \} $.

Question

The Problem:

Consider a Poisson process with $\lambda = 2$. Let $S_k$ denote the waiting time until the $k$th event. So, $S_k = \sum_{i=1}^{k} T_i $, where each $T_i$ represents the interarrival time between the $(i-1)th$ and $i$th event. It follows that $T_k, k = 1, 2, 3, ... ,$ are iid exponential random variables having mean $1/\lambda$.

Find $ \mathbb P \{ 2S_1 + 3S_2 \gt 2 \} $.

SOLUTION

I have the solution, but I'm wondering why the following approach doesn't work. I tried just decomposing the $S_i$'s into $T$'s. Observe:

$$ 2S_1 + 3S_2 = 2(T_1) + 3(T_1 + T_2) = 2\text{Expo}(2) + 3[\text{Expo}(2) + \text{Expo}(2)] $$

$$ = \text{Expo}(1) + 3[2(\text{Expo}(2)] = \text{Expo}(1) + 3[\text{Expo}(1)] $$

$$ = \text{Expo}(1) + \text{Expo}(1/3). $$

If this were the case, I could easily find the CDF of this sum of exponential random variables and proceed accordingly. However, this does not yield the correct solution, and I'm not sure why it doesn't. Is it, perhaps, because $T_1$ is (clearly) not independent of $T_1$, or something?

Answer 1

Your notation is not good. You claim that $$\text{Expo}(2)+\text{Expo}(2) = 2\text{Expo}(2)$$ which is false. If $X,Y$ are iid Exp($\lambda$), then $$X+Y\sim\text{Gamma}(2,\lambda),$$ not $$X+Y = 2X$$ nor $$X+Y\sim\,2\text{Expo}(\lambda).$$

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Further it will be a good exercise for you to show that if $c>0$, $X\sim \text{Exp}(\lambda)$, $Z\sim \text{Gamma}(n,\lambda)$, then $$cX\sim\text{Exp}(\lambda/c)$$ and $$cZ\sim\text{Gamma}(n,\lambda/c).$$

Answer 2

Comment: Here is a simulated answer for the case $\lambda = 5,$ against which you can check your analytic answer (2 or 3 place accuracy).

  t1 = rexp(10^6, 5);  t2 = rexp(10^6, 5)
  mean(5*t1 + 3*t2 > 2)
  ## 0.284454
  s1 = t1;  s2 = t1 + t2
  mean(2*s1 + 3*s2 > 2)
  ## 0.284454
  y = 2*s1 + 3*s2
  hist(y, prob=T, col="wheat")
  abline(v = 2, col="green4")