# Exponential distribution of 2 independent events (expected value)

X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?

I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?

• Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve? – Did Dec 8 '18 at 16:20
• How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right? – Jun Dec 8 '18 at 16:26
• So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right? – Did Dec 8 '18 at 16:29
• I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with. – Jun Dec 8 '18 at 16:31

The next phone call is the minimum of $$X$$ and $$Y$$; write $$Z=X\wedge Y$$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $$X\sim\mathsf{Exp}(\lambda)$$ and $$Y\sim\mathsf{Exp}(\mu)$$, then for any $$t>0$$ we have \begin{align} \mathbb P(Z>t) &= \mathbb P(X>t, Y>t)\\ &= \mathbb P(X>t)\mathbb P(Y>t)\\ &= e^{-\lambda t}e^{-\mu t}\\ &= e^{-(\lambda+\mu)t}, \end{align} so that $$Z\sim\mathsf{Exp}(\lambda+\mu)$$. Plugging in $$\lambda = \frac1{10}$$, $$\mu=\frac18$$, and $$t=5$$, we have $$\mathbb P(Z\leqslant 5) = 1 - e^{-\left(\frac1{10}+\frac18 \right)\cdot5} = 1-e^{-\frac98}.$$ The expected value of $$Z$$ is simply $$\left(\frac1{10}+\frac18 \right)^{-1}=\frac{40}9.$$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote. However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $$P((X < 5) \lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)\land(Y<5))$$
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