What will be the distribution of the difference of the longest and shortest waiting time?

Question

John would like to eat at one of the fast food restaurants around him. There are $n$ fast food restaurants around him, and the the waiting time at the $i$-th has exponential distribution: $p(T_i) \sim \exp(λ_i)$. What will be the distribution of the difference of the longest and shortest waiting time, if $λ_i=λ>0 ∀i$?
I think it will also be an exponential distribution, because we are just taking the difference of 2 exponential distribution. Is this answer good? Or should I / Can I calculate something else too?

Answer 1

Let's find the distribution of the longest waiting time $X_L$:

$$f_i(x) = \lambda_i e^{-\lambda_i x}$$

$$P(x_i > X) = 1-F_i(X) = e^{-\lambda_i X}$$

$$1- F_L(X) \equiv P (\forall i \,: \,x_i>X) = \prod_i e^{-\lambda_i X} = e^{-\sum_i \lambda_i X}$$ $$F_L(X) = 1-e^{-\sum_i \lambda_i X}$$ So the desired distribution is $$ f_L(X) = \frac{d}{dX}F_L(X) = \sum_i \lambda_i e^{-\sum_i \lambda_i X} $$ So the distribution of the longest waiting time is indeed exponential, with $\lambda$ being the sum of the individual $\lambda_i$s.

The distribution of the shortest waiting time $F_S(X)$ is a bit tougher to calculate: $$P(x_i < X) = F_i(X) = 1-e^{-\lambda_i x}$$

$$ F_S(X) \equiv P (\forall i \,: \,x_i<X) = \prod_i \left( 1-e^{-\lambda_i X}\right) $$ $$ f_S(X) = \frac{d}{dX}(F_S(X)) = \sum_i \frac{\lambda_i e^{-\lambda_i X}}{\left( 1-e^{-\lambda_i X}\right) } \prod_k \left( 1-e^{-\lambda_k X}\right) $$ That is not a very clean expression, but in fact the answer is simply not clean.