I am in a stochastic processes course, and I am trying to apply a result about the minimum of iid exponentials.
Here is the result:
Let $X_1, \ldots, X_n$ be independent exponential random variables with rates $\lambda_i$. $I = j$ if $X_j < min_{k \neq j}X_k$
Then $P(I = j) = \frac{\lambda_j}{\sum_\lambda}$
Here is the problem to which I am applying it.
Two patients are waiting for kidney transplants; each will die in exponential time, with the respective rates being $\mu_A, \mu_B$. Kidneys become available as a Poisson process with parameter $\lambda$. A will receive the first kidney unless A dies before the first kidney becomes available. What is the probability that neither A nor B gets a kidney?
I am approaching the problem the following way.
If neither A nor B gets a kidney, then $S_1$, the arrival time of the first kidney, must be greater than $D_A$ and $D_B$, the times of death of $A$ and $B$.
All three are exponential, and independent.
Therefore, $Pr(D_A < S_1, D_B < S_1)$ can be treated as $D_A$ being the minimum of exponential random variables $\{D_A, S_1\}$, and similarly for $\{D_B, S_1\}$.
$Pr(D_A < S_1, D_B < S_1) = Pr(D_A~ \text{min of} \{D_A, S_1\}) Pr(D_B ~\text{min of} \{D_B, S_1\})$
$= \big{(}\frac{\mu_A}{\mu_A + \lambda}\big{)}\big{(}\frac{\mu_B}{\mu_B + \lambda}\big{)}$
Is this reasoning sound?
$\big{(}\frac{\mu_A}{\mu_A + \lambda}\big{)}\big{(}\frac{\mu_B}{\mu_B + \lambda}\big{)}$ is unlikely to be the probability that $S_1$ is the maximum. For example, if $\mu_A = \mu_B = \lambda$ that expression would give $\frac14$ when by symmetry it should be $\frac13$
If $S_1=s$, then the conditional probability it is the maximum is $\left(1-e^{-\mu_A s}\right)\left(1-e^{-\mu_B s}\right)$
So the overall probability that $S_1$ is the maximum is $$\int_0^\infty \lambda e^{-\lambda s}\left(1-e^{-\mu_A s}\right)\left(1-e^{-\mu_B s}\right)\, ds \\ = 1 - \frac{\lambda}{\lambda+ \mu_A} -\frac{\lambda}{\lambda+ \mu_B} +\frac{\lambda}{\lambda+ \mu_A+ \mu_B}$$
and this is $\frac13$ when $\mu_A = \mu_B = \lambda$