I'm interested in whether there is a closed-form distribution of the time it takes two Poisson processes to output counts to have a fixed difference. For example, let
$k_1$ ~ Poisson($\lambda_1t$)
$k_2$ ~ Poisson($\lambda_2t$)
Is there a closed-form solution for $p(t|k_1 - k_2 = n)$?
It's tricky because presumably that equality could happen multiple times while observing the processes. I'm specifically looking for the first time.
I know that $k_1-k_2$ ~ Skellam($\lambda_1,\lambda_2$) but I can't work out a distribution for time.