Odds that a doctor has moved given that a doctor with that name is active at his postal code OK. I have a complex math problem at work that I've kind of gone down the rabbit hole with.
Assume: there is a background "doctor stops working" rate at $5\%/10$ years ($P(DS)$)
Assume: there is a background "new doctor with that name moved there" rate at $1\%/10$ years ($P(ND)$)
Also: $A$ means "this record is active", and it must be $100\%$ as we wouldn't be asking the question if the record were not active.
These are example values that I'll try to find otherwise. (I'm guessing finding good values for these will be a question on its own).
So, here's a situation. We know that a doctor filed a report 10 years ago. A doctor with that name and postal code filed a report today. What are the odds that he is the same person?
The reason that I'm having difficulty here is because I'm getting into circular logic. I can't use the $5\%$ background rate for a doctor's leaving, because the record is active. if we were to assume that nobody new ever came, it would have to be him, so the odds would be $100\%$. So I really need to find $P(DS\mid A)$. But $A = 1$, so that means that $P(DS\mid A)$ is $P(DS)$. Am I wrong? I tried other formulations where $P(DS\mid A)$ is predicated on $P(ND)$ but that's gone nowhere. Any hint as to how to proceed or even if this problem is solvable would be great.
Perhaps $A \neq 1$ - do I need to find the background rate of $A$, and I'm really using $A\mid(DS \cup ND)$?
Assuming that $A\neq 1$ by default, and it should be the background rate: $A$ would then be $(DS \cup ND)$. Does this not mean that I'm in circular logic again?
 A: Some of your constants seem a little wonky (a $1/20$ chance of retiring every $10$ years works out to an average career of $200$ years), and the problem set-up seems a little askew, but I'll make some assumptions and see if they work for you.
Your doctor's name, say, is John Doe.  Let's suppose that existing Dr John Doe's retire annually at rate $\mu$, and new Dr John Doe's appear annually at rate $\lambda$.  That is, the number $k$ of Dr John Doe's has a time-dependent distribution governed by the differential equations
$$
\frac{d}{dt} P_k(t) = \lambda P_{k-1}(t)+(k+1)\mu P_{k+1}(t)-(\lambda+k\mu)P_k(t) \qquad k > 0
$$
$$
\frac{d}{dt} P_0(t) = \mu P_1(t) - \lambda P_0(t)
$$
If the distribution is stationary, and you know the number $k_0$ of Dr John Doe's today, you can take advantage of the fact that this is a pure birth-death process and is therefore reversible.  The arrival and retirement of Dr John Doe's is independent, so we consider the Dr John Doe who reported today separately from everyone else.  Start with the distribution $P_k(0) = 1$ if $k = k_0-1$, $P_k(0) = 0$ otherwise.  Find the distribution $P_k(10)$.  We also let $p_0 = e^{-10\mu}$ be the probability that the Dr John Doe who reported today was working ten years ago.  Then the conditional probability that the Dr John Doe who reported today was the same one who reported ten years ago, given that he worked ten years ago and there were $k$ other Dr John Doe's working at that time, is
$$
P(\text{same Dr Doe} \mid \text{working ten years ago, $k$ others})
    = \frac{1}{k+1}
$$
Since it can't be the same Dr John Doe if he wasn't working ten years ago, we can write
$$
P(\text{same Dr Doe} \mid \text{$k$ others}) = \frac{p_0}{k+1}
$$
and then unconditioning, we get
$$
P(\text{same Dr Doe}) = p_0 \sum_{k=0}^\infty \frac{P_k(10)}{k+1}
$$
The analytical solution for $P_k(10)$ is probably pretty messy, involving some kind of Bessel function.
