Number of deaths in birth-death process conditioned on start and end points

Question

Say I have a simple linear continuous time birth-death process with state space the non-negative integers, where there are parameters $b$ and $d$, with the rate (as you'd see in a $Q$ matrix) of going from state $k$ to $k + 1$ is $kb$ and the rate of going from state $k$ to $k-1$ is $kd$.

Now suppose that at time 0 I'm in state $i$ and at time $t$ I'm in state $j$. I want to know the probability that $h$ of the original $i$ individuals have "died" in the meantime. So we consider a state to refer to a collection of homogeneous individuals, each of which can "give birth" to a new identical individual (with rate $b$) or "die" (with rate $d$).

My first thought was just to consider the death process embedded in the original system — just ignore births, and consider only the death rates of the original $i$ individuals and find the probability that $h$ are lost in time $t$. The problem with that is that it doesn't take into account the total size of the final population. So if we go from 10 to 5, the conditional probability that only 2 of the original 10 are lost should be 0, which it won't be if we just imagine a death process ignoring the destination state.

How could I go about calculating these conditional probabilities?

Answer 1

Let $X(t)$ be the total population of individuals alive at time $t$ and $Y(t)$ the number of original individuals (present at $t=0$) surviving at time $t$. Thus we have $(X(0), Y(0)) = (i,i)$ and the following transitions:

$$ \matrix{ (X(t), Y(t)) \to (X(t)+1, Y(t)+1) & \text{with rate } & b X(t) \cr (X(t), Y(t)) \to (X(t)-1, Y(t)) & \text{with rate } & d (X(t)-Y(t))\cr (X(t), Y(t)) \to (X(t)-1, Y(t)-1) & \text{with rate } & d Y(t)\cr} $$

If $\phi(x,y,t) = E[x^{X(t)} y^{Y(t)}]$ is the probability generating function for this process at time $t$, I get the PDE

$$ \dfrac{\partial \phi}{\partial t} = b x (x-1) \dfrac{\partial \phi}{\partial x} + d \left(\dfrac{1}{xy} - 1\right) y \dfrac{\partial \phi}{\partial y} + d \left(\dfrac{1}{x}-1\right) \left(x \dfrac{\partial \phi}{\partial x} - y \dfrac{\partial \phi}{\partial y}\right)$$ with initial condition $\phi(x,y,0) = x^i y^i$. Maple gives me the solution

$$ \phi(x,y,t) = \left( {\frac {d \left( x-1 \right) {{\rm e}^{t \left( -d+b \right) } }-x \left( y-1 \right) \left( -d+b \right) {{\rm e}^{-td}}-bx+d}{b \left( x-1 \right) {{\rm e}^{t \left( -d+b \right) }}-bx+d}} \right) ^{i} $$

For small $i$ and $j$ you can explicitly extract the relevant Taylor series coefficients of $\phi(x,y,t)$ to find the probabilities of $X(t) = j, Y(t) = k$, $0 \le k \le \min(i, j)$. They get rather complicated as $i$ and $j$ increase.

For example, with $i=3$ and $j=2$, the coefficient of $x^2$ in $\phi(x,y,t)$ is

$$ \eqalign{&-6\,{\frac { \left( -{{\rm e}^{t \left( -d+b \right) }}d+d \right) ^{3 }{b}^{2} \left( {{\rm e}^{t \left( -d+b \right) }}-1 \right) ^{2}}{ \left( b{{\rm e}^{t \left( -d+b \right) }}-d \right) ^{5}}}\cr-9\,&{ \frac { \left( -{{\rm e}^{t \left( -d+b \right) }}d+d \right) ^{2} \left( {{\rm e}^{t \left( -d+b \right) }}d- \left( y-1 \right) \left( -d+b \right) {{\rm e}^{-td}}-b \right) b \left( {{\rm e}^{t \left( -d+b \right) }}-1 \right) }{ \left( b{{\rm e}^{t \left( -d+b \right) }}-d \right) ^{4}}}\cr-3\,&{\frac { \left( -{{\rm e}^{t \left( -d +b \right) }}d+d \right) \left( {{\rm e}^{t \left( -d+b \right) }}d- \left( y-1 \right) \left( -d+b \right) {{\rm e}^{-td}}-b \right) ^{2 }}{ \left( b{{\rm e}^{t \left( -d+b \right) }}-d \right) ^{3}}}} $$ and we can extract the coefficients of $y^0$, $y^1$, $y^2$ to get $P(X(t)=2, Y(t) = k)$ for $k=0,1,2$.