Find probability: $P(X_2(t)=0|X_0(t)>0)$ where $X_i(t)$ model the population count at time $t$.

Question

We have $3$ types of cells: $i=0,1,2$. Each cell type can replicate at rate $b_i$, die at $d_i$ with the net growth rate $r_i=b_i-d_i$

The number of cells at time $t$ is given by $X_0(t),X_1(t),X_2(t)$ respectevely.

The evolution pattern where $\mu_i$ are exponential rates (So at rate $\mu_0$ type $0$ becomes type $1$): $$\text{Type 0} \xrightarrow{\mu_0} \text{Type 1} \xrightarrow{\mu_1} \text{Type 2} $$

I need to find the probability of no cells of type $2$ at time $t$ such that there are cells in the system: $$P(X_2(t)=0|X_0(t)>0)$$

I'm given the approximation that $X_0(t)\approx Ve^{r_0t}$ Edit: I realised $X_0(t)\approx Ve^{r_0t}$ is with the condition on not dying out (e.g. $X_0(t)>0$)

They find the number of cells of type $1$ at time $t$ to be: $$X_1(t)\approx \int_0^t Ve^{r_0s} \mu_0 e^{r_1(t-s)} ds=\mu_0 V\frac{e^{r_0t}(1-e^{(r_1-r_0)t})}{r_0-r_1} \approx \mu_0 V\frac{e^{r_0t}}{r_0-r_1} $$

Now the next part, I don't understand at all, how do they get expectation from probability? (I think the distibution (not sure since I derived it) of $V$, is $f_V(x)=(\frac{r_0}{b_0}\exp \{-r_0x/b_0\}$)

$$P(X_2(t)=0|X_0(t)>0)=E \exp \{-\mu_1 \int_0^t X_1(s) \frac{r_2}{b_2-d_2e^{-r_2(t-s)}} ds\}$$

My thoughts: ----------------------------------------------------------------------------------

I think they are conditioning on the value of $V$ and then integrating:

$$P(X_2(t)=0|X_0(t)>0)=\int^\infty_0 P(X_2(t)=0|Ve^{r_0t}=\sigma) f_V(\sigma)$$

Answer 1

Populations are often modeled using point processes with stochastic intensity $\lambda(t)$ (also called Cox processes). The intensity $\lambda(t)$ describes the probability that an event (death birth or migration) occurs in an infinitesimal time interval around $t$.

The probability of extinction of $X_2(t)$ conditional on $\left\{X_0>0\right\}$ can be computed by conditioning on the intensity and then integrating (taking expectation) \begin{eqnarray}P(X_2(t)=0\vert X_0>0 )&=&E[\mathbf{1}_{\{X_2(t)=0\vert X_0>0\}}] \\ &=&E[E[\mathbf{1}_{\{X_2(t)=0\vert X_0>0\}}\vert \lambda]] \end{eqnarray} Keep in mind that the value of a (time-i homogeneous-)poisson process is poisson distributed(conditional on $\lambda$) with parameter $\int_0^t \lambda(s) ds$. So you get for the above probability $$=E[\exp(-\int_0^t \lambda(s) ds )].$$

If you plug in the intensity that the authors computed for $X_2$ you get the result.