I need to solve a problem where I am asked to calculate $M=P(X(0)=2|X(1)=3,X(2)=4,X(3)=5)$ in a Yule pure birth process where $\lambda=1$, so $\lambda_n=\lambda.n=n$ and $\mu_i=0$ (the death rate is 0) for a CTMC: Continuous Time Markov Chain.
The first thing that comes to mind is to reformulate the problem using $P(B|A)=P(A,B)/P(A)$ where making:
$A:\{X(1)=3,X(2)=4,X(3)=5\}$ $B:\{X(0)=2\}$
I get:
$M=\frac{P(X(0)=2,X(1)=3,X(2)=4,X(3)=5)}{P(X(1)=3,X(2)=4,X(3)=5)}$
and decomposing the terms and using the markovian property: $M=\frac{P(X(3)=5|X(2)=4)P(X(2)=4|X(1)=3)P(X(1)=3|X(0)=2)}{P(X(3)=5|X(2)=4)P(X(2)=4|X(1)=3)}=P(X(1)=3|X(0)=2)$
However this doesn't seem right to me, and I don't know where I'm making the mistake. Suppose I solve the problem using Bayes rule: $P(A|B)P(B)=P(B|A)P(A)$
then $M=\frac{P(X(1)=3,X(2)=4,X(3)=5|X(0)=2)}{P(X(1)=3,X(2)=4,X(3)=5)}P(X(0)=2)$,
Here I can simplify the numerator and denominator again and get in the most reduced form:
$M=P(X(1)=3|X(0)=2)P(X(0)=2)$.
As you can see the P(X(0)=2) term popped up which is new. This answer makes more sense to me, but the problem is how am I supposed to calculate P(X(0)=2)? I'm assuming I can get it by computing the stationary distribution with $\pi.G=0$, and get $\pi_j=\pi_2$. However given that $\pi_j \rightarrow P_{ij}(t)$, when $t \rightarrow \infty$; $P_{ij}(t)\rightarrow 0$, given that $P_{ij}(t)=\binom{j-1}{i-1}e^{-i\lambda t}(1-e^{-\lambda t})^{j-1}, j\geq i$.
I can't seem to find the mistake, my math says that the M should be zero, but my intuition suspects that this isn't the case.
This is a strange question, because usually in a CTMC the value at $t=0$ is specified. The transition rates don't determine the probabilities by themselves, you have to say how the process starts out.
You could get the result in terms of the probabilities $P(X(0)=n)$: I get $$ P(X(0)=2 | X(1)=3, X(2)=4, X(3) = 5) = \dfrac{P(X(0)=2) P_{23}(1)}{P(X(0)=0) P_{03}(1) + P(X(0)=1) P_{13}(1) + P(X(0)=2) P_{23}(1) + P(X(0)=3) P_{33}(1)}$$
You are correct that for a pure birth process with $\lambda$ bounded below, there is no stationary distribution: the population just keeps increasing indefinitely.