Let X(t) be a Poisson process with parameter $\lambda$. Calculate: $$ \mathbb{P}(3X(2) = 2X(4) | X(1) \leq 3) $$
Now, I tried to tweak the left side: $$ 3X(2) = 2X(4) \\ X(2) + X(2) + X(2) = X(4) + X(4) \\ X(4) + X(4) - X(3) - X(3) + X(3) + X(3) - X(2) - X(2) - X(2) = 0 \\ $$ Denoting $Y,Z$ as $X(4)-X(3), X(3)-X(2)$ respectively: $$ 2Y+2Z-X(2) = 0 $$ X(2),Y,Z are independent Poisson variables and $2Y+2Z-X(2)$ has a Poisson distribution with $3 \lambda$.
Back to main equation: $$ \mathbb{P}(3X(2) = 2X(4) | X(1) \leq 3) = \frac{\mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))} $$ The denominator $\mathbb{P}(X(1) \leq 3))$ is easy to calculate.
Unfortunately, $2Y+2Z-X(2)$ is not independent from $X(1)$ so I cant just do:
$$ \frac{\mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))} = \frac{\mathbb{P}(2Y+2Z-X(2)=0)\mathbb{P}( X(1) \leq 3)}{\mathbb{P}(X(1) \leq 3))} \\ = \mathbb{P}(2Y+2Z-X(2)=0) = e^{-3\lambda} $$
Can anyone take it from here? What do I do? I thought of doing stuff like(denote $X(2)-X(1)$ as Q) $$ \mathbb{P}(2Y+2Z-X(2)=0, X(1) \leq 3) =\mathbb{P}(2Y+2Z-(X(2)-X(1))-X(1)=0, X(1) \leq 3) = \mathbb{P}(2Y+2Z-Q=X(1), X(1) \leq 3) $$ and now I can just iterate: $$ \mathbb{P}(2Y+2Z-Q=X(1), X(1) \leq 3) = \mathbb{P}(2Y+2Z-Q=0), X(1)=0) + \mathbb{P}(2Y+2Z-Q=1), X(1)=1) + \mathbb{P}(2Y+2Z-Q=2), X(1)=2) + \mathbb{P}(2Y+2Z-Q=3), X(1)=3) $$ where $2Y+2Z-Q$ is just Poisson with $3\lambda$ and $X(1)$ Poisson with $\lambda$. Would that be correct?
Your method is correct, but using the independence of $X(t_2)-X(t_1)$ and $X(t_4)-X(t_3)$ when $t_3\ge t_2$ we can write your solution even more structured. $$ \Pr\{3X(2)=2X(4),X(1)\le 3\}{= \sum_{k=0}^3\Pr\{3X(2)=2X(4),X(1)=k\} \\= \sum_{k=0}^3\sum_{u=0}^\infty\Pr\{X(2)=2u,X(4)=3u,X(1)=k\} \\= \sum_{k=0}^3\sum_{u=0}^\infty\Pr\{X(2)-X(1)=2u-k,X(4)-X(2)=u,X(1)=k\} \\= \sum_{k=0}^3\sum_{u=0}^\infty\\\Pr\{X(2)-X(1)=2u-k\}\Pr\{X(4)-X(2)=u\}\Pr\{X(1)=k\} \\= \sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty\\\Pr\{X(2)-X(1)=2u-k\}\Pr\{X(4)-X(2)=u\}\Pr\{X(1)=k\} \\= \sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty e^{-4\lambda}\frac{\lambda^{2u-k}(2\lambda)^u\lambda^k}{(2u-k)!u!k!} \\= \sum_{k=0}^3\sum_{u\ge{k\over 2}}^\infty e^{-4\lambda}\frac{(2\lambda^3)^u}{(2u-k)!u!k!} } $$ I wonder if the last summation has a closed form.