Let's say that for $E([0,w), k)$ denotes the event of observing generated observation in the Poisson process k times in the interval $[0,w)$.
I know that I can write $$P(E([0,w],k)) = \frac{(\lambda w)^{k}}{k!} e^{-\lambda w}$$
How can I write the Poisson distribution of $P(E([0,a_{1} +a_{2}),k_{1} +k_{2}))$ similar to the equation above. I need to look at the behavior of $\lambda$ so I need that version.
Also I need some idea on the conditional probability of $P(E([0,a_{1}),k_{1})|E([0,a_{1} +a_{2}),k_{1} +k_{2}))$ if you can give some advice I would appreciate it.
Assuming $k$ is a constant you can write this as $$\tag{1} \mathbb P\Big(E\big([0,a_{1} +a_{2}),k_{1} +k_{2}\big)\Big)=\frac{(\lambda(a_1+a_2))^{k_1+k_2}}{(k_1+k_2)!}e^{-\lambda(a_1+a_2)}\,. $$ To tackle the second question we only need to find out what $$\tag{2} \mathbb P\Big(E\big([0,a_1),k_1\big)\cap E\big([0,a_{1} +a_{2}),k_{1} +k_{2}\big)\Big) $$ is and then divide this by (1). A simpler notation would be to use the Poisson process $N_t$ and write $E([0,w),k)$ as $\{N_w=k\}\,.$ Then (2) can be written as \begin{align} &\mathbb P\Big(N_{a_1}=k_1,N_{a_1+a_2}=k_1+k_2\Big)=\mathbb P\Big(N_{a_1}=k_1,N_{a_1+a_2}=N_{a_1}+k_2\Big)\\ &=\mathbb P\Big( N_{a_1}=k_1,N_{a_1+a_2}-N_{a_1}=k_2\Big)\,.\tag{3} \end{align} By the independence of the increments of $N_t$ this becomes \begin{align}\tag{4} &\mathbb P\Big( N_{a_1}=k_1\Big)\mathbb P\Big(N_{a_1+a_2}-N_{a_1}=k_2\Big)=\frac{(\lambda a_1)^{k_1}}{k_1!}e^{-\lambda a_1}\frac{(\lambda a_2)^{k_2}}{k_2!}e^{-\lambda a_2}\, \end{align} where we have used that $N_T-N_t$ is Poisson distributed with parameter $\lambda(T-t)\,.$ The final result is then \begin{align} &\mathbb P\Big(E\big([0,a_1),k_1\big)\Big|E\big([0,a_1+a_2),k_1+k_2\big)\Big)=\mathbb P\Big(N_{a_1}=k_1\,\Big|\,N_{a_1+a_2}=k_1+k_2\Big)\\ &=\frac{(k_1+k_2)!}{k_1!k_2!}\frac{a_1^{k_1}a_2^{k_2}}{(a_1+a_2)^{k_1+k_2}}={k_1+k_2\choose k_1}\frac{a_1^{k_1}a_2^{k_2}}{(a_1+a_2)^{k_1+k_2}}\,\tag{5} \end{align} which is indeed independent of $\lambda\,.$