So I have few questions on poisson Stochastic processes. I have these Probabilities to find:
1)$$P(X(1)=2,X(3)=6)$$
2)$$P(X(1)=2|X(3)=6)$$
3)$$P(X(3)=6|X(1)=2)$$
with $$P(X(t)=n)=\frac{exp(-\lambda*t)(\lambda*t)^n}{n!}$$
The problem is, my lecturer doesn't care if I have the right answer, I need to say what increments I use and I don't really know. I would say :
1)$$P(X(1)=2,X(3)=6)=P(X(1)=2,X(3-1)=6-2)=P(X(1),X(2)=4)=P(X(1)=2)*P(X(2)=4)$$
2)$$P(X(1)=2|X(3)=6)=\frac{P(X(3)=6|X(1)=2)P(X(1)=2)}{P(X(3)=6)}[by BAYES]$$
and then I would find $$P(X(3)=6|X(1)=2)=P(X(3)-X(1)=6-2)=P(X(2)=4)$$
and finish the calculations
3) I effectively found by doing 2. Could anyone tell me at which lines I should state which increments do I use? Thank you
\begin{eqnarray*} (1)\quad P\{X(1)=2,X(3)=6\}&=&P\{X(1)=2,\overline{X(3)-X(1)=4}\}\\ &=&P\{X(1)=2\}\times P\{X(3)-X(1)=4\}\quad\mbox{by ind. inc.}\\ &=&P\{X(1)=2\}\times P\{X(2)=4\}\quad \mbox{by stationary prop.} \end{eqnarray*}
(2) If $\{N(t),t\geq 0\}$ is a Poisson process with intensity parameter $\lambda>0$ and if $s$ and $t$ are two arbitrary time points such that $s<t$, then \begin{equation*} P\left\{N(s)=k|N(t)=n\right\} = \dfrac{n!}{k!(n-k)!}\left(\frac{s}{t}\right)^{k}\left(1-\frac{s}{t}\right)^{(n-k)} \end{equation*}