How could I calculate this conditional probability
${\large\mathsf P}\Big(N(5)-N(3)=6 \mid N(4) - N(3)=2\Big) $, N is a poisson process. Here is my try.
$P(A|B) = \frac{P(A \cap B)(*)}{P(B)}$.
but in the above case, (*) becomes ${\large\mathsf P}(N(4)-N(3)=2)$, then dividing it by itself, I get $P(1)$, but the answer is $P(N(1)=4)$. How?
Regards,
On
${\large\mathsf P}\Big(N(5)-N(3)=6 \mid N(4) - N(3)=2\Big) = {\large\mathsf P}\Big(N(5)-N(4)=4\Big)$
By reason that since you know that two of the six events occurred in the first half of the period, and Poison Processes are memoriless, so you only need to find the probability that the four remaining events occur in the second half of the period.
Alternatively:
$$\require{cancel}\begin{align} \mathsf P\Big(N(5)-N(3)=6 \mid N(4) - N(3)=2\Big) & = \frac{ \mathsf P\Big(N(5)-N(3)=6\cap N(4)-N(3)=2\Big) }{ \mathsf P\Big(N(4)-N(3)=2\Big) } \\[2ex] & = \frac{ \mathsf P\Big(N(5)-N(4)=4\cap N(4)-N(3)=2\Big) }{ \mathsf P\Big(N(4)-N(3)=2\Big) } \\[2ex] & = \frac{ \mathsf P\Big(N(5)-N(4)=4\Big)\;\cancel{\mathsf P\Big( N(4)-N(3)=2\Big)} }{ \cancel{\mathsf P\Big(N(4)-N(3)=2\Big)} } \\[2ex] \mathsf P\Big(N(5)-N(3)=6 \mid N(4) - N(3)=2\Big) & = \mathsf P\Big(N(5)-N(4)=4\Big) \end{align}$$
After some thought, as in the hint from Andre, the intersection tells you there are 2 events in $(3,4)$ and that there are 4 events in $(4,5).$ But these are non-overlapping intervals and therefore associated events are independent. (You are not thinking clearly about the intersection.)
Is the probability of 4 events in $(4,5)$ the same as the probability of 4 events in $(0,1)?$ If so, what two properties of the Poisson process come into play in establishing that? And what is another way to write that?