Consider a Poisson process {X(t), t ≥ 0} with constant rate λ > 0. Let X(a, b) denote the number of events in the time interval a < t ≤ b, and 0 < t1 < t2 < t3 are three given time points.
Let given integers k1 ≥ 0, k2 ≥ 0, k3 ≥ 0 and n ≥ 1 satify the relationship k1 + k2 + k3 = n. Find the the conditional probability of {X(0, t1) = k1, X(t1, t2) = k2, X(t2, t3) = k3} given {X(0, t3) = n}, that is,
P{X(0, t1) = k1, X(t1, t2) = k2, X(t2, t3) = k3 | X(0, t3) = n} .
i think im making it more difficult than it needs to be but if i could get some help on how to go about it i would be grateful
We know $k_1, k_2, k_3 \ge 0$ i.e $k_1,k_2,k_3 \in N$ $\cup$ {0}.
By using the definition of conditional probability $$P(A | B) = \frac {P(A, B)}{P(B)}$$
on the you're given we get $$ \frac{P(X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3, X(0, t_3) = n)}{P(X(0, t_3) = n)} $$
If we use Lack of Memory property we can simplify the top probability since we know $n = k_1 + k_2 + k_3$. Hence, $$ \{ X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3, X(0, t_3) = n \} = \{ X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3\} $$
Now, once we have the Lack of Memory applied on the set, we can enforce the property on the probability as well,
$$ P\{ X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3, X(0, t_3) = n \} = P\{ X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3\} $$
At this point, you have a solid ground from which you can continue, i.e
$$ \frac{P(X(0, t_1) = k_1, X(t_1, t_2) = k_2, X(t_2, t_3) = k_3)}{P(X(0, t_3) = n)} $$
Next hint would be to show independence of the random variables using the fact that the intervals (0, $t_1$) (0, $t_2$) and (0, $t_3$) do not overlap. Applying independece on the aforementioned probability you get something similar to this,
$$ \frac{P(X(0, t_1) = k_1) \cdot P(X(t_1, t_2) = k_2) \cdot P(X(t_2, t_3) = k_3)}{P(X(0, t_3) = n)} $$
Last step that should be applied you know from the following fact,
$$ X(a,b) \text{~} Poisson(\lambda \cdot(b-a)) $$
Using the PDF of Poisson Distribution you have your answer.