Let $N(t)$ be an inhomogeneous Poisson process with a intensity function: $$\lambda(t)=3t+1,\ for\ \ 0\leq t\leq 5,$$ $$\lambda(t)=3\ for\ \ t>5$$ I need to calculate:
- $E(N(6)-N(3)|N(2)=3)$
- $P(N(6)-N(4)=k),\ k=0,1,\dots$
Here's my approach: $$E(N(6)-N(3)|N(2)=3)=E(N(6)-N(3)|N(2)-N(0)=3)=E(N(6)-N(3))=\int_3^6\lambda(y)dy=$$ $$=\int_3^5 3y+1\ dy+ \int_5^6 3\ dy= \frac{3y^2}{2}+y|_{y=3}^{y=5}+3y|_{y=5}^{y=6}=26+3=29$$ And for the second one: $$P(N(6)-N(4)=k)=e^{-(m(6)-m(4))}\frac{(m(6)-m(4))^k}{k!}$$ We have $m(t)=\int_0^t\lambda(y)dy$, so: $$m(4)=\int_0^43y+1\ dy=\frac{3y^2}{2}+y|_{y=4}=\frac{3\cdot 16}{2}+4=28$$ $$m(6)=\int_0^53y+1\ dy+\int_5^63\ dy=\frac{3y^2}{2}+y|_{y=5}+3y|_{y=5}^{y=6}=\frac{3\cdot 36}{2}+6+(18-15)=54+6+3=63.$$ Threfore we have $$P(N(6)-N(4)=k)=e^{-(63-28)}\frac{(63-28)^k}{k!}=e^{-35}\frac{35^k}{k!}$$
Is that correct?