Can someone tell me if I'm correct ?
Let $T > 0$ be a constant. Let $N(t)$ be a non-homogeneous Poisson process with rate $λ > 0$ if $t ∈ [0, T[$ and rate $2λ$ if $t ∈ [T, 2T]$.
Compute $P(N (2T) = 1)$ and $P(N (T ) = 1|N (2T ) = 1)$.
To tackle this, I used the fact that on separate interval $[0, T[$ and $[T, 2T[$ we have in fact two homogeneous Poisson process. So I have :
$ \begin{align} P(N(2T) = 1) &= P(\{N(T) = 1\}) + P(\{N(2T) - N(T) = 1\}) \\ &= (\lambda T)e^{-\lambda T} + (2\lambda T)e^{-2\lambda T} \\ &= \lambda T e^{-\lambda T}(1 + 2 e^{-\lambda T}). \end{align} $
and
$ \begin{align*} P(N(T)=1|N(2T)=1) &= \dfrac{P(N(T) = 1, N(2T) = 1)}{P(N(2T) = 1)} \\ &= \dfrac{P(N(T) = 1)P(N(2T) - N(T) = 0)}{P(N(2T) = 1)} \\ &= \dfrac{\lambda T e^{-\lambda T} e^{-2\lambda T}}{\lambda T e^{-\lambda T}(1 + 2 e^{-\lambda T})} \\ &= \dfrac{e^{-2\lambda T}}{1 + 2 e^{-\lambda T}}. \end{align*} $
Can someone knows if this is correct and if not where is the error ? Am I understanding the problem well ?
Any help is welcome, thanks !
For the first: $$\begin{aligned}P(N_{2T}=1)&=P(\{N_{2T}-N_T=1,N_{T}=0\}\cup\{N_{2T}-N_T=0,N_T=1\})=\\ &=P(N_{2T}-N_T=1)P(N_T=0)+P(N_{2T}-N_T=0)P(N_T=1)=\\ &=2\lambda(2T-T)e^{-2\lambda(2T-T)}e^{-\lambda T}+e^{-2\lambda(2T-T)}\lambda T e^{-\lambda T}=\\ &=2\lambda T e^{-3\lambda T}+e^{-3\lambda T}\lambda T=3\lambda T e^{-3\lambda T} \end{aligned}$$ For the second: $$\begin{aligned} P(N_T=1|N_{2T}=1)&=\frac{P(N_{2T}=1|N_T=1)P(N_T=1)}{P(N_{2T}=1)}=\\ &=\frac{P(N_{2T}-N_T=0)(\lambda T)e^{-\lambda T}}{3\lambda T e^{-3\lambda T}}=\\ &=\frac{e^{-2\lambda T}(\lambda T)e^{-\lambda T}}{3\lambda T e^{-3\lambda T}}=\frac{1}{3} \end{aligned}$$