About non homogeneous Poisson Process Prove that $\left \{ N(t+s)-N(t),\geq t \right \} $ is a nono-homogeneous Poisson process with respective intensity function $\lambda(t)$ and pdf $$P(N(t+s)-N(t)=n)=\frac{e^{m(t+s)-m(t)}[m(t+s)-m(t)]^n}{n!}$$ where $m(t)=\int_{0}^t\lambda(v)dv$, if $N(t)$ verifies
i)$N(0)=0$
ii)$\left \{ N(t+s),\geq t \right \}$ have independent increments
iii)$P(N(t+h)-N(t)\geq2)=o(h)$ 
iv)$P(N(t+h)-N(t)=1)=\lambda(t)h+o(h)$
I tried to prove in analogy with a proof in the case of homogeneous Poisson Process that i found in "Introduction to probability models" (Ross).
Ill share a dropbox link
 A: Variables $s$ and $t$ are the other way around in Ross's book so I'll go with that. So we're trying to prove:
$$P(N(s+t)-N(s)=n)=e^{-(m(s+t)-m(s))}\dfrac{[m(s+t)-m(s)]^n}{n!},\quad n\geq 0\qquad\qquad(*)$$
For easier notation, let $P_n(u,v) := P(N(v)-N(u)=n),\;$ for $n\geq 0\;$ and $u\leq v$.
Firstly, we take the case $n=0$.
\begin{eqnarray*}
P_0(s,s+t+h) &=& P(N(s+t+h)-N(s)=0) \\
&=& P(N(s+t)-N(s)=0)\; P(N(s+t+h)-N(s+t)=0) \\
&=& P_0(s,s+t)\; P_0(s+t,s+t+h) \\
&=& P_0(s,s+t)\; (1-\lambda(s+t)h+o(h)) \\
P_0(s,s+t+h) - P_0(s,s+t) &=& -\lambda(s+t)hP_0(s,s+t)+o(h) \\
P_0^{'}(s,s+t) &=& -\lambda(s+t)P_0(s,s+t) \qquad\quad\text{(divide by $h$ and let $h\to 0$)} \\
\therefore\quad P_0(s,s+t) &=& Ce^{-m(s+t)} \qquad\text{since $\frac{d}{dt}e^{-m(s+t)} = -\lambda (s+t)e^{-m(s+t)}$}. \\
&& \\
P_0(s,s)=1\quad   \text{ so }\quad  1 &=& Ce^{-m(s)} \\
C &=& e^{m(s)} \\
&& \\
\therefore\quad P_0(s,s+t) &=& e^{-(m(s+t)-m(s))}.
\end{eqnarray*}
Now we take the general case, $n\gt 0$.
\begin{eqnarray*}
P_n(s,s+t+h) &=& P(N(s+t+h)-N(s)=n) \\
&=& \sum_{i=0}^{n} P(N(s+t)-N(s)=i)\; P(N(s+t+h)-N(s+t)=n-i) \\
&=& \sum_{i=0}^{n} P_i(s,s+t)\; P_{n-i}(s+t,s+t+h) \\
&=& P_n(s,s+t)\; (1-\lambda(s+t)h+o(h)) + P_{n-1}(s,s+t)\; (\lambda(s+t)h+o(h)) + o(h) \\
\end{eqnarray*}
\begin{eqnarray*}
P_n(s,s+t+h) - P_n(s,s+t) &=& -\lambda(s+t)hP_n(s,s+t) + \lambda(s+t)hP_{n-1}(s,s+t) + o(h) \\
P_n^{'}(s,s+t) &=& \lambda(s+t)(P_{n-1}(s,s+t) - P_n(s,s+t)) \\
\end{eqnarray*}
\begin{eqnarray*}
\therefore\quad e^{m(s+t)}\left[ P_n^{'}(s,s+t) + \lambda (s+t)P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} P_{n-1}(s,s+t) \\
\frac{d}{dt}\left[ e^{m(s+t)} P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} P_{n-1}(s,s+t).
\end{eqnarray*}
From here we can prove the result by induction on $n$. The initial case for $P_0(s,s+t)$ has already been proved. So we assume that $(*)$ holds for $n-1$.
\begin{eqnarray*}
\frac{d}{dt}\left[ e^{m(s+t)} P_n(s,s+t) \right] &=& \lambda (s+t) e^{m(s+t)} e^{-(m(s+t)-m(s))}\dfrac{[m(s+t)-m(s)]^{n-1}}{(n-1)!} \\
&=& \lambda (s+t) e^{m(s)}\dfrac{[m(s+t)-m(s)]^{n-1}}{(n-1)!} \\
e^{m(s+t)} P_n(s,s+t) &=& C + e^{m(s)} \dfrac{[m(s+t)-m(s)]^{n}}{n!}.
\end{eqnarray*}
Setting $t=0$ to evaluate the constant $C$ we get $C=0$ and the proof is complete as we arrive at
$$P_n(s,s+t) = e^{-(m(s+t)-m(s))} \dfrac{[m(s+t)-m(s)]^{n}}{n!}.$$
