Integrating Poisson Process with 'Little-o' notation to derive Poisson Distribution I understand how the Poisson Distribution is derived from the the Poisson Process with 'Little-o' notation. However I am unable to do the final integration that gives the formula for the Poisson distribution, where:


integrates into:




I haven't touched integration in a long time and I would greatly appreciate it if anyone could show me how it is done. Thank you!

 A: We prove by induction the following claim:
$$\text{For } x\geq 0,\qquad p_x(t) = \dfrac{(\lambda t)^x e^{-\lambda t}}{x!} \qquad\qquad\qquad\qquad\text{(1)}$$
For the initial case, $x=0$, we have:
\begin{eqnarray*}
p_0^{'}(t) &=& -\lambda p_0(t) \\
\therefore\quad \dfrac{p_0^{'}(t)}{p_0(t)} &=& -\lambda \\
\text{Integrating,}\qquad \ln{p_0(t)} &=& -\lambda t + c_1 \\
p_0(t) &=& c_2 e^{-\lambda t} \\
p_0(0) = 1 & \implies& c_2 = 1 \\
\therefore\quad p_0(t) &=& e^{-\lambda t}\qquad\qquad\qquad\text{completing the initial case.}
\end{eqnarray*}
Now assume $(1)$ holds for some $x \geq 0$ and consider the case for $x+1$. We have
\begin{eqnarray*}
p_{x+1}^{'}(t) &=& -\lambda p_{x+1}(t) + \lambda p_{x}(t) \\
\lambda e^{\lambda t}p_{x+1}(t) + e^{\lambda t}p_{x+1}^{'}(t) &=& \lambda e^{\lambda t}p_{x}(t) \qquad\qquad\qquad\text{(multiply by $e^{\lambda t}$ and re-arrange)} \\
\dfrac{d}{dt}\bigg[e^{\lambda t}p_{x+1}(t)\bigg] &=& \lambda e^{\lambda t}p_{x}(t) \\
&=& \lambda e^{\lambda t} \dfrac{(\lambda t)^x e^{-\lambda t}}{x!} \qquad\qquad\qquad\text{(by inductive assumption)} \\
&=& \dfrac{\lambda (\lambda t)^x} {x!} \\
\text{Integrating,}\qquad e^{\lambda t}p_{x+1}(t) &=& \dfrac{(\lambda t)^{x+1}} {(x+1)!} + c_1 \\
p_{x+1}(0) = 0 & \implies& c_1 = 0 \\
\therefore\quad p_{x+1}(t) &=& \dfrac{(\lambda t)^{x+1}e^{-\lambda t}} {(x+1)!} \qquad\qquad\qquad\text{completing the proof.}
\\
\end{eqnarray*}
