pure Poisson birth process ordinary differential equations Consider the pure Poisson process defined by
\begin{align}
P_n'(t) &= -\lambda_n P_n(t) + \lambda_{n-1}P_{n-1}, \quad n \geq 1,\\
P'_0(t) &= -\lambda_0 P_0(t).
\end{align}
with $P_0(0) = 1$. Let $\lambda_n > 0$ for all $n$, prove that for every fixed $n \geq 1$, the function $P_n(t)$ first increases, then decreases to $0$. If $t_n$ is the place of the maximum, then $t_1 < t_2< t_3<\dots$
The hint suggests to use induction and differentiate these set of equations, but I'm not sure how that gives arise to the answer.
 A: It follows from the differential equation and initial condition for $\ P_0\ $ that $\ P_0(t)=e^{-\lambda_0t}\ $. Thus, $\ \lambda_0P_0\ $ is the density of an exponential distribution with parameter $\ \lambda_0\ $.
From the differential equation and initial condition for $\ P_n\ $ for $\ n\ge1\ $ (namely, $\ P_n(0)=0\ $), we get
\begin{align}
P_n(t)&=\lambda_{n-1}e^{-\lambda_nt}\int_0^te^{\lambda_ns}P_{n-1}(s)ds\\
&=\int_0^te^{-\lambda_n(t-s)}\lambda_{n-1}P_{n-1}(s)ds\\
&=\int_0^te^{-\lambda_n\sigma}\lambda_{n-1}P_{n-1}(t-\sigma)d\sigma
\end{align}
Thus, if $\ \lambda_{n-1}P_{n-1}\ $ is a probability density function, then $\ \lambda_nP_n\ $ is the convolution of this density function with that of an exponential distribution with parameter $\ \lambda_n\ $, and is hence also a probability density function—namely that of the sum of two independent random variables $\ T_n\sim\text{Exp}\big(\lambda_n\big)\ $ and $\ X_n\sim\lambda_{n-1}P_{n-1}\ $.  Since $\ \lambda_0P_0 $ is the density function of a random variable $\ T_0\sim\text{Exp}\big(\lambda_0\big)\ $, it follows by induction that $\ \lambda_nP_n\ $ is the probability density function of a sum
$$
\sum_{i=0}^nT_i
$$
of independent random variables $\ T_0,T_1,\dots, T_n\ $ with $\ T_i\sim\text{Exp}\big(\lambda_i\big)\ $, and hence that $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $.
Evaluating the above integral for $\ P_1\ $ gives
$$
P_1(t)=\cases{\frac{\lambda_0\big(e^{-\lambda_1t}-e^{-\lambda_0t}\big)}{\lambda_0-\lambda_1}&if $\ \lambda_1\ne\lambda_0\ $,\\
\lambda_0t e^{-\lambda_0t} &if $\ \lambda_1=\lambda_0\ $.}
$$
In the first case, $\ P_1(t)\ $ is strictly increasing for $\ 0<t<t_1=\frac{\ln\lambda_0-\ln\lambda_1}{\lambda_0-\lambda_1}\ $, reaches a maximum at $\ t=t_1\ $ and is strictly decreasing for $\ t>t_1\ $.  In the second case, $\ P_1(t)\ $ is strictly increasing for $\ 0<t<t_1'=\frac{1}{\lambda_0}\ $, reaches a maximum at $\ t=t_1'\ $ and is strictly decreasing for $\ t>t_1'\ $.
Suppose now that $\ n\ge2\ $ and $\ P_{n-1}(t)\ $ is strictly increasing for $\ 0<t<t_{n-1}\ $, reaches a maximum at $\ t=t_{n-1}\ $, and is strictly decreasing for $\ t>t_{n-1}\ $. Differentiating the last of the integral expressions above for $\ P_n(t)\ $ gives
\begin{align}
P_n'(t)&= \lambda_{n-1}\int_0^t e^{-\lambda_n\sigma}P_{n-1}'(t-\sigma)d\sigma+\lambda_{n-1} e^{-\lambda_nt}P_{n-1}(0)\\
&=\lambda_{n-1}\int_0^t e^{-\lambda_n\sigma}P_{n-1}'(t-\sigma)d\sigma\\
&=\lambda_{n-1}\int_0^t e^{-\lambda_n(t-s)}P_{n-1}'(s)ds\ .
\end{align}
For $\ t<t_{n-1}\ $, the integrand of this last integral is strictly positive over the interval $\ [0,t]\ $, and hence so is the integral itself.  That is, $\ P_n'(t)>0\ $ for $\ t<t_{n-1}\ $, and so $\ P_n(t)\ $ is strictly increasing over the interval $\ \big[0,t_{n-1}\big]\ $.
Since $\ P_n\ $ is continuously differentiable for $\ t>0\ $, and $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $, there must exist a point at which $\ P_n(t)\ $ reaches a maximum and $\ P_n'(t)\ $ vanishes. Let
$$
t_n=\inf\big\{\,t\,\big|\,t>0\ \&\ P_n'(t)=0\,\big\}\ .
$$
Then $\ t_n>t_{n-1}\ $, and at that point, we have
$$
P_n'(t_n)=0= \lambda_{n-1}\int_0^{t_n} e^{-\lambda_n(t_n-s)}P_{n-1}'(s)ds\ ,
$$
and $\ P_n\ $ must be strictly increasing over the interval $\ \big[0,t_n\big]\ $.
For $\ t>t_n\ $, we have
\begin{align}
P_n'(t)&=\lambda_{n-1}\int_0^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\\
&=\lambda_{n-1}\left(\int_0^{t_n}e^{-\lambda_n(t-s)}P_{n-1}'(s)ds+\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\right)\\
&=\lambda_{n-1}\left(e^{\lambda(t_n-t)}\int_0^{t_n}e^{-\lambda_n(t_n-s)}P_{n-1}'(s)ds+\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\right)\\
&=\lambda_{n-1}\int_{t_n}^te^{-\lambda_n(t-s)}P_{n-1}'(s)ds\\
&<0\ ,
\end{align}
because the integrand of the final integral is strictly negative over the interval $\ \big[t_n,t\big]\ $. It follows that $\ P_n(t)\ $ is strictly decreasing for $\ t>t_n\ $.  This completes the proof that $\ P_n(t)\ $ is strictly increasing over an interval $\ \big[0,t_n\big]\ $, with  $\ t_n>t_{n-1}\ $, strictly decreasing for $\ t>t_n\ $, and $\ \lim_\limits{t\rightarrow\infty}P_n(t)=0\ $. It follows by induction that these properties hold for all $\ n\ge1\ $.
