Integrating a poisson rate

Question

I am completing some practice questions, in which I have been asked the following.

My query is with the solution, in which we integrate our poisson process rate. There is no explanation in my notes provided as to why we perform this. The answer begins as follows:

I understand how to derive my solution, but I am a bit unsure as to why we integrate the poisson rate we have? I can't find anything explaining the integral of a poisson rate.

TIA.

Answer 1

In a regular Poisson process, the rate is a fixed real number $\lambda$, such that $\mathbb P(N(t) = n) = e^{-\lambda t}\frac{(\lambda t)^n}{n!}$ for $n=0,1,2,\ldots$. In a non-homogeneous Poisson process, the rate is a function of time $\lambda(t)$. So to find the distribution of $N(t)$, we integrate the rate function with respect to time to obtain the cumulative rate function. In this case, $$ \Lambda(t) = \int_0^t \lambda(u)\ \mathsf du = \int_0^t k\rho e^{-\rho t}\ \mathsf dt = k(1-e^{-\rho t}). $$ Note that in a regular Poisson process the rate function is $\lambda(t)=\lambda$, so integrating yields $\int_0^t \lambda\ \mathsf du = \lambda t$. Back to the problem at hand, we wish to find $t_0$ such that $\mathbb E[N(t_0)] = \lim_{t\to\infty}\mathbb E[N(t)] - \mathbb E[N(t_0)]$. Since \begin{align} \mathbb E[N(t)] &= \sum_{n=1}^\infty n\cdot \mathbb P(N(t) = n)\\ &= \sum_{n=1}^\infty n\cdot e^{-\Lambda(t)}\frac{\Lambda(t)^n}{n!}\\ &= \sum_{n=1}^\infty n\cdot e^{-k(1-e^{-\rho t})}\frac{(k(1-e^{-\rho t}))^n}{n!}\\ &= e^{-k(1-e^{-\rho t})}\sum_{n=1}^\infty n\cdot \frac{(k(1-e^{-\rho t}))^n}{n!}\\\\ &= e^{-k(1-e^{-\rho t})} k(1-e^{-\rho t})\sum_{n=0}^\infty \frac{(k(1-e^{-\rho t}))^n}{n!}\\ &= e^{-k(1-e^{-\rho t})} k(1-e^{-\rho t})e^{k(1-e^{-\rho t})}\\ &= k(1-e^{-\rho t}), \end{align} we have $$ \lim_{t\to\infty} \mathbb E[N(t)] = \lim_{t\to\infty} k(1-e^{-\rho t}) = k, $$ hence $$ \mathbb E[N(t_0)] = k - \mathbb E[N(t_0)]. $$ It follows that $\mathbb E[N(t_0)] = \frac k2$, from which $$ k(1-e^{-\rho t_0}) = \frac k2\implies t_0 = \frac{\log 2}\rho. $$

Answer 2

I believe that the Poisson rate is expected number of decays per second as a function of time, and you are trying to determine the expected number of decays over some time period in order to calculate the half-life according to the definition in your problem.

Therefore, you need to integrate the $\lambda(t)$ to determine the number of expected decays. From this result, you can determine the half-life, $t_{0}$, at which 1/2 of your expected decays have occurred.

I hope this helps.