Poisson process, distribution of life time

Let $\tau_d$ be the death time of a particle, and suppose the rate of death at time $t$ is given as $\lambda(t)$, that is, $$P\{\tau_d\in [t,t+\delta t]\mid F_t\}=\lambda(t)\delta t+o(\delta t), \text{ for } t <\tau_d,$$ where $F_t$ is the sigma algebra of the history up to time $t$.

Show that $\int_0^{\tau_d}\lambda(t) \, dt$ is exponentially distributed.

This is similar to the counting process, but I can not get the answer by mimicking the method do for counting process. I am curious about if we can actually derive the distribution of $\tau_d$ or we can only have that of the integral.

On the other hand, I have no intuition for it to be exponential too. Is there a way to understand formally?

• Do you perhaps mean $o(\delta t)$ rather than $o(t)$? – Marcus M Sep 29 '17 at 22:52
• Yes, thank you! – Simo Sep 29 '17 at 22:56

I think you need the assumption that $\int_0^\infty \lambda(t) \,dt = \infty$ and I think also $\lambda(t) > 0$, otherwise this doesn't happen.

To get an intuition for why this is true (and to prove it), try the following steps:

• Convince yourself that one way to define an exponential random variable with mean $\mu$ is a random variable satisfying $$P[X \in [x,x+dx] \mid X \geq x] = \mu \, dx + o(dx).$$
• Try to calculate $$P\left[\int_0^{\tau_d} \lambda(t)\,dt \in [x,x+dx] \,\middle|\, \int_0^{\tau_d} \lambda(t)\,dt \geq x\right]$$
• To deal with the above, define $y$ so that $x = \int_0^y \lambda(t) \,dt$, and then use the fundamental theorem of calculus to find $dy$ so that $x + dx = \int_0^{y + dy} \lambda(t) \,dt$.

Edit: I made a comment originally about changing the assumption on the $\sigma$-algebra, but I've gotten rid of it now.

Edit 2: Some details on the second step. Note that the fundamental theorem of calculus implies $$\int_0^{y + \,dy} \lambda(t)\,dt = \int_0^y \lambda(t)\,dt + \lambda(y)\,dy + o(dy).$$

Choosing $dy = \frac{dx}{\lambda(y)}$ shows $$x + \,dx = \int_0^{y + dx/\lambda(y)} \lambda(t)\,dt + o(dx).$$ Thus, we have \begin{align}P\left[\int_0^{\tau_d} \lambda(t)\,dt \in [x,x+dx] \,\middle|\, \int_0^{\tau_d} \lambda(t)\,dt \geq x\right] &= P\left[ \tau_d \in \left[y, y + \frac{dx}{\lambda(y)}\right] \,\middle|\, \tau_d \geq y \right] \\ &= dx + o(dx).\end{align}

• I can show that for the first step, $X$ is exponential, but for the second step, I don't know how to use the third hint. What I am thinking is that for the integral to be in $[x,x+dx]$, $\tau_d$ should not grow more than "$dy$". But I have no way of handling "$dy$"? – Simo Sep 29 '17 at 23:51
• @Simo, Okay, I added the rest of the argument. Try to use the fundamental theorem of calculus to find what $dy$ is. – Marcus M Sep 30 '17 at 0:06
• The assumption you made is needed. Thank you. Is this related to the change of time and where can I find any reference of this? – Simo Sep 30 '17 at 0:24
• @Simo, I'm not really sure, to be totally honest. This is some sort of time change, but I'm not as well-versed in the literature on this as I'd like to be. – Marcus M Sep 30 '17 at 15:18