Distribution of cumulated intensity for Poisson process

Question

I've come across a statement in a textbook that is not proved, and I have a hard time coming up with the proof myself.

Let $\tau$ be the first jump of a (time inhomogeneous) Poisson process with intensity $\lambda(t)$. Define the cumulated intensity: $\Lambda(t):=\int_0^t\lambda(u) \, du$

The claim the authors make (which I want to see a proof of) is:

One of the important facts about Poisson processes is a property of the jump time $\tau$ according to its own cumulated intensity $\Lambda$. We have

$\Lambda(\tau)=:\xi \sim $ exponential standard random variable.

(If anyone is interested, the book is Interest Rate Models - Theory and Practice 2nd ed. by Brigo and Mercurio, where this statement is found on page 698)

Answer 1

For inhomogeneous Poisson process the probability that there are no jumps on the interval $[0,y]$ is $$ \mathbb P(\tau >y) = e^{-\Lambda(y)} $$ Recall that the distribution of $\tau$ is absolutely continuous and the pdf can be obtained from the above equality.

Since the function $\Lambda(t)$ is continuous and non-decreasing, we can define inverse of it as $$ \Lambda^{-1}(x) = \inf\{t: \Lambda(t)\geq x\}. $$ Note that $\Lambda(\tau)\geq x$ $\iff$ $\tau\geq\Lambda^{-1}(x)$.

Then $$ \mathbb P(\Lambda(\tau)\geq x) = \mathbb P(\tau\geq\Lambda^{-1}(x))=\mathbb P(\tau >\Lambda^{-1}(x)) = e^{-\Lambda(\Lambda^{-1}(x))} = e^{-x}. $$ So the distribution of $\xi=\Lambda(\tau)$ is standard exponential.

Answer 2

\begin{align} & \Pr(\xi >x) \\[10pt] = {}& \Pr(\Lambda(\tau) > \Lambda(t)) \\[10pt] = {} & \Pr\big((\text{number of jumps before time $t$}) = 0\big) \\[10pt] = {} & \frac{\Lambda(t)^0 e^{-\Lambda(t)}}{0!} \\[10pt] = {} & e^{-\Lambda(t)} \\[10pt] = {} & e^{-x}. \end{align}