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)
Best Answer
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.