What rule is used in this derivation of the interarrival time for the Poisson process?

Question

I'm working on calculating the probability distribution of the interarrival time of the Poisson process. The method used in my textbook is very strange I don't understand how the probabilities are calculated. I am familiar with the method used eg. here but not the working below. $ N(t) $ is the Poisson process: $$ P\{N(s) = 1, N(t) = 2\} = P\{\xi_1 \leq s < \xi_2 \leq t < \xi_3\} $$

$$ = P\{\eta_1 \leq s < \eta_1 + \eta_2 \leq t < \eta_1 + \eta_2 + \eta_3\}$$

At this point we have converted the Poisson process into 3 independent, exponentially distributed random variables. But now we have the first conversion of the probability to integrals and I don't understand how it was done:

$$ = \int_{0}^{s} P\{s < u + \eta_2 \leq t < u + \eta_2 + \eta_3\} \lambda e^{-\lambda u} \, du$$

What rule of probability was used? The first inequality has now become the integral and the random variable $ \eta_1 $ has become u in the probability. Is this a well known rule of probablity (and what is it called if so?) or is this somehow a specific property of this problem? And likewise for the next two steps: $$ = \int_{0}^{s} \left( \int_{s-u}^{t-u} P\{t < u + v + \eta_3\} \lambda e^{-\lambda v} \, dv \right) \lambda e^{-\lambda u} \, du$$

$$ = \int_{0}^{s} \left( \int_{s-u}^{t-u} e^{-\lambda(t-u-v)} \lambda e^{-\lambda v} \, dv \right) \lambda e^{-\lambda u} \, du$$

$$ = \lambda^2 e^{-\lambda s} (t - s). $$

I've been trying to understand this for a few days now and haven't managed to make any progress.

Answer 1

After reading more I've found that the rule used in this example is the law of total probability: $$ P\{A\}=\int_{-\infty}^{\infty} P\{A|X=x\}f_X(x) dx $$ First, we notice that the event can be written as: $$ \{\eta_1 \leq s < \eta_1 + \eta_2 \leq t < \eta_1 + \eta_2 + \eta_3\} = \{\eta_1 \leq s\} \cap \{ s < \eta_1 + \eta_2 \leq t < \eta_1 + \eta_2 + \eta_3\} $$ Since $\eta_1$ is exponentially distribution it is non-negative, and the event $\{\eta_1 \leq s\}$ has 0 probability for $\eta_1 > s$ which we can use to adjust the bounds of the integral, therefore we find that: $$= P\{\eta_1 \leq s < \eta_1 + \eta_2 \leq t < \eta_1 + \eta_2 + \eta_3\}$$ $$=\int_{-\infty}^{\infty} P(\{\eta_1 \leq s < \eta_1 + \eta_2 \leq t < \eta_1 + \eta_2 + \eta_3\}|\eta_1=u)f_{\eta_1}(u) du$$ $$=\int_{-\infty}^{\infty} P(\{u \leq s < u + \eta_2 \leq t < u + \eta_2 + \eta_3\})f_{\eta_1}(u) du$$ $$=\int_{0}^{s} P(\{s < u + \eta_2 \leq t < u + \eta_2 + \eta_3\})\lambda e^{-\lambda u} \, du$$

The rule is applied again to the next two equalities in this way as well.

Answer 2

Remember that for a continuous random variable $X$ with probability distribution function $f_X(x)$, the cumulative distribution function $F_X(x)$ is given by $F_X(x) = P(X \leq x) = \int_{-\infty}^x f_X(\chi) d\chi$.

In the given expression, we have applied that definition to the random variable $\eta_1$ along with the usual decomposition for independent events $P(A \cap B) = P(A) P(B)$.