I am trying to understand the inverse transform method for simulating random processes and have managed to completely confuse myself. Consider a Poisson process whose conditional intensity is
$$\lambda(t) = \alpha e^{-t}$$
starting at time $t=0$ for some parameter $\alpha>0$.
To apply the inverse transform method I need to start with the cdf (called $F$ in the notes that I have linked to) and then compute its inverse. What is the relevant cdf $F$ for this process?
My overall goal is to be able to simulate the arrival times from this process.
Looking in the obvious place, the non-homogeneous Poisson process uses the intensity function to construct a parameter for a Poisson process, which is a discrete probability distribution. \begin{align*} \lambda_{a,b} &= \int_a^b \lambda(t) ~\mathrm{d}t \\ &= \int_a^b \alpha \mathrm{e}^{-t} ~\mathrm{d}t \\ &= \alpha ( \cosh(a) - \sinh(a) - \cosh(b) + \sinh(b) ) \end{align*} Inserting this value in the following doesn't result in any opportunities for simplification, so you should just compute it as some numerical value for a given $\alpha$, $a$, and $b$
The probability of $k$ events between times $a$ and $b$ is $$P[N(b)-N(a) = k] = \frac{\mathrm{e}^{-\lambda_{a,b}}(\lambda_{a,b})^k}{k!}$$
The cumulative distribution of up to $k$ events between times $a$ and $b$ is $$P[N(b)-N(a) \leq k] = \sum_{j=0}^k\frac{\mathrm{e}^{-\lambda_{a,b}}(\lambda_{a,b})^j}{j!}$$
For your purpose you want to start with $j=0$ and then extend this sum by one term at a time until it exceeds your $U[0,1]$ variate and then return that value of $k$ as your Poisson variate. (Note that exceeding the uniform variate can occur on the first, $k=0$, term and this is a valid value of a Poisson variate.)