Questions regarding this derivation of the Poisson Distribution from exponential densities

Question

On page 217 - 218 of the pdf of this book, the author derives the Poisson Distribution using gamma and exponential densities. The author defines $S_n$ to be the sum of a sequence of independent exponentially distributed random variables ($X_1 + X_2 .... + X_n$) with parameter $\lambda$ and $Y$ to count the number of emissions in a time interval $t$, and then states the following:

$$P(Y = n) = P(S_n < t \space and \space S_{n+1} > t) = P(S_n < t) - P(S_{n+1} <t)$$

The reasoning being that $S_{n+1} < t$ is a subset of the event $S_{n} < t$

1. I don't understand why $P(Y = n)$ is equal to $P(S_{n} < t < S_{n+1})$, why the event $S_{n+1} < t$ is a subset of $S_{n} < t$ (and not the other way around?), and why we take the difference of the two?

2. Is the $\lambda$ defined for the sequence of exponential densities, in the gamma density (used to model $S_n$), and the final Poisson distribution, all the same?

3. To simulate a Poisson random variable ($W$) with parameter $\lambda$ using exponential densities, the author proposes that $-\frac{1}{\lambda}log(rnd)$ (which simulates an exponential density with parameter $\lambda$) be run sequentially until $S_{n} < 1 < S_{n+1}$ and the $n$ returned is the simulated value for $W$. Would I replace the $1$ with $t$ if I want to simulate a Poisson random variable with parameter $\lambda t$?

Answer 1

If $Y(t) = \sum_{n=1}^\infty \mathsf 1_{(0,t]}(S_n)$ is the counting process associated with the renewal times $\{S_n\}$, then by definition $$\{Y(t) = n\} = \{S_n\leqslant Y(t) < S_{n+1}\}. $$ This is simply because $n$ renewals have occured by time $t$, and the $(n+1)^{\mathrm{th}}$ renewal has yet to occur. As for $\{S_{n+1}<t\}\subset\{S_n<t\}$, we have $$ \{S_{n+1}<t\} = \left\{\sum_{i=1}^{n+1}X_i<t \right\} = \left\{X_{n+1}+\sum_{i=1}^n X_i<t\right\}\subset\left\{\sum_{i=1}^n X_i<t\right\} = \{S_n<t\} $$ since $\mathbb P(X_{n+1}\geqslant 0)=1$.

$\lambda$ is a constant, and is the same everywhere.

To simulate a Poisson random variable with rate $\lambda t$, you would use $-\frac1{\lambda t}\log U$ where $U$ is uniformly distributed over $(0,1)$.