Deriving exponential distribution from geometric

Question

Let $\lambda$ be the expected number of events in a unit time interval $[s,s+1]$ (events are independent of each other and of the time interval), and $T$ a continuous random variable that represents the time between two events.
If we divide the time interval $[0,1]$ into $n$ intervals of length $\displaystyle \frac{1}{n}$, and set the probability of an event occurring in any interval to be $\displaystyle \frac{\lambda}{n}$, then $T$ can be approximated by a discrete geometric random variable $X$ which represents the number of time intervals of length $\displaystyle \frac{1}{n}$ between two events.
Let $t\in \mathbb{R}$, then $F_T(t)$ can be approximated by $F_X(nt)$

$$ P(T<t) \approx P(X < tn) $$

Where $P(X<nt)$ is the sum of geometric distributions \begin{align} P(X<nt) &= \sum_{k=1}^{\lfloor{t} n\rfloor} P(X=k) \\ &= \sum_{k=1}^{\lfloor{t} n\rfloor} \left(1 - \frac{\lambda}{n}\right)^{k-1}\left( \frac{\lambda}{n} \right) \\ &= \frac{\lambda}{n} \sum_{i=0}^{\lfloor{t} n\rfloor -1} \left(1 - \frac{\lambda}{n}\right)^i \\ &= \frac{\lambda}{n} \left( \frac{1 - (1 - \frac{\lambda}{n})^{\lfloor{t}n\rfloor }}{\lambda /n} \right) \underset{n \to \infty}{\longrightarrow} 1 - e^{-\lambda t } \end{align}

But how to justify

$$ \lim_{n \to \infty} \left( 1 - \frac{\lambda}{n} \right)^{\lfloor{t}n\rfloor } = e^{-\lambda t} $$

How can I get rid of the floor function, in a formal way?

Answer 1

For any fixed $t$, pick any $K$ such that $t K \geq 1$. Then, we have

$t (n - K) \leq \lfloor tn \rfloor \leq tn$, and so

$\left(1 - \frac{\lambda}{n}\right)^{t(n - K)} \leq \left(1 - \frac{\lambda}{n}\right)^{\lfloor tn \rfloor} \leq \left(1 - \frac{\lambda}{n}\right)^{tn}$

The limit of the first expression is

$\lim_{n \to \infty} \left(1 - \frac{\lambda}{n}\right)^{t(n - K)} = \lim_{n \to \infty} \left(1 - \frac{\lambda}{n}\right)^{tn} \cdot \left(1 - \frac{\lambda}{n}\right)^{-tK} = e^{-\lambda n}$, and so the limit follows from applying the Squeeze Theorem to the above inequalities.

Answer 2

If $0 \le x_n \le 1$ and $\lambda \ge 0$, we have $$ 1 = \left( 1 - \frac{\lambda}n \right)^0 \ge \left( 1 - \frac{\lambda}n \right)^{x_n} \ge \left( 1 - \frac{\lambda}n \right)^1 = 1 - \frac{\lambda}n. $$ for $n$ large enough so that $0 < 1 - \frac{\lambda}n \le 1$ (the function $x \mapsto a^x$ is decreasing on the interval $[0,1]$ when $0 < a < 1$ because its derivative is $a^x \log a < 0$). Letting $n \to \infty$ and using the sandwich theorem, we see that whatever the sequence $x_n$ is, the middle term goes to $1$.

Now put $x_n = tn - \lfloor tn \rfloor$.

Hope that helps,