# About the relationship between the exponential and the geometric

I have in my notes a possible interpretation of the exponential random variable X with parameter $$\lambda$$, as the moment an event of the kind of the Poisson r.v happens for the first time, when the time passes starting at zero. For instance, the moment we receive the first phone call.

We can relate the exponential with the geometric by dividing the interval $$[0,x]$$ in $$n$$ subintervals of length $$\Delta t = \frac{x}{n}$$.

If $$n$$ is very large $$\Delta t$$ is very small, and we can assume that the event can happen once in each subinterval $$n$$ , with probability $$p= \lambda \Delta t$$.

Now we can consider that we have a geometric experiment, and using the geometric cdf, $$F_x(n)= 1-q^n$$:

$$1-F_x(x) = P(X > x) = \lim_{n \to \infty} (1-p)^n = \lim_{n \to \infty} (1-\frac{\lambda x}{n})^n = e^{- \lambda x}$$

Why does $$P(X > x) = \lim_{n \to \infty} (1-p)^n$$ ?

Because $$P(X > x) = 1 - P(X \leq x) = 1 - (1 - q^n) = q^n = (1-p)^n$$
• Yes, but here it takes the limit to infinity: $P(X > x) = \lim_{n \to \infty} (1-p)^n$. – roy212 May 2 at 6:18
• You already stated that you take $n$ to be very large, that's why the limit – ab123 May 2 at 6:20
• I made a mistake, instead of $F_x(x)= 1-q^n$ is $F_x(n)= 1-q^n$. So in the proof there's a mix of "x's" and "n's" that I don't understand. – roy212 May 2 at 7:26
• Does my mistake affect your answer? As in $F_x(n)$ $n$ is a discrete number, while $x$ is continous. – roy212 May 3 at 19:23