How is the mode of an exponential distribution zero?

Question

The exponential distribution has a mode of $0$, which, according to Wikipedia, means that $0$ "is the value that is most likely to be sampled". This is not what I would expect, given that the exponential distribution "describes the time between events in a Poisson process".

So, say me getting phone calls is a Poisson process and I get a call every hour on average. How is it that $0$ is the most likely sample? My intuition tells me that $0$ is one of the most unlikely samples, or at least much less likely than some number around $1$ hour.

Answer 1

But the chance of any continuous distribution taking any particular value is 0? The chance it is 3.5 or $\pi$ or $e$ is also 0.

All this means is that: let $X$ have exponential distribution. take any $x>0$, I can find an $\epsilon>0$ such that for all $y<\epsilon$

$$P(X\in (x,x+y)) < P(X\in(0,y))$$

What it says is that the chance the distribution takes a value close to 0, is bigger than the chance it take a value close to some other $x$, if we define close to be small enough. I believe that.

Why do you not try to show this and convince yourself?

Answer 2

An exponential distribution is that of a continuous random variable.   All particular values it can take have probability mass of zero.

The mode of a continuous random variable is not the point where its probability is most massive.   It is where it is most dense.

The probability density function of an exponential random variable, $X$ with rate $\lambda$, is:

$$f_X(x)=\lambda \mathsf e^{-\lambda x} \mathbf 1_{x\in[0;\infty)}$$

This is at its maximum when $x=0$.   Hence the mode of the distribution is: $0$.