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.
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$\begingroup$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: $0$.
$\endgroup$ 2 $\begingroup$If it's counter intuitive for you, then exponential/Poisson is likely inadequate for the applications that you have in mind. As explained elsewhere:
Unfortunately, assuming that state sojourn times are exponentially distribute in things like a disease model really isn’t very realistic. (...) A much better probability distribution for sojourn times in compartmental biological and epidemiological models is the Gamma distribution.
So for generalizations, check for example the Gamma distribution and Gamma processes.
$\endgroup$ $\begingroup$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?
$\endgroup$ 0 $\begingroup$In a histogram (as opposed to a continuous distribution) the mode is clearly location of the highest bar.
If you construct a histogram (i.e. a step function) that approximates a decaying exponential the highest bar will be the leftmost - the one that starts at $0$.
For skewed distributions like this you will always have mode < median < mean (think family income distribution in the U.S.). The mode may be counterintuitive (to you) because you would rather think about the other two "averages".
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