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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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    $\begingroup$ This about it in these terms. Poisson count happens in the interval $dt$ w/ probability $dt$. So for the first count to happen around $x$, you need no counts to happen previously - so probability is $e^{-x}dx$. Clearly it's maximized at $0$. $\endgroup$ – A.S. Mar 23 '16 at 1:58
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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: $$f_X(x)=\lambda \mathsf e^{-\lambda x} \mathsf 1_{x\in[0;\infty)}$$

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

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  • $\begingroup$ Thanks for the answer. I understand that the exponential distribution is continuous and that every particular value has zero probability. But that is beside the point. To put it with your terms, the point is that I find it unintuitive that the probability is most dense at $0$. To go back to the phone call example, I would not expect a phone call just after another one, but mode $= 0$ seems to say that this is the most likely thing to happen. $\endgroup$ – rolve Mar 22 '16 at 13:53
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    $\begingroup$ @rolve obviously nothing is ever really exponentially distributed... $\endgroup$ – Lost1 Mar 23 '16 at 1:47
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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?

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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.

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