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When we have a random variable that follows an exponential distribution, the mode is at 0, the probability $P(\frac{1}{\lambda}<X<\frac{1}{\lambda}+\epsilon)$ is lower than $P(\frac{1}{\lambda}-\epsilon < X < \frac{1}{\lambda})$, so in some sense if $X$ is interpreted as the waiting time between two events, there is more chance that the waiting time is slightly lower than $\frac{1}{\lambda}$, than slightly bigger than $\frac{1}{\lambda}$. I was wondering if this comes from (linked) to the fact that the exponential distribution is memoryless?

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Yes, it follows as a consequence of that property: Interpreting the exponential random variable $T$ as describing the time until an event, the "memoryless property" means that the distribution has constant hazard rate:

$$\lambda (t) = \lim_{\Delta \downarrow 0} \frac{\mathbb{P}(t \leqslant T \leqslant t + \Delta| T \geqslant t)}{\Delta} = \lambda.$$

This property means that the instantaneous probability of the event at any time $t$ is constant, conditional on the fact that the event has not occurred prior to that time. In order for the event to occur at time $t$, it must not occur prior to this, and then it must occur at that time. Hence, using the rules for joint probability we have:

$$f(t) = \lim_{\Delta \downarrow 0} \frac{\mathbb{P}(t \leqslant T \leqslant t + \Delta)}{\Delta} =\mathbb{P}(T \geqslant t) \times \lambda(t) = \lambda \exp (-\lambda t).$$

This gives the form of the density function which is a strictly decreasing function. But why is it decreasing if the hazard rate is constant? Well, to see this, consider two times $t_0 < t_1$. In order to get $T = t_0$ the event must "survive" up to time $t_0$ and then occur. In order to get $T = t_1$ the event must "survive" up to time $t_1$ and then occur. The instantaneous probability of occurrence is the same in both cases (constant hazard), but with constant hazard, the probability of "surviving" to time $t_0$ is larger than the probability of "surviving" up to the later time $t_1$.

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