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I'm trying to wrap my head around the Exponential distribution and the meaning of its parameter. The parameter is the rate, right? So take, e.g., $$X\sim \exp(0.05)\,.$$ Now the probability of failure during the first time period is: $$P(X\le1)=1-P(X>1)=1-e^{-0.05}=0.04877\,.$$ Now I can do the math and get the correct result and so on, but cannot wrap my head around why the result should be slightly less than 5%, rather than exactly 5%. I don't get the intuition behind it.

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The rate $\lambda$ has nothing to do with the pdf value in $1$, the correspondence is simply a first order approximation $1-exp(-x) \approx x$... –  Xi'an Nov 14 '12 at 11:06
    
A better "intuition" about the Exponential distribution is the link between this distribution and the Poisson process, as explained in this Stack Exchange question. –  Xi'an Nov 14 '12 at 11:11

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up vote 3 down vote accepted

Suppose you have some mildly radioactive substance so that you expect to wait $20$ seconds between decays. That is a rate of $1/20$ per second. The average number of decays in one second is $1/20$.

One way to get an average count of $1/20$ would be if the count were $1$ with probability $1/20$ and $0$ with probability $19/20$. However, sometimes there are $2$ or more decays in that second. For the average count to be $1/20$, when it is sometimes $2$ or more, the probability that you get a count of $0$ must be greater than $19/20$. Therefore, the probability that you wait more than one second before the first decay is greater than $19/20$, and the probability that the first decay occurs within the first second is less than $1/20$.

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