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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.

Xi'an
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Mr Alpha
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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
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    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
  • Suppose the average rate was $0.05$ events per second, which is an average of $3$ events per minute. The probability of a failure in the first minute is $1-e^{-3} \approx 0.95$ i.e. $95%$ and likely though not certain. This seems intuitively plausible. You would not accept a probability of $3$ or $300%$, since probabilities cannot exceed $1$, but a rate of $3$ events per minute is totally acceptable. – Henry Sep 28 '23 at 23:05

2 Answers2

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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 $1/20$.

Douglas Zare
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There is a well known approximation for $e^x$ when $x$ is very close to zero:

$$ e^x \approx 1 + x $$

If we take $ x = -0.05 $, we get $ e^{-0.05} \approx 1 - 0.05 = 0.95 $.

So we can say:

$$ 1 - e^{-0.05} \approx 1 - 0.95 = 0.05 $$

You make a mistake when you think that a good approximation is the same as the exact value: the approximation is close to the exact value but rarely is the same number. The exact value is:

$$ 1 - e^{-0.05} = 1 - 0.951229 = 0.0487705 $$

The exact value of a calculus is difficult to reach. For an engineer or a scientific is better to use a good approximation.

Rafael Marazuela
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