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Mean of exponential distribution is $$\frac{1}{\lambda}$$

What is mean of all samples greater than some value $S$?

Some context: A DC power supply (as used in a telecommunications installation) has a battery of limited capacity $S$. A failure of AC supply for less than $S$ hours does not drop power to equipment. A failure in excess of $S$ hours will cause whole site to go down.

AC supply Mean Time Between Outage is about 4000 hours. Mean outage duration is 2 hours.

Say a site's battery capacity is 8 hours so $S=8$. AC outages shorter than 8 hours do not impact site function, but AC outages longer than 8 hours cause a DC outage and hence a site outage.

To calculate effective DC unavailability I need to calculate mean outage time for all AC outages longer than 8 hours in this case.

Mean of exponential distribution is found by this integral: $$\int_{x=0}^\infty x\lambda e^{-\lambda x}.dx$$

I think I want this: $$\int_{x=S}^\infty x\lambda e^{-\lambda x}.dx$$

But this finds area under exponential curve from $S$ to $\infty$ - a smaller value than $S$ so I have confirmed that I have no idea what I am doing.

philcolbourn
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  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Matthew Conroy Jan 15 '14 at 03:30

3 Answers3

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Say, for example, you want to determine the centre of mass of a set of masses. There's a 5kg weight at x = 2, 4kg at x = 3 and 1kg at x = 10.

First, you 'weight' each mass by its position (multiply by x) and sum the values, then you divide by the total weight. Here it's $2*5$ kg $+3*4$ kg $+10*1$ kg $= 32$ kg divided by $5$ kg $+4$ kg $+1$ kg $= 10$ kg, giving $3.2$.

Now extend this to a continuous case. You want to find the centre of mass - the mean - of some section of a function.

You weight it with its position by multiplying in x and sum (integrate) over all the values. Then, the next step is to sum the values of the function over that section and divide them out.

Note that this is consistent with your definition of the mean of the exponential distribution - it's just that in that case, the integral you divide out has value $1$.

Edit: Making my analogy between sums/integrals and $C_m$/mean more explicit, the center of mass $C_m$ of some masses $m_i$ at positions $x_i$ is:

$$ C_m = \frac{\sum\limits_{i} x_i m_i}{\sum\limits_{i} m_i} $$

As I detailed in my example. When we extend this to the continuous case with some function $m(x)$ over an interval $[a, b]$ it becomes:

$$ mean = \frac{\int\limits^{b}_{a}x \space m(x) \space dx}{\int\limits^{b}_{a} m(x) \space dx} $$ Compute the integrals for your function on $[S, \infty)$ and this gives $S+{1\over \lambda} \to 8+2 = 10$

G. H. Faust
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  • I think this calculates area under curve from $S$: $$\int_{x=S}^\infty x\lambda e^{-\lambda x}.dx$$ as you suggest. But what is this 'division' you are referring to? – philcolbourn Jan 16 '14 at 02:22
  • @philcolbourn I was trying to lead/hint you towards the solution yourself, but I'll edit my post to be more explicit. – G. H. Faust Jan 16 '14 at 02:52
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    I appreciate your lead and your solution. – philcolbourn Jan 16 '14 at 04:20
  • Wolfram Alpha seems to agree with you "E[X|X>8] of exp dist rate=0.5 -> 10". I understand how this works, but I can not replicate result after doing integrations. I get $$ mean = \frac{e^{-\lambda x}(S+\frac{1}{\lambda})}{e^{-\lambda x}(\frac{S}{\lambda}+\frac{1}{\lambda^2})} $$ My mistake. I get $$ mean = S+\frac{1}{\lambda} $$ – philcolbourn Jan 22 '14 at 23:09
  • It is really hard to edit formula in comments and get it right first time (no preview). Then I get an error message about editing too quickly. Anyway, my intermediate step: $$ mean = \frac{e^{-\lambda x}(S+\frac{1}{\lambda})}{e^{-\lambda x}} $$ – philcolbourn Jan 22 '14 at 23:20
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To answer your question, we need to establish some assumptions about the probability distribution of both the length of an outage, and the time between outages. If both are assumed to be exponential and independent with respective means $\mu_o = 2$ and $\mu_b = 4000$, then we can reasonably assume that the probability that there is more than one outage in an $8$ hour period is so small as to be negligible. (In a real-world situation, however, this is clearly not necessarily a valid model.)

In any case, suppose we are interested in the probability distribution of the amount of downtime of the site: i.e., if $X$ is the random variable indicating the length of outage, then we are interested in $Y = (X - 8)_+ = \max\{X-8, 0\}$. It is quite easy to compute the expected value of $Y$: $${\rm E}[Y] = \int_{x=0}^\infty \max\{X-8,0\} f_X(x) \, dx = \int_{x=8}^\infty (x-8) \frac{1}{2}e^{-x/2} \, dx = 2e^{-4}.$$ The actual probability distribution of $Y$ is more complicated to write: Clearly, $$\Pr[Y = 0] = \Pr[X \le 8] = F_X(8) = 1-e^{-8/2} = 1-e^{-4},$$ which is a discrete probability mass that represents a positive probability of observing no site downtime. For $Y > 0$, we have a continuous density, since $X$ is itself continuous: $$f_Y(y) = \frac{1}{2}e^{-(y+8)/2}, \quad y > 0.$$ But again, these computations have the caveat that because $\mu_0 \ll \mu_b$, the chance that there is more than one outage whose total time might exceed $8$ hours is vanishingly small.

heropup
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  • E[Y]=2e^-4~0.037. To get actual outage time, is 8 added to this? I derived a different E[Y]=(8+2)*e^(-8/2)=10e^(-4). This turns out to be area under curve from 8 to inf. It too is less than 8 so I felt I was in error. – philcolbourn Jan 16 '14 at 00:53
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    No, you cannot add 8 to get the expected value of $X$, the outage time. This is because if we observe $Y = 0$, we have insufficient information to determine $X$: an outage not leading to any downtime could be of any length less than 8 hours. Thus it cannot be the case that the probability distribution of $X$ is simply $Y$ plus 8 hours. – heropup Jan 16 '14 at 01:40
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    The expected downtime ${\rm E}[Y]$ is as I wrote it. It is not $$\int_{x=8}^\infty x \cdot \frac{1}{2} e^{-x/2} , dx$$ because that counts the expected outage time GIVEN that the outage exceeded 8 hours; i.e., ${\rm E}[X \mid X > 8]$, which is not the same as ${\rm E}[Y]$. – heropup Jan 16 '14 at 01:44
  • (I did not know you could use formatting in comments). – philcolbourn Jan 16 '14 at 01:58
  • I don't understand then how I can get E[X|X>8]. Both E[X|X>8] and E[Y] give answers that are less than 8 which makes no sense to me. – philcolbourn Jan 16 '14 at 02:04
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    Sorry, I meant to write $${\rm E}[X \mid X > 8] = \frac{1}{S_X(8)} \int_{x=8}^\infty x \cdot \frac{1}{2} e^{-x/2} , dx.$$ $S_X(8)$ is the probability of surviving to $X > 8$, which is just $e^{-8/2} = e^{-4}$. So ${\rm E}[X \mid X > 8] = 10$, which makes sense because $X$ is exponential and therefore memoryless: given $X > 8$, then $X$ will be expected to survive for another $2$ hours. – heropup Jan 16 '14 at 02:46
  • The expectation of $Y$ is smaller than 8 because very few outages will result in any downtime: we would have to observe $X > 8$ in order to see $Y > 0$, and for an exponential distribution with mean $2$ hours, the probability that this happens is quite small; thus, ${\rm E}[Y] < 8$ should not be surprising. – heropup Jan 16 '14 at 02:47
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The mean of the exponential distribution with parameter $\lambda$ is $1/λ$. The mean of the same exponential distribution conditioned to be at least $s\geqslant0$ is $(1/λ)+s$.

This result is related to the so called lack of memory property of the exponential distributions.

Did
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