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Exponential distribution
By definition, the exponential distribution f(x) is:
* f(x) = 0 for x < 0
* and for x
0
:
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f(x) = λe-λx |
where
is a positive
parameter (verify that this is indeed a probability density function).
We'll show that the mean µ of the distribution
is equal to 1/
, and the distribution is often written
as :
f(x) = (1/µ).e- x/µ
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The exponential distribution plays a central role in a large class of problems related to the concept of "lifetime". For example, an electronic component might be known to have a lifetime of, say, 10.000h. This means that the component is expected to fail after about 10.000h of use. But of course, this is an average value, and some components from the same batch will last less than 10.000 hours, while others will last longer. So the lifetime of a component is a random variable.
We'll establish the following basic properties of the exponential distribution :
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µ = 1/λ |
σ² = 1/λ² |
You'll find here an interactive animation that illustrates these basic properties of the exponential distribution.
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We'll show that the order n moment of the exponential distribution is :
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We establish here that the sum of n iid Exp(λ) distributed random variables is distributed as Gamma(n, 1/λ).
We address here the important question of the distribution of the sum of a random number of iid exponential variables when the number of variables in the sum is geometrically distributed.
Some lifetime problems exhibit a very important property called the "memoryless property". For example, consider a very large number of identical radioactive atoms, and observe their decay.
* During the first t seconds, a proportion p of these atoms disintegrate.
* Now leave the lab, and come back any time later. Consider the atoms that have not disintegrated yet, and observe the radioactive process for another t seconds. During this time, the proportion of these atoms that will disintegrate is also p.
Now translate these observations into terms of probabilities for individual atoms.
* We first state that the probability for an atom to disintegrate in the first t seconds is p.
* We return to the lab after s seconds. Then the probability for a (still not disintegrated) atom to disintegrate within the next t seconds is also p.
So the probability that an s second old atom will last another t seconds is the same as the probability for a "new" atom to last t seconds. It is as if the atom had absolutely no memory of how long it has been around. For radioactive atoms, there is no aging. As long as such an atom is alive, it remains absolutely identical to itself. And then, without any warning, it decays.
Physicists have long given up the hope of finding a more
deterministic model of radioactivity.
The memoryless property is expressed mathematically by writing that the two following probabilities are equal :
* The probability that a new atom will live for at least t seconds.
* The probability for an
atom that is still alive at time s to live at least an extra t
seconds, that is to be still alive at time t + s.
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P{X > t}= P{X > s + t | X > s} |
We show here that the exponential distribution has the memoryless property. This means that if a device's lifetime is exponentially distributed, then it will fail exactly the same way as a radioactive atome does.
We also show that the memoryless property is unique to the exponential distribution. More specifically, we show that the exponential distribution is the only continuous distribution with the memoryless property, which explains why it is so important in practice.
In the discrete domain, the geometric distribution also has the memoryless property, and is the only one to have this property.
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We'll also show that, as a consequence of the memoryless property of the exponential distribution, the lifetime distribution of those atoms that survived up to time s is, from this point on, identical to their original lifetime distribution. In other words, the new lifetime distribution of these "survivors" is just their original lifetime distribution right-shifted by a quantity s.
The foregoing property can be generalized as follows. The "weak" memoryless property refers to survival after a reference date s that is arbitrary but fixed. We show here that the property is still true if the reference date s is not fixed, but instead is itself an exponentially distributed random variable. More precisely, we'll show that if :
then whenever X2 > X1, the distribution of the excess lifetime of X2 over that of X1 does not depend on the value of X1.
This translates into the expression :
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P{X2 > t } = P{X2 > X1 + t | X2 > X1} |
which is identical to the expression describing the weak memoryless property, except that the fixed date s is now replaced by the random variable X1.
This property is called the strong memoryless property of the exponential distribution.
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We'll also show that the distribution of the difference X2 - X1 conditionally to X2 > X1 is identical to that of X2, and we'll illustrate this result by an interactive animation.
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In practical terms, the strong memoryless property describes the following situation. Let M1 and M2 be two machines whose lifetimes are both exponentially distributed and independent. Then, after one machine fails, the survival time of the other machine beyond this failure has the same distribution as the original lifetime of the machine
The memoryless property is not universal
Of course, not all lifetimes problems can be expressed in terms of the memoryless property. For example, all living organisms go through the aging process, and although their lifetimes are more or less random variables, we should not expect their probability density functions to be the same as that of radioactive atoms. Still, many complex devices have lifetimes that behave in a nearly memoryless fashion, which explains why the exponential distribution is so important in many engineering applications..
If a machine is still operational at time t, what is the (infinitely small) probability dP that it will fail during the next dt time interval ? This probability is :
So, whatever the nature of the distribution of the lifetime of the machine (exponential or not), we have :
dP = h(t).dt
The function h(t) is called the hazard rate function, or "failure rate function" of the distribution.
We describe below the properties of the hazard rate function, and show that the hazard rate function of the exponential distribution is constant (does not depend on t). In fact, this is a characteristic property of the exponential distribution, and is therefore equivalent to the (weak) memoryless property.
We identify here a sufficient statistic for the parameter λ of the exponential distribution, then show here that this statistic is minimal sufficient, and finally that it is complete.
From this result, we'll deduce a Minimum Variance Unbiased Estimator of λ, which will be shown not to be efficient (its variance is larger than the Cramér-Rao lower bound).
We show here
that the exponential distribution belongs to the exponential family. From this result,
we'll deduce that the sample mean
is an efficient estimator
of the mean µ.
The exponential distribution plays a central role in studying record values, because it may well be the only distribution for which the distributions of the record values can be calculated directly.
The Probability Integral Transformation then allows linking the distributions of the record values of any other absolutely continuous distribution (i.e. a distribution with a probability density) to those of the exponential distribution.
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Tutorial 1 |
In this first Tutorial, we describe the elementary
properties of the exponential distribution.
We also clarify the relationship between the exponential and the geometric distributions by showing that an exponential r.v. may be considered as the limit of a series of geometric r.v. (the reverse relation is addressed here).
BASIC PROPERTIES OF THE EXPONENTIAL DISTRIBUTION
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Probability density function Cumulative distribution function Quantile function Q(p) General Median Exponential r.v. as the limit |
Moment generating function Moments Mean Direct calculation Moment generating function Variance Direct calculation Moment generating function All moments |
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TUTORIAL |
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Tutorial 2 |
We then calculate the Maximum
Likelihood estimator of the parameter
of the
exponential distribution. It will turn out to be easier to find the Maximum
Liklihood estimator of µ = 1/
, that
is, of the mean of the distribution. We also identify the distribution of this
estimator, which is closely related to the Gamma
distribution.
We illustrate this result with a two-fold interactive animation :
MAXIMUM LIKELIHOOD ESTIMATION OF THE
MEAN OF THE EXPONENTIAL DISTRIBUTION
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Maximum Likelihood estimation of the mean of the exponential distribution The Likelihood The Log-Likelihood Maximum of the Log-Likelihood The Maximum Likelihood estimator Distribution of the estimator Expectation of the estimator _________________________________________________
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Tutorial 3 |
In this Tutorial, we demonstrate that the exponential
distribution is memoryless, and that it is the only continuous distribution
with this property. This groundbreaking result is illustrated by an interactive
animation.
THE MEMORYLESS PROPERTY
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The exponential distribution is memoryless The exponential distribution is the only continuous memoryless distribution Distribution of the observations whose values are larger than a threshold value s ____________________________________________________________ |
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TUTORIAL |
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Tutorial 4 |
In the next Tutorial, we demonstrate the strong
memoryless property of the exponential distribution as described here.
This demonstration is more difficult than that of the weak property, but
it exploits useful probabilistic techniques, and we made it detailed enough
to be accessible with only a basic background in probability theory.
The strong memoryless property generalizes to n independent exponential distributions. We give this generalized result without demonstration.
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We illustrate this important result with a twin interactive animation :
STRONG MEMORYLESS PROPERTY
OF THE EXPONENTIAL DISTRIBUTION
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The strong memoryless property Theory Consequences Strong memoryless property Independence with respect to X1 Distribution Generalization to n exponential distributions (No demonstration) ___________________________________________
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Tutorial 5 |
In this Tutorial, we define and determine the properties
of the Hazard Rate Function (HRF) (also known as "Failure Rate Function"), a fundamental quantity in survival issues.
We show that it is a characteristic of a distribution (just as is the moment
generating function).
The HRF of the exponential distribution is constant,
a circumstance that is therefore equivalent to the memoryless property. Real
devices (not to mention people) do wear out, and a constant HRF is then an irrealistic
assumption. Various assumptions about the time evolution of the HRF lead to defining
lifetime probability distributions other than exponential, and that are here only briefly
touched upon.
THE HAZARD RATE FUNCTION
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Hazard rate function The general case Special case : the exponential distribution The hazard rate function uniquely determines a distribution The general case Increasingly varying hazard rates Constant rate : the exponential distribution Rate increasing linearly with time : the Rayleigh distribution Rate increasing as a power of time : the Weibull distribution Rate increasing exponentially with time : the Gompertz distribution
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Tutorial 6 |
In the next Tutorial, we consider a set of independent
devices, whose lifetimes are all exponentially distributed. At time 0, all the
devices are operational. But sooner or later, one device will fail. We calculate
the probability for any of these devices to be the first one to fail.
We'll also use this result in a different context when we
study the superposition
of Poisson processes.
FIRST DEVICE TO FAIL
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Probability for any device to be the first one to fail Theory __________________________________________
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Tutorial 7 |
We now examine three classical ways to assemble devices
or components into one sytem :
and we calculate the lifetime distribution and expected lifetime of each of these three set-ups.
DEVICES IN SERIES, PARALLEL AND STAND-BY
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Minimum of exponentials, devices in series The problem Distribution of min(X1, X2) Expectation of min(X1, X2) Interactive animation Maximum of exponentials, devices in parallel The problem Distribution of max(X1, X2) Expectation of max(X1, X2) |
One device on line, another one on stand-by The problem Identical lifetimes Distribution Expected lifetime Different lifetimes Distribution Expected lifetime |
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The next two Tutorials are dedicated to applications of the strong memoryless property.
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Tutorial 8 |
In this Tutorial, we revisit the problem of components
in parallel. We analyzed
the setting made of only two components in parallel. We didn't go any further
because, although straightforward in principle, the calculation of the
distribution of the max of independent exponential r.v. is intractable
in practice.
We now consider any number of independent devices
in parallel, but that are constrained to have identical distributions : they
are all ~Exp(
) for some
.
Even with this simplification, we won't attempt to calculate the distribution of the lifetime of the
system, but only concentrate on a more modest goal : calculating the expected
lifetime of the system.
The strong memoryless property, in its most general form relative to n exponential distribution, will lead us to the result. We'll then discover that increasing the number of identical components in parallel is a very inefficient way of increasing the lifetime of a system, as this expected lifetime increases only very slowly with the number of components, and all the more so that the number of components is already large. This sad fact is often called "the law of diminishing returns", as investing in more components brings about only a vanishingly small increase in expected lifetime.
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In the process, we'll also calculate the distribution of the difference between two consecutive order statistics of the exponential distribution, a quantity known as a "spacing" in the Theory of Distributions.
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We illustrate these two results with an interactive animation.
THE "LAW OF DIMINISHING RETURNS"
SPACINGS OF THE EXPONENTIAL DISTRIBUTION
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Expectation of the max Zn of n iid exponentials The "law of diminishing returns" Warm standbys Cold standbys Distribution of spacings ___________________________________________ |
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Tutorial 9 |
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We now spend some time on a little problem
that we dub "Two identical components in parallel versus one component",
or "2// vs. 1" for short.
The question is :
"What is the probability for A to fail before B ?"
The problem looks deceptively simple. We
propose three solutions :
* The first solution is short, elegant and requires virtually no calculation. Yet it calls twice on the strong memoryless property, and on two important results established in the preceeding Tutorials, so we find it useful to go over it in some detail.
* The second solution is more straightforward, although it requires some pedestrian calculations. Because it calls on material developed elsewhere on this site, we only outline the solution and leave the actual calculations as an exercise.
* The third solution is often perceived
by beginners as the most intuitive one. Unfortunately, it is wrong. We
give the "solution" but leave it as a teaser to find the flaw in the
reasoning.
THE "2// vs. 1" PROBLEM
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The problem First solution Second solution (Outline only) Third solution (WRONG !) |
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Tutorial 10 |
We now establish the distribution
of the r.v. Z, defined as the sum of a random number of iid exponential variables. The number
of variables in the sum is assumed to be geometrically
distributed. The problem looks complicated and in fact, the solution relies
on somewhat advanced material but the final solution is very simple. This is
fortunate because, although the problem looks rather academic, it
accurately describes a realistic situation encountered in Reliability Theory.
1) We'll also use this result in a different context when we
study the splitting of
a Poisson process.
2) The distribution of the sum of a fixed number
of iid exponential rvs is addressed here.
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Although the mean and variance of the distribution
were already calculated somewhere else, recall that we established
expressions for the mean and variance of a random sum of iid rvs in a more
general context. We now use these expressions on the problem at hand, but just as
an exercise.
DISTRIBUTION OF THE SUM OF A
RANDOM NUMBER OF EXPONENTIAL R.V.
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Distribution of Z Example of application Calculating the mean and the variance of Z by the general method |
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Tutorial 11 |
We conclude with two illustrated exercises.
The strong memoryless property states that :
* If X1 and X2 are two independent exponential r.v.,
* Then the distribution of X2 - X1 conditionally to X2 > X1 is identical to the (unconditional) distribution of X2.
But it says nothing about the respective distributions of X1 and of X2 under the same condition.
The goal of the following two exercises is to calculate these distributons..
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The upper frame of the following animation displays two exponential distributions : a red one (that of X1) and a green one (that of X2). You can change the value or the parameter of the distribution of X1 by sliding horizontally the small ball at the upper end of the vertical segment marking the mean of the red distribution with your mouse.
* By default, the lower frame displays the distribution of X1 conditionally to X1 < X2.
* Click on "Go" and observe the build-up of the histogram of this distribution. Note that only draws with X1 < X2 contribute to this histogram, the other draws being ignored (click on "Pause", then several times on "Next").
* The question is : "What is this distribution ?".
We answer the question (as well as the next one) in the Tutorial below.
Click on "Reset", then select "X2".
The lower frame now displays two curves :
* The green curve is the distribution of X2 conditionally to X1 < X2.
* The blue curve is the distribution of max(X1, X2). It plays no active role in the animation, and is displayed for the sole purpose of convincing you that the distribution we are trying to identify, although similar to that of max(X1, X2), is in general not identical to it.
* Also, although "Gamma looking", it's not a Gamma distribution.
* Click on "Go" and observe the build-up of the histogram of the distribution. As before, only those draws resulting in X1 < X2 are retained.
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In this Tutorial, we give two solutions to this second exercise :
* The first one is a direct calculation.
* The second one calls on the result of the first exercise, and on the strong memoryless property.
SOLUTIONS OF THE TWO EXERCISES
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Distribution of X1 conditionally to X1 < X2 Animation Distribution of X2 conditionally to X1 < X2 |
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