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Why the mean of this distribution is undefined.
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==Why the mean of the Cauchy distribution is undefined==
==Why the mean of the Cauchy distribution is undefined==


If a probability distribution has a density function ''f''(''x'') then the mean or expected value is
If a [[probability distribution]] has a [[probability density function|density function]] ''f''(''x'') then the mean or [[expected value]] is


:<math>\int_{-\infty}^\infty x f(x)\,dx.\qquad\qquad (1)</math>
:<math>\int_{-\infty}^\infty x f(x)\,dx.\qquad\qquad (1)</math>
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:<math>\int_0^\infty x f(x)\,dx-\int_{-\infty}^0 x f(x)\,dx\quad\rm{?}\qquad\qquad (2)</math>
:<math>\int_0^\infty x f(x)\,dx-\int_{-\infty}^0 x f(x)\,dx\quad\rm{?}\qquad\qquad (2)</math>


If both the positive and negative terms in (2) are finite, then (1) is the same as (2). If either the positive term or the negative term is finite, then (1) is the same as (2). But in the case of the Cauchy distribution, both are infinite. This means (2) is undefined, and then:
If both the positive and negative terms in (2) are finite, then (1) is the same as (2). If either the positive term or the negative term is finite, then (1) is the same as (2) (and is infinite, with either a positive or a negative sign). But in the case of the Cauchy distribution, both are infinite. This means (2) is undefined, and then:


* If (1) is construed as a [[Lebesgue integral]], then (1) is also undefined, since (1) is then defined simply as the difference (2) between positive and negative parts; however
* If (1) is construed as a [[Lebesgue integral]], then (1) is also undefined, since (1) is then defined simply as the difference (2) between positive and negative parts; however

Revision as of 23:58, 24 August 2004

The Cauchy distribution is a probability distribution with probability density function

where t is the location parameter and s is the scale parameter. The special case when t = 0 and s = 1 is called the standard Cauchy distribution with the probability density function

The Cauchy distribution is often cited as an example of a distribution which has no mean, variance or higher moments defined, although its mode and median are well defined and both zero.

When U and V are two independent normally distributed random variables with expected value 0 and variance 1, then the ratio U/V has the standard Cauchy distribution.

If X1, ..., Xn are independent random variables, each with a standard Cauchy distribution, then the sample mean (X1 + ... + Xn)/n has the same standard Cauchy distribution. This example serves to show that the hypothesis of finite variance in the central limit theorem cannot be dropped (although it can be replaced with other, in some cases weaker, assumptions). To see that this is true, compute the characteristic function

where is the sample mean.

The Cauchy distribution is an infinitely divisible probability distribution.

The Cauchy distribution is the Student's t-distribution with just one degree of freedom.

The Cauchy distribution is sometimes called the Lorentz distribution, because it is based on the Lorentzian function.

Why the mean of the Cauchy distribution is undefined

If a probability distribution has a density function f(x) then the mean or expected value is

Is this the same thing as

If both the positive and negative terms in (2) are finite, then (1) is the same as (2). If either the positive term or the negative term is finite, then (1) is the same as (2) (and is infinite, with either a positive or a negative sign). But in the case of the Cauchy distribution, both are infinite. This means (2) is undefined, and then:

  • If (1) is construed as a Lebesgue integral, then (1) is also undefined, since (1) is then defined simply as the difference (2) between positive and negative parts; however
  • If (1) is construed as an improper integral rather than a Lebesgue integral, then (2) is undefined, and (1) is not necessarily well-defined. We may take (1) to mean
and this is its Cauchy principal value, and it is zero, but we could also take (1) to mean, for example,
and this is not zero, as you will see easily if you compute the integral.

Various results in probability theory about expected values, such as the strong law of large numbers, will not work in such cases.