When an expectation does not exist

The expected value page required an absolute sum or integral. Why do we need that condition when a symmetric distribution appears to balance perfectly around its center? The Cauchy family gives a case in which the visual center exists but the probability-weighted mean does not.

The Cauchy density

For x_0\in\mathbb R and \gamma>0, write

X\sim\operatorname{Cauchy}(x_0,\gamma)

when X has density

f_X(x) \equiv \frac{1}{\pi\gamma \left[1+\left(\frac{x-x_0}{\gamma}\right)^2\right]}, \qquad x\in\mathbb R.

The standard Cauchy distribution has x_0=0 and \gamma=1.

Code

x <- seq(-6, 6, length.out = 1000)
plot(x, dcauchy(x, location = 0, scale = 1),
     type = "l", lwd = 3, col = "#2166AC",
     xlab = "x", ylab = "Density",
     main = "PDF of Cauchy(0, 1)", bty = "l")

The density is symmetric around zero, so the tempting move is to pair each positive contribution with a negative one and declare the mean zero. The next calculation shows why that move is not licensed.

Checking the two sides before combining them

For the standard Cauchy density,

x f_X(x)=\frac{x}{\pi(1+x^2)}.

Consider the positive part:

\begin{aligned} \int_0^b \frac{x}{\pi(1+x^2)}\,\mathrm dx &=\frac{1}{2\pi}\log(1+b^2). \end{aligned}

As b\to\infty, this quantity diverges to +\infty. By symmetry, the integral over the negative half-line diverges to -\infty. An expectation exists only when these two parts can be combined without making the result depend on how the limits are taken. Here they cannot, so

\mathbb E[X]

is undefined.

Principal value is not expectation

For every finite b>0,

\int_{-b}^{b}x f_X(x)\,\mathrm dx=0.

Taking the symmetric limit produces a Cauchy principal value of zero. An expectation is not defined by forcing the positive and negative truncation bounds to approach infinity at the same rate. It requires the probability-weighted integral itself to exist. Thus the principal value is not \mathbb E[X].

Finite samples do not repair the model expectation

Every finite sample of Cauchy draws has a finite arithmetic mean, but that fact does not imply that the distribution has an expected value. As the sample grows, an extreme draw can move the sample mean by an arbitrarily large amount. The model supplies no finite population mean to which those sample means must converge.

Check your understanding

Evaluate the antiderivative of x/[\pi(1+x^2)].
Explain why symmetry does not define the Cauchy mean.
Distinguish the Cauchy principal value from an expected value.
Explain why every finite sample can have a finite arithmetic mean even when the distributional expectation is undefined.

The exact upshot is that symmetry identifies a geometric center but does not guarantee a finite expected value. The later summaries in this module, including variance, likewise require their defining expectations to exist.

--- title: "When an expectation does not exist" execute: enabled: true echo: true warning: false message: false --- The [expected value page](expected-values.qmd) required an absolute sum or integral. Why do we need that condition when a symmetric distribution appears to balance perfectly around its center? The Cauchy family gives a case in which the visual center exists but the probability-weighted mean does not. ## The Cauchy density For $x_0\in\mathbb R$ and $\gamma>0$, write $$ X\sim\operatorname{Cauchy}(x_0,\gamma) $$ when $X$ has density $$ f_X(x) \equiv \frac{1}{\pi\gamma \left[1+\left(\frac{x-x_0}{\gamma}\right)^2\right]}, \qquad x\in\mathbb R. $$ The standard Cauchy distribution has $x_0=0$ and $\gamma=1$. ```{r} x <- seq(-6, 6, length.out = 1000) plot(x, dcauchy(x, location = 0, scale = 1), type = "l", lwd = 3, col = "#2166AC", xlab = "x", ylab = "Density", main = "PDF of Cauchy(0, 1)", bty = "l") ``` The density is symmetric around zero, so the tempting move is to pair each positive contribution with a negative one and declare the mean zero. The next calculation shows why that move is not licensed. ## Checking the two sides before combining them For the standard Cauchy density, $$ x f_X(x)=\frac{x}{\pi(1+x^2)}. $$ Consider the positive part: $$ \begin{aligned} \int_0^b \frac{x}{\pi(1+x^2)}\,\mathrm dx &=\frac{1}{2\pi}\log(1+b^2). \end{aligned} $$ As $b\to\infty$, this quantity diverges to $+\infty$. By symmetry, the integral over the negative half-line diverges to $-\infty$. An expectation exists only when these two parts can be combined without making the result depend on how the limits are taken. Here they cannot, so $$ \mathbb E[X] $$ is undefined. ## Principal value is not expectation For every finite $b>0$, $$ \int_{-b}^{b}x f_X(x)\,\mathrm dx=0. $$ Taking the symmetric limit produces a Cauchy principal value of zero. An expectation is not defined by forcing the positive and negative truncation bounds to approach infinity at the same rate. It requires the probability-weighted integral itself to exist. Thus the principal value is not $\mathbb E[X]$. ## Finite samples do not repair the model expectation Every finite sample of Cauchy draws has a finite arithmetic mean, but that fact does not imply that the distribution has an expected value. As the sample grows, an extreme draw can move the sample mean by an arbitrarily large amount. The model supplies no finite population mean to which those sample means must converge. ## Check your understanding 1. Evaluate the antiderivative of $x/[\pi(1+x^2)]$. 2. Explain why symmetry does not define the Cauchy mean. 3. Distinguish the Cauchy principal value from an expected value. 4. Explain why every finite sample can have a finite arithmetic mean even when the distributional expectation is undefined. The exact upshot is that symmetry identifies a geometric center but does not guarantee a finite expected value. The later summaries in this module, including [variance](variance-and-covariance.qmd), likewise require their defining expectations to exist.