Tutorials › Statistical Theory I - Probability and Distribution Theory › Symmetry and Expectations of Odd Functions

Home  /  Tutorials  /  Expectation and LOTUS

Expectation and LOTUS · Tutorial 315 of 994

Symmetry and Expectations of Odd Functions

Use reflection symmetry to identify zero expectations and reduce calculations, while checking the integrability conditions that make the argument valid.

Intermediate 9 min read

What You'll Learn

  • Define symmetry of a distribution about zero or a general center
  • Prove that an integrable odd function has expectation zero under zero-centered symmetry
  • Use reflection to simplify expectations of functions that are not themselves odd
  • Apply the result to discrete, continuous, and centered distributions
  • Recognize why symmetry alone does not define an expectation

Why Symmetry Can Simplify an Expectation

When a distribution gives matching probability to values equally far on either side of a center, contributions from an odd function can cancel in pairs. The important qualifications are that the function must be odd about the relevant center and that its expectation must exist. Without the latter condition, an apparent cancellation of positive and negative contributions may not define an expectation at all.

The reflection argument applies to discrete, continuous, and mixed distributions. It is therefore more general than pairing terms in a sum or canceling areas under a symmetric density. We first state it in terms of the distribution itself, so that no density is required.

Definition: A random variable \(X\) is symmetric about zero if \(X\) and \(-X\) have the same distribution. More generally, \(X\) is symmetric about \(c\) if \(X-c\) and \(-(X-c)\) have the same distribution; equivalently, \(X\) and \(2c-X\) have the same distribution. A function \(g\) is odd if \(g(-x)=-g(x)\) wherever both sides are defined. It is odd about \(c\) when \(g(c-t)=-g(c+t)\) for every relevant \(t\).

For a distribution function, symmetry about zero can also be expressed as \(F(x)=1-F((-x)-)\), where \(F(y-)=P(X<y)\). The left limit handles atoms correctly: the probability at a reflected endpoint must be included. If \(X\) has a density \(f\), symmetry about zero is equivalent to \(f(x)=f(-x)\) almost everywhere. If \(X\) is discrete, it means \(P(X=x)=P(X=-x)\) for each \(x\).

The Zero-Expectation Result

Theorem: Suppose \(X\) is symmetric about zero, \(g\) is measurable and odd, and \(E[|g(X)|]<\infty\). Then \(E[g(X)]=0\).

The integrability condition is absolute: the positive and negative parts of \(g(X)\) must have finite expectations. It makes \(E[g(X)]\) a finite, well-defined number and permits the reflection argument without relying on a rearrangement of divergent quantities.

1
Reflect the random variable.
Symmetry says \(-X\) has the same distribution as \(X\), so applying the same measurable function to either variable gives the same expectation whenever it exists.
2
Use oddness.
Pointwise, \(g(-X)=-g(X)\). The two equal expectations are therefore negatives of one another.
3
Conclude they vanish.
A finite real number equal to its own negative is zero.

Here is the complete argument. Since \(X\) and \(-X\) have the same distribution, the law of \(g(X)\) is the same as the law of \(g(-X)\). Also, \(|g(-X)|=|g(X)|\), so both expectations below are absolutely integrable. Equality in distribution therefore gives equality of expectations, while oddness gives the second equality:

$$ E[g(X)] = E[g(-X)] = E[-g(X)] = -E[g(X)]. $$

Thus \(2E[g(X)]=0\), and \(E[g(X)]=0\). The proof uses only invariance of the distribution under reflection, not a particular formula for its density or mass function. \(\square\)

Worked Example: A Discrete Symmetric Distribution

Suppose \(X\) has probabilities \(P(X=-2)=0.1\), \(P(X=-1)=0.2\), \(P(X=0)=0.4\), \(P(X=1)=0.2\), and \(P(X=2)=0.1\). Find \(E[X^3-2X]\).

The probabilities at each pair of reflected values match, so \(X\) is symmetric about zero. The function \(g(x)=x^3-2x\) is odd because \(g(-x)=(-x)^3-2(-x)=-x^3+2x=-g(x)\). Since \(X\) has finite support, \(E[|g(X)|]<\infty\), and the theorem already gives \(E[g(X)]=0\).

For a direct check, evaluate the function at each support point: \(g(-2)=-8+4=-4\), \(g(-1)=-1+2=1\), \(g(0)=0\), \(g(1)=1-2=-1\), and \(g(2)=8-4=4\). Thus

$$ E[X^3-2X] =0.1(-4)+0.2(1)+0.4(0)+0.2(-1)+0.1(4) =-0.4+0.2+0-0.2+0.4 =0. $$

The terms cancel in reflected pairs: the contribution from \(-2\) cancels the one from \(2\), and the contribution from \(-1\) cancels the one from \(1\). The atom at zero contributes zero.

Worked Example: A Bounded Odd Function Under a Continuous Distribution

Let \(X\sim \operatorname{Uniform}(-2,2)\), and find \(E\left[\frac{X}{1+X^2}\right]\). The density is \(f(x)=1/4\) for \(-2<x<2\), and zero elsewhere. The distribution is symmetric about zero. The function \(g(x)=x/(1+x^2)\) is odd, since its denominator is unchanged by reflection and its numerator changes sign. It is bounded on this interval, so it is absolutely integrable.

The theorem gives an expectation of zero. Direct integration verifies the result:

$$ E\left[\frac{X}{1+X^2}\right] =\frac14\int_{-2}^{2}\frac{x}{1+x^2}\,dx =\frac14\left[\frac12\log(1+x^2)\right]_{-2}^{2} =\frac18\bigl(\log 5-\log 5\bigr) =0. $$

In this calculation the antiderivative takes the same value at the two endpoints. Equivalently, the integrand is odd and the integration interval is symmetric about zero. The distribution-based result explains why the same conclusion does not depend on having a convenient antiderivative.

Reflection About a General Center

If \(X\) is symmetric about \(c\), the zero-expectation result applies to \(X-c\). In particular, if \(g\) is odd, measurable, and \(E[|g(X-c)|]<\infty\), then \(E[g(X-c)]=0\). This includes the familiar conclusion \(E[X-c]=0\) when \(E[|X-c|]<\infty\), and hence \(E[X]=c\).

A broader simplification applies even when the function of interest is not odd. For an integrable function \(h\), pair its value at \(x\) with its value at the reflected point \(2c-x\). The average of those two values is symmetric about \(c\); the difference is odd about \(c\).

Theorem: Suppose \(X\) is symmetric about \(c\), \(h\) is measurable, and \(E[|h(X)|]<\infty\). Then \(h(2c-X)\) is integrable and
$$ E[h(X)] = E\left[\frac{h(X)+h(2c-X)}{2}\right]. $$

To prove the identity, first note that \(2c-X\) has the same distribution as \(X\). Consequently, \(E[|h(2c-X)|]=E[|h(X)|]<\infty\), and \(E[h(2c-X)]=E[h(X)]\). The sum \(h(X)+h(2c-X)\) is integrable because its absolute value is at most \(|h(X)|+|h(2c-X)|\), which has finite expectation. By linearity for integrable random variables,

$$ E\left[\frac{h(X)+h(2c-X)}{2}\right] =\frac{E[h(X)]+E[h(2c-X)]}{2} =\frac{E[h(X)]+E[h(X)]}{2} =E[h(X)]. $$

This proves the identity. In effect, it discards the odd-about-\(c\) part of \(h\), because that part has expectation zero under the symmetric distribution. \(\square\)

Worked Example: A Cubic Moment About a Nonzero Center

Let \(X\sim \operatorname{Uniform}(1,5)\). This distribution is symmetric about \(c=3\). Find \(E[X^3]\) by centering the variable, rather than integrating the cubic directly.

Set \(Y=X-3\). Then \(Y\sim\operatorname{Uniform}(-2,2)\), so \(Y\) is symmetric about zero. Its density is \(1/4\) on \((-2,2)\). By symmetry and integrability, \(E[Y]=0\) and \(E[Y^3]=0\). Its second moment is

$$ E[Y^2] =\frac14\int_{-2}^{2}y^2\,dy =\frac14\left[\frac{y^3}{3}\right]_{-2}^{2} =\frac14\left(\frac83+\frac83\right) =\frac43. $$

Since \(X=3+Y\), expand the cube and take expectations term by term:

$$ E[X^3] =E[(3+Y)^3] =E[27+27Y+9Y^2+Y^3] =27+27(0)+9\left(\frac43\right)+0 =39. $$

The centered odd terms vanish, leaving only the constant and the second-moment term. This illustrates how reflection can simplify a calculation even when \(x^3\) itself is not odd about the distribution’s center.

Common Mistake: Cancellation Without an Expectation

Symmetry does not by itself guarantee that an expectation exists. The theorem requires absolute integrability; without it, writing equal positive and negative contributions and canceling them can be invalid.

For example, a standard Cauchy random variable has the symmetric density \(f(x)=1/(\pi(1+x^2))\). It is tempting to say that \(E[X]=0\) because the density is even and \(x\) is odd. But

$$ E[|X|] =\frac{2}{\pi}\int_0^\infty \frac{x}{1+x^2}\,dx =\infty, $$

because the integral up to \(b\) equals \(\frac{1}{\pi}\log(1+b^2)\), which tends to infinity. In fact, the positive and negative parts of \(X\) both have infinite expectation. Thus \(E[X]\) is undefined: symmetry does not turn this into a valid zero expectation. The correct reasoning is to check \(E[|g(X)|]<\infty\) before applying the odd-function result.

A second common error is to use oddness around the wrong point. If \(X\) is symmetric about \(c\ne 0\), then \(g(X)\) need not have expectation zero merely because \(g\) is odd about zero. Instead, check whether the relevant function is odd in the centered variable \(X-c\), or pair values at \(x\) and \(2c-x\).

Key takeaway: Under reflection symmetry, integrable odd contributions cancel in expectation. Center at the distribution’s symmetry point, and verify absolute integrability before concluding that an expectation is zero.

Check Your Understanding

For each question, identify the symmetry and integrability conditions that justify your conclusion.

  1. If \(X\) is symmetric about zero and \(E[|X|^5]<\infty\), what is \(E[X^5]\)? Explain why the existence condition matters.
  2. A variable is symmetric about \(c=4\), and \(E[|X-4|]<\infty\). Find \(E[X]\) and state the reflection argument.
  3. Suppose \(X\sim\operatorname{Uniform}(-1,1)\). Is \(g(x)=x^2\) odd? Does symmetry force \(E[g(X)]=0\)? Explain.
  4. For a distribution symmetric about \(c\), write the reflected-point average of a function \(h(x)\), and state what condition makes the expectation identity valid.
  5. Why does symmetry of a standard Cauchy distribution not imply that its mean is zero?

Previous

Next

Back to Tutorials