What Is the Expectation of a Function Symmetric About Zero?

“Symmetric about zero” can describe either a function or a random variable, and the distinction matters. A function symmetric about zero is even: $g(-x)=g(x)$. A random variable symmetric about zero has the same distribution as its negative: $X$ and $-X$ have the same distribution.

If both are symmetric in these respective senses, the expected value of the function is generally not zero. Instead, evenness lets you combine the contributions from positive and negative values. By contrast, an integrable odd function has expectation zero under a zero-symmetric distribution.

What evenness does to an expectation

For a continuous random variable with density $f_X$, the expectation is

when the integral is defined. If $X$ is symmetric about zero and has a density, then its density is even almost everywhere: $f_X(-x)=f_X(x)$ for almost every $x$. If $g$ is even too, then the product $g(x)f_X(x)$ is even almost everywhere, so

This formula doubles the contribution from the positive half of the distribution; it does not make that contribution cancel. The same idea works for discrete distributions: the contributions from $X=x$ and $X=-x$ add when $g$ is even.

For any random variable $X$ and even function $g$, $g(X)=g(|X|)$ pointwise. Thus, when the expectation exists, $\mathbb{E}[g(X)]=\mathbb{E}[g(|X|)]$ for even $g$.

Worked examples with a standard normal variable

Let $X$ have the standard normal distribution. Its density is symmetric about zero.

First take $g(x)=x^2$. This function is even, and the standard normal has variance $1$ and mean $0$, so

The result is positive, not zero. The density-and-symmetry calculation says the same thing:

For another even function, $g(x)=|x|$, the expectation is

Now take the odd function $h(x)=x^3$, for which $h(-x)=-h(x)$. The contributions from $x$ and $-x$ cancel. Since a standard normal has a finite third absolute moment, this cancellation is valid and $\mathbb{E}[X^3]=0$.

Why odd functions have expectation zero

Suppose $X$ and $-X$ have the same distribution, and $h$ is odd. If $\mathbb{E}[|h(X)|]<\infty$, then

The only finite number equal to its own negative is zero. Therefore $\mathbb{E}[h(X)]=0$.

The integrability condition matters. For example, a standard Cauchy random variable is symmetric about zero, but its mean does not exist. Symmetry does not make its expectation zero: the positive and negative parts both have infinite integral, so they cannot be subtracted to form an expectation. A symmetric improper integral taken as a principal value is not the same as an expected value.

Common mistakes

• Confusing even and odd functions: $g(-x)=g(x)$ means even; $h(-x)=-h(x)$ means odd. Even contributions pair by addition, while odd contributions pair by cancellation.
• Assuming an even function has mean zero: For example, $X^2$ is nonnegative, and for a standard normal variable its expectation is $1$.
• Using symmetry of the function alone: An even function does not guarantee a simplified expectation unless the distribution of $X$ is also symmetric about zero. Without that distributional symmetry, use the usual expectation formula.
• Ignoring existence: Cancellation proves a zero expectation only when the relevant expectation exists, such as when $\mathbb{E}[|h(X)|]$ is finite.

So, if “symmetric about zero” means the function is even, there is no general rule that its expectation is zero. If the random variable is also symmetric about zero, use evenness to reduce the calculation to the nonnegative half-line; use oddness, with integrability, to conclude an expectation is zero.