Symmetry And Moments From Characteristic Functions

Symmetry Leaves a Signature

For a real-valued random variable $X$, its characteristic function is $\phi_X(t)=\mathbb{E}[e^{itX}]$, for real $t$. It always exists, because $|e^{itX}|=1$. Its values are generally complex, but a symmetric distribution has a particularly simple signature: its characteristic function is real-valued and even. Derivatives of the same function can encode moments, provided the corresponding absolute moments are finite.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $X:\Omega\to\mathbb{R}$ be a random variable. We say that $X$ is symmetric about zero if $X$ and $-X$ have the same distribution. Symmetry about a center $a\in\mathbb{R}$ means that $X-a$ and $-(X-a)$ have the same distribution. Equality in distribution lets us replace one random variable by the other inside an expectation whenever the expectation exists; for the bounded function $x\mapsto e^{itx}$, it always exists.

Proposition: If $X$ is symmetric about zero, then $\phi_X(t)$ is real for every real $t$, and $\phi_X(-t)=\phi_X(t)$. If $X$ is symmetric about $a$, then $e^{-iat}\phi_X(t)$ is real and even as a function of $t$.

Proof. Write $e^{itX}=\cos(tX)+i\sin(tX)$. If $X$ and $-X$ have the same distribution, then $\mathbb{E}[\sin(tX)]=\mathbb{E}[\sin(-tX)]=-\mathbb{E}[\sin(tX)]$, so this expectation is zero. Thus $\phi_X(t)=\mathbb{E}[\cos(tX)]$, which is real. Since cosine is even, $\phi_X(-t)=\mathbb{E}[\cos(-tX)]=\mathbb{E}[\cos(tX)]=\phi_X(t)$. For symmetry about $a$, put $Y=X-a$. Its characteristic function is $\phi_Y(t)=\mathbb{E}[e^{it(X-a)}]=e^{-iat}\phi_X(t)$. The first part applied to $Y$ proves the claim. $\square$

This proposition gives a necessary signature of symmetry, not a converse proved here. Establishing that a characteristic function determines a distribution requires a uniqueness result. Also, a real-valued characteristic function need not be interpreted as evidence of symmetry without that extra argument.

Worked Example: A Symmetric Two-Point Distribution

Let $X$ take the values $-2$ and $2$, each with probability $\tfrac12$. Its distribution is symmetric about zero. Compute its characteristic function and use it to check the first two moments.

By summing over the two possible values,

\phi_X(t)=\tfrac12 e^{-2it}+\tfrac12 e^{2it} =\cos(2t).

Differentiate: $\phi_X'(t)=-2\sin(2t)$ and $\phi_X''(t)=-4\cos(2t)$. Therefore $\phi_X'(0)=0$ and $\phi_X''(0)=-4$. The moment formulas proved below give $\mathbb{E}X=0$ and $\mathbb{E}X^2=-\phi_X''(0)=4$. Directly, $\mathbb{E}X=\tfrac12(-2)+\tfrac12(2)=0$ and $\mathbb{E}X^2=\tfrac12(4)+\tfrac12(4)=4$.

Check: $\cos(2t)$ is real and even, as symmetry predicts; the direct moments agree with the derivatives.

Derivatives at Zero and Moments

The derivative formula is a statement about ordinary derivatives of a complex-valued function of the real variable $t$. Its condition matters: to take $r$ derivatives, we require the $r$th absolute moment to be finite. Absolute integrability controls the difference quotients inside the expectation.

Theorem: Let $m\ge 1$ be an integer and suppose $\mathbb{E}|X|^m<\infty$. Then $\phi_X$ has derivatives through order $m$, and, for $0\le r\le m$, $$ \phi_X^{(r)}(t)=\mathbb{E}\!\left[(iX)^r e^{itX}\right]. $$ In particular, $\phi_X^{(r)}(0)=i^r\mathbb{E}[X^r]$, so $\mathbb{E}[X^r]=\phi_X^{(r)}(0)/i^r$.

Proof. Since $\mathbb{E}|X|^m<\infty$, every lower absolute moment is finite: for $0\le r\le m$, $|x|^r\le 1+|x|^m$. The case $r=0$ is the defining formula $\phi_X(t)=\mathbb{E}[e^{itX}]$. Suppose the formula has been proved at order $r$, where $0\le r<m$. For a nonzero real increment $h$, the difference quotient of the integrand is $$ (iX)^r e^{itX}\frac{e^{ihX}-1}{h}. $$ For every real $u$, $|e^{iu}-1|\le |u|$; this follows, for example, by writing $e^{iu}-1=\int_0^u i e^{iv}\,dv$ and using $|ie^{iv}|=1$. The absolute value of the displayed quotient is therefore at most $|X|^{r+1}$, which is integrable because $r+1\le m$. As $h\to0$, the quotient converges pointwise to $(iX)^{r+1}e^{itX}$. The dominated convergence theorem now permits passage of the limit through the expectation, proving the derivative formula at order $r+1$. Induction proves the formula through order $m$. Setting $t=0$ gives $\phi_X^{(r)}(0)=i^r\mathbb{E}[X^r]$, and division by the nonzero number $i^r$ gives the moment identity. $\square$

For example, when the second moment is finite, $\mathbb{E}X=\phi_X'(0)/i$ and $\mathbb{E}X^2=-\phi_X''(0)$, since $i^2=-1$. If the variance is also being computed, then $$ \operatorname{Var}(X)=-\phi_X''(0)-\left(\frac{\phi_X'(0)}{i}\right)^2. $$ This is the raw second moment minus the square of the mean. More generally, the theorem gives raw moments; converting them into central moments or cumulants is a separate calculation. The theorem does not say that a derivative at zero exists only when the corresponding absolute moment is finite. It gives a sufficient condition that justifies the expectation formula.

Worked Example: Uniform Distribution on a Symmetric Interval

Let $X$ be uniform on $[-1,1]$, with density $f(x)=\tfrac12$ for $-1\le x\le1$. The distribution is symmetric about zero. Find its characteristic function and obtain its second and fourth moments from derivatives.

Integrating the density gives, for $t\ne0$,

\phi_X(t)=\tfrac12\int_{-1}^{1}e^{itx}\,dx =\frac{e^{it}-e^{-it}}{2it} =\frac{\sin t}{t}.

At $t=0$, the defining expectation gives $\phi_X(0)=1$; this also equals the limit of $\sin(t)/t$. The Taylor expansion at zero is $\sin(t)/t=1-t^2/3!+t^4/5!+\cdots$. Hence $\phi_X''(0)=-2/3!=-1/3$, and $\phi_X^{(4)}(0)=4!/5!=1/5$. The derivative theorem applies because $|X|\le1$, so all absolute moments are finite. It yields $\mathbb{E}X^2=-\phi_X''(0)=1/3$ and $\mathbb{E}X^4=\phi_X^{(4)}(0)=1/5$.

Check by integration: $\mathbb{E}X^2=\tfrac12\int_{-1}^{1}x^2\,dx=\tfrac12(2/3)=1/3$, and $\mathbb{E}X^4=\tfrac12\int_{-1}^{1}x^4\,dx=\tfrac12(2/5)=1/5$. Both direct calculations agree with the derivatives.

Odd Moments and Symmetry About a Center

Symmetry also directly determines odd moments when they exist. If $X$ is symmetric about zero and $\mathbb{E}|X|^r<\infty$ for an odd positive integer $r$, then $$ \mathbb{E}[X^r]=\mathbb{E}[(-X)^r]=-\mathbb{E}[X^r], $$ so $\mathbb{E}[X^r]=0$. The absolute-moment condition ensures the expectation is finite; without it, the equality does not establish a finite moment. The same idea applies to $Y=X-a$ when $X$ is symmetric about $a$: every existing odd centered moment $\mathbb{E}[(X-a)^r]$ is zero. Correspondingly, the centered characteristic function $e^{-iat}\phi_X(t)$ is real and even.

Worked Example: A Distribution Symmetric About One

Let $X$ take values $0$ and $2$, each with probability $\tfrac12$. This distribution is symmetric about $1$, rather than about zero. Compute $\phi_X$, its first two derivatives at zero, and the centered characteristic function.

From the two possible values,

\phi_X(t)=\tfrac12+\tfrac12e^{2it} =e^{it}\cos t.

Differentiating the finite sum gives $\phi_X'(t)=i e^{2it}$ and $\phi_X''(t)=-2e^{2it}$. Thus $\phi_X'(0)=i$ and $\phi_X''(0)=-2$. The derivative identities give $\mathbb{E}X=i/i=1$ and $\mathbb{E}X^2=(-2)/i^2=2$. Directly, $\mathbb{E}X=\tfrac12(0)+\tfrac12(2)=1$, and $\mathbb{E}X^2=\tfrac12(0)+\tfrac12(4)=2$. Consequently, $\operatorname{Var}(X)=2-1^2=1$.

Centering removes the phase: $e^{-it}\phi_X(t)=\cos t$, which is real and even. Equivalently, $X-1$ takes the values $-1$ and $1$ with equal probabilities. Check: its mean is $\tfrac12(-1)+\tfrac12(1)=0$, and its second moment is $\tfrac12(1)+\tfrac12(1)=1$, matching the variance of $X$.

Common Mistakes and Practical Use

A frequent error is to use the moment-generating-function derivative rule without its factors of $i$. For a characteristic function, the factor $e^{itX}$ differentiates to $iXe^{itX}$, not $Xe^{itX}$. This changes signs and complex factors, even when the final moments are real.

Common mistake—dropping the $i$. For the two-point distribution on $0$ and $2$, one might incorrectly write $\mathbb{E}X=\phi_X'(0)=i$. But the derivative calculation gives $\phi_X'(0)=i$, while the actual mean is $\tfrac12(0)+\tfrac12(2)=1$. The correction is $\mathbb{E}X=\phi_X'(0)/i=i/i=1$. Likewise, writing $\mathbb{E}X^2=\phi_X''(0)$ would give $-2$; the correct formula is $\mathbb{E}X^2=\phi_X''(0)/i^2=(-2)/(-1)=2$.

Another mistake is to treat “symmetric” as meaning “symmetric about zero.” A shift changes the characteristic function by a phase: if $X-a$ is symmetric about zero, then $e^{-iat}\phi_X(t)$, rather than necessarily $\phi_X(t)$, is real and even. Finally, symmetry alone does not guarantee that every moment exists. For an odd-moment conclusion or a derivative formula, check the relevant absolute-moment assumption first.

These facts make characteristic functions useful for quick structural checks. A proposed characteristic function for a zero-centered symmetric distribution should be real and even; if the distribution has a finite mean, its first derivative at zero must be zero. If it also has a finite second moment, its second derivative at zero must be nonpositive, since it equals $-\mathbb{E}X^2$. Such checks can reveal a sign error or a mistaken centering before a longer calculation is attempted.

Check Your Understanding

Use the definitions and formulas above, and state the moment assumptions needed for each derivative-based answer.

If $X$ is symmetric about zero and has a finite third absolute moment, what is $\mathbb{E}X^3$, and why?
A random variable has $\phi_X'(0)=2i$ and a finite first moment. Find $\mathbb{E}X$.
For a distribution symmetric about $a$, which function of its characteristic function must be real and even?
Suppose $\phi_X''(0)=-9$ and $\mathbb{E}|X|^2<\infty$. Find $\mathbb{E}X^2$.
Explain why a real-even characteristic function is a necessary symmetry signature in the result proved here, but why that proof alone does not establish the converse.

Symmetry And Moments From Characteristic Functions

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