Criteria for Moment Determinacy

When Do Moments Identify a Distribution?

A sequence of moments can summarize important features of a distribution, but it is natural to ask whether it contains enough information to recover the whole distribution. The answer is not always yes: distinct distributions can share every moment. The question of moment determinacy is whether a given distribution is the only probability distribution with its complete moment sequence.

Two useful sufficient conditions are an MGF finite in a neighborhood of zero and Carleman’s criterion. Neither is a necessary condition: passing either test guarantees uniqueness, but failing both does not show that uniqueness fails.

Definition: A probability distribution $F$ on the real line is moment determinate if all its moments $m_k=E[X^k]$, for integers $k\geq 0$, are finite and every probability distribution on the real line with the same moments is equal to $F$. Otherwise, $F$ is moment indeterminate. The sequence $(m_k)_{k\geq 0}$ is called its moment sequence.

Determinacy is about uniqueness, not existence. A criterion for determinacy assumes there is a distribution with the given moments and asks whether another distribution can have them too. In particular, merely calculating many moments, or even finding a formula for all of them, does not by itself establish uniqueness.

A Finite MGF Near Zero Guarantees Determinacy

For a random variable $X$, its moment-generating function is $M_X(t)=E[e^{tX}]$ wherever this expectation is finite. The important condition here is that it is finite on some open interval containing zero. In that case, the moments determine the MGF near zero; the uniqueness theorem for MGFs then identifies the distribution.

Theorem: If $M_X(t)$ is finite for every $t$ in some open interval $(-a,a)$, where $a>0$, then the distribution of $X$ is moment determinate on the real line.

Here is a proof that addresses a subtle point: a second distribution with the same moments is not assumed in advance to have a finite MGF.

Start with a competing distribution.
Let $Y$ be any real-valued random variable whose moments satisfy $E[Y^k]=E[X^k]$ for every integer $k\geq 0$. Choose $r$ with $0<r<a$.

Establish exponential integrability for $Y$.
Since the power series for $\cosh$ has nonnegative terms, Tonelli’s theorem gives $E[\cosh(rY)]=\sum_{j=0}^{\infty}r^{2j}E[Y^{2j}]/(2j)!$. These even moments equal those of $X$, and $E[\cosh(rX)]\leq (M_X(r)+M_X(-r))/2<\infty$. Thus $E[\cosh(rY)]<\infty$. The pointwise inequality $e^{r|y|}\leq 2\cosh(ry)$ implies $E[e^{r|Y|}]<\infty$.

Match the MGFs on an interval.
For $|t|<r$, $e^{tY}$ has an absolutely integrable power series, because $E[\sum_{k=0}^{\infty}|tY|^k/k!]=E[e^{|t||Y|}]<\infty$. Therefore the series can be integrated term by term, giving $M_Y(t)=\sum_{k=0}^{\infty}t^kE[Y^k]/k!=\sum_{k=0}^{\infty}t^kE[X^k]/k!=M_X(t)$.

Use uniqueness of MGFs.
Both MGFs are finite on an open interval around zero and agree there. The MGF uniqueness theorem says that probability distributions with MGFs finite on such an interval and equal there are equal. Hence $Y$ and $X$ have the same distribution. Since $Y$ was arbitrary, $X$ is moment determinate. $\square$

Worked Example: An Exponential Distribution Is Moment Determinate

Let $X\sim\operatorname{Exponential}(1)$, with density $f(x)=e^{-x}$ for $x>0$. Its MGF is $M_X(t)=1/(1-t)$ for $t<1$, so it is finite on the open interval $(-1,1)$, which contains zero.

For each integer $k\geq 0$, integration by parts gives $E[X^k]=\int_0^\infty x^ke^{-x}\,dx=k!$. The boundary term vanishes at infinity because $x^ke^{-x}\to0$; at zero it vanishes for $k>0$, and the $k=0$ integral is 1. The MGF criterion applies, so no other probability distribution on the real line can have all these same moments.

The conclusion is uniqueness, not a recipe for recovering a density by simply summing the moment sequence. The theorem establishes that a distribution with these moments exists and is the only one; it does not claim that every moment sequence has a corresponding distribution.

Carleman’s Criterion

The MGF condition is powerful, but some distributions have all moments while their MGFs are not finite on any open interval around zero. Carleman’s criterion gives a different sufficient test using the even moments alone.

Carleman’s criterion (Hamburger case): Suppose a probability distribution on the real line has finite moments of every nonnegative integer order, and write $m_{2n}=E[X^{2n}]$. If $$ \sum_{n=1}^{\infty}m_{2n}^{-1/(2n)}=\infty, $$ then the distribution is moment determinate. If an even moment is zero, interpret its term in the sum as $+\infty$.

The criterion is sufficient, not necessary. Its divergence certifies determinacy. A convergent sum only means that this particular test does not settle the question. Carleman’s theorem is a deeper result than the MGF argument above; its conclusion applies without requiring the distribution to have an MGF near zero.

Worked Example: Checking Carleman’s Criterion for a Normal Distribution

Let $X\sim N(0,1)$. Its even moments are $m_{2n}=(2n)!/(2^n n!)$. By Stirling’s formula, $(2n)!\sim\sqrt{4\pi n}(2n/e)^{2n}$ and $n!\sim\sqrt{2\pi n}(n/e)^n$. Substitution gives $m_{2n}^{1/(2n)}\sim\sqrt{2n/e}$.

Consequently $m_{2n}^{-1/(2n)}\sim\sqrt{e/(2n)}$. The terms are asymptotic to a positive constant times $n^{-1/2}$, whose sum diverges. Carleman’s criterion therefore shows that the standard normal distribution is moment determinate. Its MGF is also finite near zero, so this example passes both tests.

A Second Determinacy Result: Bounded Support

A useful related result gives a direct criterion when the distribution is supported on a bounded interval. It also illustrates how moment equality can lead to equality of distributions through polynomial approximation.

Theorem: If a probability distribution is supported on $[-K,K]$ for some finite $K\geq0$, then it is moment determinate among all probability distributions on the real line.

To prove the result, let $X$ have the stated distribution and let $Y$ be any random variable with the same moments. If $K=0$, then $E[Y^2]=E[X^2]=0$, which implies $Y=0$ almost surely. Now suppose $K>0$. For any $\varepsilon>0$ and integer $n\geq1$, Markov’s inequality and moment equality give $$ P(|Y|\geq K+\varepsilon)\leq \frac{E[Y^{2n}]}{(K+\varepsilon)^{2n}} =\frac{E[X^{2n}]}{(K+\varepsilon)^{2n}} \leq\left(\frac{K}{K+\varepsilon}\right)^{2n}. $$ The last bound tends to zero as $n\to\infty$. Thus $P(|Y|\geq K+\varepsilon)=0$ for every $\varepsilon>0$, and taking a countable sequence of positive $\varepsilon$ decreasing to zero shows $P(|Y|>K)=0$. Both variables are supported on $[-K,K]$.

Their moments agree, so their integrals agree for every polynomial. By the Weierstrass approximation theorem, every continuous function on $[-K,K]$ is a uniform limit of polynomials. Since both distributions are probability measures, uniform approximation implies that their integrals agree for every continuous function on this interval. Probability measures on a compact interval are uniquely determined by their integrals against all continuous functions. Hence $X$ and $Y$ have the same distribution, proving the claim. $\square$

Worked Example: A Uniform Distribution on a Bounded Interval

Let $X$ be uniform on $[-1,1]$. Its support is contained in $[-1,1]$, so the bounded-support theorem guarantees that its moment sequence determines its distribution among all probability distributions on the real line.

For clarity, its odd moments are zero by symmetry, and for integers $n\geq0$, $E[X^{2n}]=\frac12\int_{-1}^1x^{2n}\,dx=1/(2n+1)$. The bounded-support argument applies to the entire sequence, not just to a finite number of these values. In this case Carleman’s sum also diverges, since $(1/(2n+1))^{-1/(2n)}=(2n+1)^{1/(2n)}\to1$, so its terms do not tend to zero.

Worked Example: What a Failed Test Does Not Tell You

Let $X$ be lognormal with $X=e^Z$, where $Z\sim N(0,1)$. Completing the square in the normal integral gives $E[X^k]=E[e^{kZ}]=e^{k^2/2}$ for every nonnegative integer $k$. In particular, $m_{2n}=e^{2n^2}$, and the $n$th Carleman term is $$ m_{2n}^{-1/(2n)}=e^{-n}. $$ The sum $\sum_{n=1}^\infty e^{-n}$ converges, so Carleman’s criterion does not certify determinacy here.

The MGF condition also fails: for $t>0$, $E[e^{tX}]=E[e^{te^Z}]=\infty$. Indeed, in the normal-density integral the integrand is proportional to $\exp(te^z-z^2/2)$, which does not have a finite integral as $z\to\infty$. But failure of the two sufficient tests alone would not establish that this distribution is indeterminate. The lognormal distribution is in fact moment indeterminate, as shown by explicit alternative distributions with the same moments. This example is a reminder to separate a test’s failure from a conclusion the test does not provide.

How to Use the Criteria

When a complete moment sequence is available, first check whether the MGF is finite on an open interval around zero. If it is, the MGF theorem gives determinacy. If not, check whether the even moments satisfy Carleman’s divergence condition. If that sum diverges, determinacy follows. A bounded-support argument can also be useful when support information is known.

These tests are particularly useful because they do not require constructing a competing distribution and ruling it out one at a time. However, a convergent Carleman sum does not imply indeterminacy, and the absence of an MGF near zero does not imply indeterminacy either. Other arguments may still establish uniqueness. Conversely, matching only a finite set of moments never establishes moment determinacy: the definition concerns equality of every moment.

Key takeaway: An MGF finite near zero and Carleman’s divergent sum are sufficient conditions for moment determinacy on the real line. They are not necessary conditions, so failure to meet them leaves the question open.

Check Your Understanding

Use the definitions and criteria to answer the following questions.

What does it mean for a distribution to be moment determinate?
Why does a competing distribution with the same moments as a variable whose MGF is finite near zero also have an MGF finite near zero on a possibly smaller interval?
State Carleman’s criterion for a distribution on the real line. What conclusion can you draw if its sum converges?
For $X\sim N(0,1)$, the even moments are $m_{2n}=(2n)!/(2^n n!)$. What is the asymptotic order of the Carleman terms, and does their sum diverge?
If a distribution is supported on $[-K,K]$, why must any distribution with the same moments also be supported there?
The lognormal distribution fails both the MGF test and Carleman’s divergence test. Does that fact alone prove moment indeterminacy? Explain.