Tutorials › Statistical Theory I - Probability and Distribution Theory › Moment-Indeterminate Lognormal Distributions

Home  /  Tutorials  /  Moments and Generating Functions

Moments and Generating Functions · Tutorial 517 of 994

Moment-Indeterminate Lognormal Distributions

Learn how a bounded sinusoidal perturbation creates distinct lognormal-type distributions with the same moments of every nonnegative integer order.

Intermediate 10 min read

What You'll Learn

  • Construct and validate a Stieltjes class based on the standard lognormal density
  • Derive the shared integer-moment sequence using a Gaussian integral
  • Show explicitly that different perturbation parameters give different distributions
  • Verify moment agreement for the endpoint members of the class
  • Explain why these distributions have no moment-generating function on an interval around zero

When Moments Do Not Identify a Distribution

A distribution can have finite moments of every nonnegative integer order and still not be determined by those moments. The lognormal distribution gives a concrete example: its moments grow rapidly, yet one can perturb its density in a controlled way to obtain distinct distributions with exactly the same moment sequence. The construction makes the source of the nonuniqueness explicit.

We begin with the standard lognormal distribution \(X\sim \operatorname{Lognormal}(0,1)\), so that \(\ln X\sim N(0,1)\). Write \(\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}\) for the standard normal density. The lognormal density is

$$ f_0(x)=\frac{1}{x}\phi(\ln x) =\frac{1}{x\sqrt{2\pi}}\exp\!\left(-\frac{(\ln x)^2}{2}\right), \qquad x>0. $$

The useful idea is to multiply this density by a bounded oscillating factor. The oscillation changes the distribution, but its integral against each power \(x^n\) will vanish.

Definition: For a real number \(\varepsilon\) with \(|\varepsilon|\leq 1\), define $$ f_\varepsilon(x)=f_0(x)\bigl[1+\varepsilon\sin(2\pi\ln x)\bigr], \qquad x>0, $$ and set \(f_\varepsilon(x)=0\) for \(x\leq 0\). The family \(\{f_\varepsilon:-1\leq\varepsilon\leq1\}\) is a Stieltjes class based on \(f_0\): its members are probability densities with the same nonnegative integer moments as \(f_0\).

Why These Functions Are Densities

The factor in square brackets is between \(0\) and \(2\), since \(-1\leq\sin(2\pi\ln x)\leq1\). Thus \(f_\varepsilon(x)\geq0\) for every \(x>0\). To verify that it integrates to one, make the change of variable \(z=\ln x\), for which \(dx/x=dz\). Then

$$ \int_0^\infty f_\varepsilon(x)\,dx =\int_{-\infty}^\infty \phi(z)\bigl[1+\varepsilon\sin(2\pi z)\bigr]\,dz =1+\varepsilon\int_{-\infty}^\infty\phi(z)\sin(2\pi z)\,dz =1. $$

The final integral is zero because \(\phi\) is even and \(z\mapsto\sin(2\pi z)\) is odd; the integrand is integrable, being bounded in absolute value by \(\phi(z)\). Therefore every \(f_\varepsilon\) is a probability density, including at \(\varepsilon=1\) and \(\varepsilon=-1\), where the density can equal zero at some points.

Worked Example: Checking the Member with \(\varepsilon=\frac12\)

For \(\varepsilon=\frac12\), the proposed density is $$ f_{1/2}(x)=\frac{1}{x\sqrt{2\pi}}\exp\!\left(-\frac{(\ln x)^2}{2}\right) \left[1+\frac12\sin(2\pi\ln x)\right], \qquad x>0. $$

Since \(-1\leq\sin(2\pi\ln x)\leq1\), its bracket lies between \(1/2\) and \(3/2\), so the function is nonnegative. Its integral is $$ \int_0^\infty f_{1/2}(x)\,dx =\int_{-\infty}^\infty\phi(z)\,dz +\frac12\int_{-\infty}^\infty\phi(z)\sin(2\pi z)\,dz =1+0=1. $$ Thus it is a probability density, and it is strictly positive wherever \(f_0\) is positive.

Deriving the Shared Integer Moments

The central calculation is the integral of the perturbation against \(x^n\). We first record the Gaussian integral needed for it. For real \(a\) and \(b\), completing the square in the real exponential gives

$$ e^{az}\phi(z)=e^{a^2/2}\phi(z-a). $$

Using the standard normal characteristic function, \(\int_{-\infty}^\infty e^{ibw}\phi(w)\,dw=e^{-b^2/2}\), and substituting \(w=z-a\), we obtain

$$ \int_{-\infty}^\infty e^{(a+ib)z}\phi(z)\,dz =e^{a^2/2}e^{iba}\int_{-\infty}^\infty e^{ibw}\phi(w)\,dw =\exp\!\left(\frac{(a+ib)^2}{2}\right). $$

Taking imaginary parts gives the particular real integral

$$ \int_{-\infty}^\infty e^{az}\phi(z)\sin(bz)\,dz =\exp\!\left(\frac{a^2-b^2}{2}\right)\sin(ab). $$

All integrals here are absolutely convergent: for fixed real \(a,b\), the absolute value of the integrand is at most \(e^{az}\phi(z)\), whose integral is \(e^{a^2/2}\).

Theorem: For every \(\varepsilon\in[-1,1]\) and every nonnegative integer \(n\), a random variable with density \(f_\varepsilon\) has moment $$ E[X^n]=\exp\!\left(\frac{n^2}{2}\right). $$ Consequently, every member of the class has the same moments as the standard lognormal distribution.

Proof. For a nonnegative integer \(n\), use \(z=\ln x\), so that \(x^n=e^{nz}\) and \(f_\varepsilon(x)\,dx=\phi(z)[1+\varepsilon\sin(2\pi z)]\,dz\). The moment is finite because $$ \int_{-\infty}^\infty e^{nz}\phi(z)\bigl|1+\varepsilon\sin(2\pi z)\bigr|\,dz \leq 2\int_{-\infty}^\infty e^{nz}\phi(z)\,dz =2e^{n^2/2}. $$ We can therefore split its integral into the unperturbed and perturbation terms:

$$ \begin{aligned} E[X^n] &=\int_{-\infty}^\infty e^{nz}\phi(z)\,dz +\varepsilon\int_{-\infty}^\infty e^{nz}\phi(z)\sin(2\pi z)\,dz\\ &=e^{n^2/2} +\varepsilon\exp\!\left(\frac{n^2-(2\pi)^2}{2}\right)\sin(2\pi n)\\ &=e^{n^2/2}. \end{aligned} $$

The last equality holds because \(n\) is an integer, so \(\sin(2\pi n)=0\). This includes \(n=0\), confirming normalization once more. \(\square\)

Worked Example: Comparing the First Three Moments

Take the member with \(\varepsilon=\frac12\). The theorem gives, for \(n=0,1,2\), $$ E[1]=e^0=1,\qquad E[X]=e^{1/2},\qquad E[X^2]=e^{4/2}=e^2. $$

For example, the first moment can also be checked directly by separating the two terms: $$ E[X]=\int_{-\infty}^\infty e^z\phi(z)\,dz +\frac12\int_{-\infty}^\infty e^z\phi(z)\sin(2\pi z)\,dz =e^{1/2}+\frac12e^{(1-4\pi^2)/2}\sin(2\pi) =e^{1/2}. $$ For the second moment the corresponding perturbation is $$ \frac12\int_{-\infty}^\infty e^{2z}\phi(z)\sin(2\pi z)\,dz =\frac12e^{(4-4\pi^2)/2}\sin(4\pi)=0, $$ so \(E[X^2]=e^2\). These calculations show how the oscillating factor disappears at integer powers, not that the density itself is unchanged.

Same Moments, Different Distributions

Having equal moments does not mean that the distributions are equal. If \(\varepsilon_1\ne\varepsilon_2\), their density difference on \(x>0\) is $$ f_{\varepsilon_1}(x)-f_{\varepsilon_2}(x) =(\varepsilon_1-\varepsilon_2)f_0(x)\sin(2\pi\ln x). $$ The sine factor is nonzero on open intervals, and \(f_0(x)>0\) throughout \(x>0\). Thus the densities differ on a set of positive length. The following example verifies a difference directly through event probabilities.

Worked Example: Distinguishing Two Members by an Interval Probability

Compare the standard lognormal member \(f_0\) with \(f_{1/2}\), using the event \(A=\{e^{1/12}<X<e^{5/12}\}\). In the coordinate \(z=\ln x\), this event is \(1/12<z<5/12\). On that interval, \(2\pi z\) lies between \(\pi/6\) and \(5\pi/6\), so \(\sin(2\pi z)>0\). The difference in probabilities is $$ P_{1/2}(A)-P_0(A) =\frac12\int_{1/12}^{5/12}\phi(z)\sin(2\pi z)\,dz>0. $$

The strict inequality follows because both factors in the integrand are strictly positive throughout the interval, which has positive length. This proves that \(f_{1/2}\) and \(f_0\) define different distributions even though all their nonnegative integer moments agree.

Worked Example: The Endpoint Densities Are Valid and Distinct

At the endpoints, \(f_1(x)=f_0(x)[1+\sin(2\pi\ln x)]\) and \(f_{-1}(x)=f_0(x)[1-\sin(2\pi\ln x)]\). Both are nonnegative because \(1+\sin(\cdot)\) and \(1-\sin(\cdot)\) lie between \(0\) and \(2\); both integrate to one by the normalization calculation. The moment theorem applies to each, so \(E[X^n]=e^{n^2/2}\) for both for every nonnegative integer \(n\).

They are not the same distribution. For the interval \(A\) in the previous calculation, $$ P_1(A)-P_{-1}(A) =2\int_{1/12}^{5/12}\phi(z)\sin(2\pi z)\,dz>0. $$ The integrand is strictly positive there. Thus the endpoint distributions are distinct as well as distinct from the unperturbed lognormal law.

Why the Moment-Generating Function Does Not Resolve This

For any member of this class, \(X>0\). If \(t\leq0\), then \(0<e^{tX}\leq1\), so \(E[e^{tX}]\) is finite. If \(t>0\), the expectation is infinite. For \(|\varepsilon|<1\), the bracket in \(f_\varepsilon\) is at least \(1-|\varepsilon|>0\); hence its moment-generating-function integral is bounded below by a positive constant times the standard lognormal integral, which diverges.

The endpoint cases require a separate argument because that uniform lower bound becomes zero. For \(\varepsilon=1\), take the disjoint intervals \(I_k=[k+1/12,k+5/12]\) in the coordinate \(z=\ln x\), for integers \(k\geq0\). The sine is nonnegative on each interval, so \(1+\sin(2\pi z)\geq1\) there. For \(\varepsilon=-1\), instead take \(J_k=[k+7/12,k+11/12]\); the sine is nonpositive there, so \(1-\sin(2\pi z)\geq1\). In either case, the contribution to \(E[e^{tX}]\) from the selected interval is at least

$$ \frac{1}{3\sqrt{2\pi}} \exp\!\left(t e^{k+c}-\frac{(k+d)^2}{2}\right), $$

where \(c\) is the left endpoint offset and \(d\) the right endpoint offset of that interval: \((c,d)=(1/12,5/12)\) for \(I_k\), and \((c,d)=(7/12,11/12)\) for \(J_k\). This bound uses \(e^{tz}\) in the transformed integral in its correct form \(e^{t e^z}\): on the interval, \(e^{t e^z}\geq e^{t e^{k+c}}\), while \(\phi(z)\geq(2\pi)^{-1/2}e^{-(k+d)^2/2}\). For \(t>0\), \(t e^{k+c}-(k+d)^2/2\) tends to \(+\infty\), since the exponential term dominates the quadratic term. The lower bounds therefore do not tend to zero, and summing the integrals over the disjoint intervals gives divergence. This establishes \(E[e^{tX}]=\infty\) at both endpoints as well.

Thus \(E[e^{tX}]\) is finite for \(t\leq0\) and infinite for \(t>0\). An MGF is defined on an open interval containing zero on which the expectation is finite; none of these distributions has such an interval. In particular, there is no MGF around zero whose derivatives could distinguish the members of the class.

Common Mistake and Key Takeaway

  • Mistake: assuming all moments determine a distribution. Having \(E[X^n]=e^{n^2/2}\) for every nonnegative integer \(n\) does not establish that \(X\) has the standard lognormal distribution. The explicitly different densities \(f_\varepsilon\) disprove that conclusion.
  • Mistake: applying an interior-parameter bound at the endpoints. The inequality \(1+\varepsilon\sin(2\pi\ln x)\geq1-|\varepsilon|\) proves a positive lower bound only when \(|\varepsilon|<1\). At \(\varepsilon=\pm1\), divergence of the positive-\(t\) exponential expectation instead follows by restricting to intervals where the relevant bracket is at least one.
  • Mistake: confusing moment agreement with density agreement. The perturbation integrates to zero against \(x^n f_0(x)\) for integer \(n\), but it does not vanish as a function. It changes probabilities of suitable intervals.
Key takeaway: The factors \(1+\varepsilon\sin(2\pi\ln x)\) preserve nonnegativity and normalization, while the oscillation has zero integral against every integer power under the lognormal base measure. This produces a continuum of distinct distributions sharing the moments \(E[X^n]=e^{n^2/2}\).

Check Your Understanding

Use the density construction and calculations above to answer the following.

  1. Why is \(f_\varepsilon\) nonnegative for every \(\varepsilon\in[-1,1]\), including the endpoints?
  2. What Gaussian integral makes the perturbation term in \(E[X^n]\) equal to a constant times \(\sin(2\pi n)\)?
  3. For \(n=3\), write the value of \(E[X^3]\) for every member of the class.
  4. Give an interval in the \(z=\ln x\) coordinate on which \(P_{1/2}(A)>P_0(A)\), and explain why the inequality is strict.
  5. Why does the lower bound \(1-|\varepsilon|\) not prove divergence of \(E[e^{tX}]\) when \(\varepsilon=1\) or \(\varepsilon=-1\)?

Previous

Next

Back to Tutorials