Existence Threshold for Pareto Moments

Why a Moment Threshold Matters

A distribution can have a probability density that integrates to one and still lack a finite mean or variance. For the Type I Pareto distribution, the decisive feature is its slowly decreasing density in the far right tail. Integrating that density against a power of the observation reveals a precise cutoff: the moment of order \(r\) is finite exactly when \(r\) is less than the tail-index parameter.

The cutoff is useful in practice because it identifies which familiar summaries are available. Depending on the parameter, a Pareto model can have a finite mean but no finite variance, or it can have neither. The calculation also makes an important distinction: divergence exactly at the cutoff is logarithmic, while above it the divergence is polynomial.

This is a probability density: since \(\alpha>0\),

Because \(x_m>0\), powers \(X^r\) are well-defined for every real \(r\). In many applications, “the \(r\)th moment” refers to a nonnegative integer \(r\); the integration result below applies more generally to every real order.

Deriving the Moment Formula

For a nonnegative random variable \(Y\), its expectation is defined as the integral of \(Y\) with respect to its distribution, with the value \(+\infty\) allowed. Since \(X^r\ge 0\), this definition applies for every real \(r\). Substituting the Pareto density reduces the question of moment existence to the convergence of an improper integral.

Proof. Using the density and the definition of expectation for a nonnegative function,

First suppose \(r<\alpha\). Then \(r-\alpha<0\), and for a finite upper endpoint \(R>x_m\),

As \(R\) tends to infinity, \(R^{r-\alpha}\) tends to zero because \(r-\alpha<0\). Therefore,

Now suppose \(r=\alpha\). The integrand in the last integral is \(x^{-1}\), so

Finally, if \(r>\alpha\), then \(r-\alpha>0\), and

The expectation therefore diverges both at \(r=\alpha\) and for \(r>\alpha\), and is finite for all \(r<\alpha\). This proves the stated equivalence and formula. \(\square\)

The lower endpoint being strictly positive matters when considering negative orders: there is no singularity near zero because \(X\ge x_m>0\). Thus every negative-order moment is finite. For example, if \(r=-1\), the formula gives \(E[X^{-1}]=\alpha/((\alpha+1)x_m)\).

Consequences for the Mean and Variance

The mean is the moment of order one, so its existence threshold is \(\alpha>1\). The variance needs more care. The standard variance \(\operatorname{Var}(X)=E[(X-E[X])^2]\) requires a finite mean; in the usual finite-variance setting, it is enough that the second moment is finite. For Pareto distributions, a finite second moment also guarantees a finite mean, because \(\alpha>2\) implies \(\alpha>1\).

Derivation. Taking \(r=1\) in the moment theorem gives, when \(\alpha>1\),

If \(0<\alpha\le 1\), the theorem says \(E[X]=\infty\). For the variance, first take \(r=2\). When \(\alpha>2\),

Both moments are finite in this case, so subtraction is valid:

For \(1<\alpha\le2\), the mean is finite, but \(E[X^2]=\infty\). The nonnegative quantity \((X-E[X])^2\) has infinite expectation as well: for \(X\ge 2E[X]\), it is at least \(X^2/4\), and the Pareto second-moment integral over that tail diverges. Thus the variance is infinite. When \(0<\alpha\le1\), the mean itself is not finite, so the standard variance about \(E[X]\) is not defined as a finite variance.

Common Mistake: Treating the Boundary as Finite

A frequent error is to infer that \(E[X^r]\) is finite whenever \(r\le\alpha\), perhaps because the power in the density seems to balance the power \(x^r\) at equality. At \(r=\alpha\), however, the powers combine to give an integrand proportional to \(1/x\), not a constant or a decaying integrable function. The integral of \(1/x\) over \([x_m,\infty)\) diverges.

A related mistake is to call the variance finite whenever the mean is finite. The mean tests the first moment; the variance requires control of the second moment. For \(1<\alpha\le2\), the first moment exists but the second does not. Conversely, when \(\alpha\le1\), it is more precise to say that no finite mean or standard finite variance exists than to use the variance formula with an infinite mean.