Lp Moments From Distribution Functions

Moments as Integrals of Tail Sizes

A positive power moment records the size of a nonnegative function, but it can be useful to compute that moment without integrating the function directly. Instead, one can measure the sets on which the function exceeds each threshold. For $p>0$, the contribution of a threshold $t$ is weighted by $p t^{p-1}$. This gives a formula for the moment in terms of the tail sizes $t\mapsto\mu(\{f>t\})$, and it remains valid when the moment is infinite.

Throughout the main result, $(X,\mathcal{M},\mu)$ is a sigma-finite measure space, and $f:X\to[0,\infty]$ is measurable. The exponent $p$ is a real number with $p>0$. The integral $\int_X f^p\,d\mu$ is the nonnegative Lebesgue integral, allowed to have value $+\infty$. Write $D_f(t)=\mu(\{x\in X:f(x)>t\})$ for $t>0$; this tail size is allowed to be infinite. The integral in the threshold variable uses Lebesgue measure on $(0,\infty)$.

Theorem (Power-moment formula): Under these assumptions, $$ \int_X f^p\,d\mu = p\int_0^\infty t^{p-1}\,\mu(\{x\in X:f(x)>t\})\,dt. $$ Both sides are nonnegative extended real numbers, and equality includes the possibility that both are infinite.

The strict inequality in $\{f>t\}$ fixes the tail convention. Replacing it by $\{f\ge t\}$ gives the same integral: for each fixed $x$, the two threshold indicators differ at most at the single threshold $t=f(x)$, which has Lebesgue measure zero. The strict-tail version will be used below.

Proof. For each $x\in X$, integrate the power function over the thresholds below $f(x)$. If $f(x)<\infty$, the elementary integral gives $$ \int_0^\infty p t^{p-1}\chi_{\{t<f(x)\}}\,dt =\int_0^{f(x)}p t^{p-1}\,dt =f(x)^p. $$ When $f(x)=0$, both sides are zero. When $f(x)=\infty$, the integral over $(0,\infty)$ is infinite because $\int_0^R p t^{p-1}\,dt=R^p$ for every $R>0$; thus the identity also holds with value $+\infty$.

We can apply Tonelli's theorem to the function $H(x,t)=p t^{p-1}\chi_{\{t<f(x)\}}$ on $X\times(0,\infty)$. To check product measurability, its indicator set can be written $$ \{(x,t):0<t<f(x)\} = \bigcup_{q\in\mathbb{Q},\,q>0}\{x:f(x)>q\}\times(0,q). $$ Indeed, if $0<t<f(x)$, there is a positive rational $q$ strictly between $t$ and $f(x)$; the reverse inclusion follows from $t<q<f(x)$. Every set in this countable union belongs to $\mathcal{M}\otimes\mathcal{B}_{(0,\infty)}$, so $H$ is measurable and nonnegative. The two spaces are sigma-finite, and Tonelli's theorem therefore permits either order of integration, including when the value is infinite. Integrating first in $t$ gives $\int_X f^p\,d\mu$. Integrating first in $x$ gives $p\int_0^\infty t^{p-1}\mu(\{f>t\})\,dt$, since the section at threshold $t$ is exactly $\{f>t\}$. The two expressions are equal. $\square$

For a nonnegative random variable $Y:\Omega\to[0,\infty]$ on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, this reads $$ \mathbb{E}[Y^p] = p\int_0^\infty t^{p-1}\mathbb{P}(Y>t)\,dt. $$ The statement holds for every $p>0$, not only for integer powers. In particular, a finite $p$-th moment is equivalent to finiteness of the weighted tail integral on the right.

Computing Moments from Tail Measures

Worked Example: A Moment on a Three-Point Space

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be the collection of all subsets of $X$, and let $\mu$ be counting measure. Define $f(a)=3$, $f(b)=2$, and $f(c)=0$. Compute the second moment using the tail formula.

For $0\le t<2$, both $a$ and $b$ satisfy $f>t$, so $D_f(t)=2$. At $t=2$, only $a$ satisfies the strict inequality; for $2\le t<3$, the same is true, so $D_f(t)=1$. For $t\ge3$, no point satisfies $f>t$, and $D_f(t)=0$. Values at the endpoints do not affect a Lebesgue integral in $t$. With $p=2$,

\begin{aligned} \int_X f^2\,d\mu &=2\int_0^\infty t\,D_f(t)\,dt\\ &=2\int_0^2 2t\,dt+2\int_2^3 t\,dt\\ &=4\left[\frac{t^2}{2}\right]_0^2+\left[t^2\right]_2^3\\ &=4\cdot2+(9-4)\\ &=13. \end{aligned}

Check: counting measure gives the direct sum $f(a)^2+f(b)^2+f(c)^2=3^2+2^2+0^2=9+4+0=13$, agreeing with the tail calculation.

Worked Example: Moments of an Exponential Function

On $X=[0,\infty)$ with Lebesgue measure $m$, let $f(x)=e^{-x}$. Find $\int_X f^p\,dm$ for any $p>0$ from the level-set measures.

If $0<t<1$, then $e^{-x}>t$ is equivalent to $x<-\ln t$. Within $X$, the level set is $[0,-\ln t)$, of measure $-\ln t$. If $t\ge1$, the level set is empty: at $t=1$, even $f(0)=1$ does not satisfy the strict inequality. Thus $D_f(t)=-\ln t$ on $(0,1)$ and $D_f(t)=0$ on $[1,\infty)$. The formula gives $$ \int_0^\infty e^{-px}\,dx =p\int_0^1 t^{p-1}(-\ln t)\,dt. $$

To evaluate the right side, put $u=-\ln t$, so that $t=e^{-u}$ and $dt=-e^{-u}\,du$. As $t$ increases from $0$ to $1$, $u$ decreases from $+\infty$ to $0$, and hence $$ \int_0^1 t^{p-1}(-\ln t)\,dt =\int_0^\infty u e^{-pu}\,du. $$ Integration by parts on $[0,R]$ gives $$ \int_0^R u e^{-pu}\,du =-\frac{R e^{-pR}}{p}+\frac{1-e^{-pR}}{p^2}. $$ As $R\to\infty$, $e^{-pR}\to0$ and $R e^{-pR}\to0$, since $p>0$. The integral is therefore $1/p^2$, and multiplying by $p$ gives $1/p$. Check: direct integration yields $\int_0^\infty e^{-px}\,dx=[-e^{-px}/p]_0^\infty=1/p$, the same value.

Worked Example: A Pareto Probability Distribution

On $X=(1,\infty)$ with Lebesgue measure $m$, define a measure $\mu$ by the density $2x^{-3}$: $\mu(A)=\int_A2x^{-3}\,dm$ for Lebesgue-measurable $A\subseteq X$. This is a probability measure because $$ \mu(X)=\int_1^\infty 2x^{-3}\,dx =\left[-x^{-2}\right]_1^\infty =0-(-1)=1. $$ Let $f(x)=x$. Find its $p$-th moment and determine precisely when it is finite.

For $0<t\le1$, every $x\in X$ satisfies $x>t$, so $D_f(t)=1$. For $t>1$, $$ D_f(t)=\mu((t,\infty)) =\int_t^\infty 2x^{-3}\,dx =\left[-x^{-2}\right]_t^\infty =t^{-2}. $$ The single threshold $t=1$ does not affect the integral. For $0<p<2$, the power-moment formula now gives $$ \begin{aligned} \int_X x^p\,d\mu &=p\int_0^1 t^{p-1}\,dt+p\int_1^\infty t^{p-3}\,dt\\ &=p\left[\frac{t^p}{p}\right]_0^1 +p\left[\frac{t^{p-2}}{p-2}\right]_1^\infty\\ &=1+\frac{p}{2-p} =\frac{2}{2-p}. \end{aligned} $$ The second integral converges here because $p-2<0$, so $t^{p-2}\to0$ as $t\to\infty$. At $p=2$, its tail is $2\int_1^\infty t^{-1}\,dt=\infty$; for $p>2$, the integral of $t^{p-3}$ also diverges at infinity. Thus the moment is finite exactly when $0<p<2$.

Check: direct integration gives $\int_1^\infty x^p\,2x^{-3}\,dx=2\int_1^\infty x^{p-3}\,dx$. For $p<2$, this equals $2[\,x^{p-2}/(p-2)\,]_1^\infty=2/(2-p)$; for $p\ge2$, it diverges. This independently verifies both the value and the finiteness condition.

A Tail Bound from a Finite Moment

Proposition (Moment tail bound): Let $(X,\mathcal{M},\mu)$ be a measure space, let $f:X\to[0,\infty]$ be measurable, and let $p>0$. For every $a>0$, $$ \mu(\{f>a\})\le \frac{1}{a^p}\int_X f^p\,d\mu. $$ The integral may be infinite, in which case the inequality is understood in the extended nonnegative reals. In particular, on a probability space it gives $\mathbb{P}(f>a)\le a^{-p}\mathbb{E}[f^p]$.

Proof. On $E=\{f>a\}$, the pointwise inequality $f^p\ge a^p$ holds. Therefore $f^p\ge a^p\chi_E$ on all of $X$, because both sides are nonnegative and the right side is zero outside $E$. Monotonicity of the nonnegative Lebesgue integral gives $$ \int_X f^p\,d\mu\ge a^p\int_X\chi_E\,d\mu=a^p\mu(E). $$ Since $a^p>0$, division yields the stated bound. This argument also covers $\mu(E)=\infty$: in that case the displayed inequality forces $\int_X f^p\,d\mu=\infty$, and the claimed extended-real inequality holds. $\square$

Common Pitfall: Check the Total Mass First

A density that resembles a probability density is not automatically normalized, and its domain matters. For example, the density $2x^{-3}$ on $(0,\infty)$ does not define a probability measure: its total mass is infinite, because $$ \int_0^\infty 2x^{-3}\,dx \ge \int_0^1 2x^{-3}\,dx =\infty. $$ The antiderivative $-x^{-2}$ also shows the divergence at the lower endpoint. Restricting the same density to $(1,\infty)$ gives total mass $1$, as calculated in the Pareto example. Using a probability tail formula before checking this normalization would assign the wrong meaning to the tail sizes.

Another useful check is that a tail moment calculation has both a weight and a factor of $p$. For a fixed finite value $c\ge0$, the relevant identity is $c^p=\int_0^c p t^{p-1}\,dt$. Omitting $p$ would give $c^p/p$, not $c^p$. The theorem's proof identifies exactly where the factor enters, and the direct computations in the examples provide independent checks.

Check Your Understanding

In each question, specify the level-set measure, including the strict threshold convention, before evaluating the moment.

On a two-point space with counting measure, let $f$ take values $4$ and $1$. Use the tail formula to compute $\int f^3\,d\mu$.
On $[0,\infty)$ with Lebesgue measure, let $g(x)=e^{-2x}$. Find $\int_0^\infty g(x)^p\,dx$ for $p>0$ using its level-set measures.
For the probability density $2x^{-3}$ on $(1,\infty)$, compute the first moment and determine whether the second moment is finite.
Prove directly from monotonicity of the integral that $\mu(\{f>a\})\le a^{-p}\int f^p\,d\mu$ for $a>0$, and identify where strictness of the level set is used.
Explain why $2x^{-3}$ on $(0,\infty)$ has infinite total mass, while its restriction to $(1,\infty)$ has total mass one.