Tail Integrals and Moment Existence

Why Tail Decay Determines Moments

A moment such as $E[|X|^r]$ weights large values of $X$ heavily when $r$ is large. Its existence therefore depends not only on how likely extreme values are, but on how quickly their probabilities decrease. The tail-integral formula makes this relationship explicit: instead of integrating the values of a random variable directly, it integrates the probabilities that the variable exceeds a threshold.

This approach works for discrete, continuous, and mixed distributions. It requires no density or mass function, which makes it useful when the tail probability is easier to describe than the full distribution.

Definition: For a random variable $X$ and $r>0$, the $r$th absolute moment is the possibly infinite expectation $E[|X|^r]$. If $X\geq 0$ almost surely, this is $E[X^r]$. For $t\geq 0$, the tail probability of a nonnegative random variable $Y$ is $P(Y>t)$.

The Tail-Integral Formula

The formula follows by writing each nonnegative number as the area under a simple indicator function. For $y\geq 0$ and $r>0$, $y^r=\int_0^\infty r t^{r-1}\mathbf{1}_{\{t<y\}}\,dt$. Indeed, the indicator equals one from $t=0$ to $t=y$, so the integral is $\int_0^y r t^{r-1}\,dt=y^r$.

Theorem: If $Y$ is a nonnegative random variable and $r>0$, then, allowing either side to be infinite, $$ E[Y^r]=r\int_0^\infty t^{r-1}P(Y>t)\,dt. $$ Consequently, for any real-valued random variable $X$, $$ E[|X|^r]=r\int_0^\infty t^{r-1}P(|X|>t)\,dt. $$

Proof. Apply the pointwise identity to $Y$ and take expectations. The function $r t^{r-1}\mathbf{1}_{\{t<Y\}}$ is nonnegative and jointly measurable in the outcome and $t$. Tonelli’s theorem therefore permits exchanging expectation and integration, even if the result is infinite:

\begin{aligned} E[Y^r] &=E\left[\int_0^\infty r t^{r-1}\mathbf{1}_{\{t<Y\}}\,dt\right]\\ &=\int_0^\infty r t^{r-1}E[\mathbf{1}_{\{t<Y\}}]\,dt\\ &=r\int_0^\infty t^{r-1}P(Y>t)\,dt. \end{aligned}

The final equality uses $E[\mathbf{1}_{\{t<Y\}}]=P(Y>t)$. Taking $Y=|X|$ gives the absolute-moment formula. The argument proves equality as an identity of extended nonnegative quantities, so it does not assume in advance that the moment is finite. $\square$

In particular, a nonnegative random variable has a finite $r$th moment exactly when the corresponding tail integral is finite. For a signed variable, the same test applies to its absolute moment, which is the relevant criterion for absolute integrability of powers.

Power-Law Tails and the Moment Threshold

A common situation is that the tail probability decreases approximately like a power of $t$. The following criterion turns that decay rate into a direct moment test.

Theorem: Let $Y\geq 0$, $r>0$, and $\alpha>0$. Suppose that for some $t_0>0$ and constants $c_1,c_2>0$, $$ c_1t^{-\alpha}\leq P(Y>t)\leq c_2t^{-\alpha} \qquad (t\geq t_0). $$ Then $E[Y^r]<\infty$ if and only if $r<\alpha$.

Proof. On the bounded interval $(0,t_0)$, the tail probability is at most one, and $\int_0^{t_0} r t^{r-1}\,dt=t_0^r<\infty$. Thus only the integral over $[t_0,\infty)$ can determine finiteness. On that interval, the theorem’s bounds compare the tail integral above and below with positive constant multiples of $\int_{t_0}^\infty t^{r-1-\alpha}\,dt$. This power integral converges exactly when $r-1-\alpha<-1$, or $r<\alpha$. If $r\geq\alpha$, the lower bound forces divergence; if $r<\alpha$, the upper bound ensures convergence. The tail-integral formula gives the claimed equivalence. $\square$

Worked Example: Moments of a Pareto Random Variable

Let $Y$ have the Pareto distribution with parameter $\alpha>0$ and support $y\geq 1$, with density $f(y)=\alpha y^{-\alpha-1}$. Its tail is $P(Y>t)=1$ for $0\leq t<1$, and $P(Y>t)=t^{-\alpha}$ for $t\geq 1$. Determine when its $r$th moment exists and calculate it when finite.

For $r>0$, the tail-integral formula gives

\begin{aligned} E[Y^r] &=r\int_0^1 t^{r-1}\,dt +r\int_1^\infty t^{r-1}t^{-\alpha}\,dt\\ &=1+r\int_1^\infty t^{r-\alpha-1}\,dt. \end{aligned}

The first integral is one. The second converges if and only if $r-\alpha-1<-1$, that is, $r<\alpha$. When $r<\alpha$, its value is $1/(\alpha-r)$, so

E[Y^r]=1+\frac{r}{\alpha-r}=\frac{\alpha}{\alpha-r}.

For $r=\alpha$, the second integral is $\alpha\int_1^\infty t^{-1}\,dt$, which diverges logarithmically. For $r>\alpha$, it also diverges. Thus this distribution has a finite $r$th moment exactly below the tail exponent.

Faster Tails and All Positive Moments

When the tail decreases faster than every power, every positive moment can be finite. A Weibull tail gives a direct calculation and illustrates how a change of variables evaluates the tail integral.

Worked Example: Moments of a Weibull-Tailed Variable

Let $Y\geq0$ have survival function $P(Y>t)=e^{-t^2}$ for $t\geq0$. Find $E[Y^r]$ for $r>0$ and determine whether any positive moment is infinite.

The tail formula yields

E[Y^r]=r\int_0^\infty t^{r-1}e^{-t^2}\,dt.

Set $u=t^2$, so $dt=\frac12u^{-1/2}du$ and $t^{r-1}=u^{(r-1)/2}$. Therefore

E[Y^r]=\frac r2\int_0^\infty u^{r/2-1}e^{-u}\,du =\frac r2\Gamma(r/2) =\Gamma(1+r/2).

Here $\Gamma(a)=\int_0^\infty u^{a-1}e^{-u}\,du$ for $a>0$, and the last equality follows from $\Gamma(a+1)=a\Gamma(a)$, obtained by integration by parts. Since this gamma integral is finite for every $r/2>0$, every positive moment exists. For example, at $r=2$, the formula gives $E[Y^2]=\Gamma(2)=1$; directly, $2\int_0^\infty t e^{-t^2}\,dt=1$ as well.

Logarithmic Factors at the Boundary

Power comparisons settle the question when the tail is bounded above and below by the same power. At the boundary $r=\alpha$, however, a slowly varying factor can change the answer. Logarithmic factors provide a simple example.

Worked Example: A Logarithmic Factor Changes the Boundary Moment

Define a nonnegative random variable $Y$ by $P(Y>t)=1$ for $0\leq t<e$ and $P(Y>t)=t^{-2}(\log t)^{-2}$ for $t\geq e$. This is a valid survival function: it decreases from one to zero, with a jump at $e$ corresponding to an atom there. Determine which positive moments exist.

For $r>0$, split the tail integral at $e$:

E[Y^r]=r\int_0^e t^{r-1}\,dt +r\int_e^\infty t^{r-1}t^{-2}(\log t)^{-2}\,dt =e^r+r\int_e^\infty t^{r-3}(\log t)^{-2}\,dt.

If $r<2$, the power factor $t^{r-3}$ makes the upper integral finite; for example, $(\log t)^{-2}\leq1$ on $t\geq e$, and $\int_e^\infty t^{r-3}\,dt<\infty$. If $r>2$, the integral diverges. To see this, for sufficiently large $t$, $(\log t)^2\leq t^{(r-2)/2}$, so the integrand is at least $t^{r-3-(r-2)/2}=t^{(r-4)/2}$, whose integral diverges for $r>2$.

At the boundary $r=2$, the tail contribution is

2\int_e^\infty \frac{dt}{t(\log t)^2} =2\int_1^\infty \frac{du}{u^2} =2,

where $u=\log t$. Thus $E[Y^2]=e^2+2<\infty$, even though a pure $t^{-2}$ tail would give an infinite second moment. The logarithmic factor changes the boundary case, but not the conclusions for $r<2$ and $r>2$.

Common Mistakes and Practical Use

Confusing a tail probability with a density. The integrand uses $P(Y>t)$, not the density at $t$. The formula applies even when no density exists.
Ignoring the weight $t^{r-1}$. A tail probability tending to zero does not by itself guarantee a finite moment. The rate of decrease must be compared with this weight.
Assuming the power-law boundary is always infinite. A tail comparable to $t^{-\alpha}$ gives divergence at $r=\alpha$, but a faster factor such as $(\log t)^{-2}$ can make that boundary moment finite.
Using a signed expectation when absolute integrability is the issue. For a general signed $X$, the tail formula tests $E[|X|^r]$. Finite positive and negative contributions cannot be used to justify a finite absolute moment if the absolute tail integral diverges.

The method matters because moment existence underlies many distributional summaries and calculations. It gives a precise way to check whether a proposed moment is finite before using it, and it works directly from tail probabilities without requiring a full density calculation. For a signed random variable, apply it to $|X|$; for a nonnegative variable, apply it directly to $X$.

Key takeaway: For $r>0$, $E[|X|^r]$ is finite exactly when $r\int_0^\infty t^{r-1}P(|X|>t)\,dt$ is finite. A tail of power order $t^{-\alpha}$ gives finite moments below $\alpha$ and infinite moments above it; the boundary requires closer inspection.

Check Your Understanding

Use the tail-integral formula and justify each conclusion.

If $P(Y>t)$ is bounded above and below by positive constant multiples of $t^{-3}$ for all sufficiently large $t$, for which $r>0$ is $E[Y^r]$ finite?
Why does the tail-integral formula apply to a discrete random variable even when it has no density?
For the Weibull tail $P(Y>t)=e^{-t^2}$, what is $E[Y^4]$?
For a tail proportional to $t^{-2}(\log t)^{-1}$ for sufficiently large $t$, does the second moment exist? Explain using the integral at the boundary.
What tail probability should be used to test $E[|X|^r]$ when $X$ can be positive or negative?