Pareto Thresholds And Nonexistent Moments

Why Pareto Thresholds Need a Moment Check

A threshold probability can be straightforward to calculate even when the expected loss is infinite. That distinction matters in heavy-tailed models: the chance of exceeding a specified amount may be finite and well defined, while a requested mean or variance may not exist. Before using an expectation formula, first identify the Pareto parameters and check the shape parameter against the order of the moment.

This tutorial uses the Pareto Type I convention, with a positive scale parameter and a positive shape parameter. Other Pareto conventions exist, so the stated model matters. The examples are hypothetical probability calculations; no interest rate or mortality assumption is relevant.

Definition: A random variable $X$ has a Pareto Type I distribution with scale $\theta>0$ and shape $\alpha>0$ if $X\geq\theta$ and its distribution satisfies $$ F_X(x)= \begin{cases} 0, & x<\theta,\\ 1-\left(\frac{\theta}{x}\right)^\alpha, & x\geq\theta. \end{cases} $$ Its survival function and density are $$ S_X(x)=P(X>x)=\left(\frac{\theta}{x}\right)^\alpha,\qquad f_X(x)=\frac{\alpha\theta^\alpha}{x^{\alpha+1}},\qquad x\geq\theta. $$

The scale $\theta$ is the smallest possible value of $X$; it is not the mean. The shape $\alpha$ controls how quickly the tail decreases. For any threshold $t\geq\theta$, the probability of exceeding that threshold follows directly from the survival function. If $t<\theta$, the probability of exceeding $t$ is 1. Since this model is continuous, $P(X>t)=P(X\geq t)$.

Threshold Probabilities and Tail Expectations

For a threshold $t\geq\theta$, the first calculation is

P(X>t)=\left(\frac{\theta}{t}\right)^\alpha.

A different quantity is the expected contribution from values above $t$, written $E[X\mathbf{1}_{\{X>t\}}]$, where the indicator is 1 when $X>t$ and 0 otherwise. It is not the same as the conditional expectation $E[X\mid X>t]$. When $\alpha>1$, both quantities are finite and related by

E[X\mathbf{1}_{\{X>t\}}] =\frac{\alpha\theta^\alpha}{\alpha-1}t^{1-\alpha}, \qquad E[X\mid X>t] =\frac{\alpha}{\alpha-1}t.

The first expression is an unconditional expected contribution, so it includes the probability of reaching the tail. The second is the average value among outcomes already known to exceed the threshold. The conditional formula requires both $P(X>t)>0$ and a finite tail expectation; under this Pareto model, the latter requires $\alpha>1$.

Derivation: For $t\geq\theta$, use the density to integrate over the tail. If $\alpha>1$, then $$ E[X\mathbf{1}_{\{X>t\}}] =\int_t^\infty x\frac{\alpha\theta^\alpha}{x^{\alpha+1}}\,dx =\alpha\theta^\alpha\int_t^\infty x^{-\alpha}\,dx =\frac{\alpha\theta^\alpha}{\alpha-1}t^{1-\alpha}. $$ Dividing by $P(X>t)=(\theta/t)^\alpha$ gives $E[X\mid X>t]=\alpha t/(\alpha-1)$. If $\alpha\leq1$, the integral of $x^{-\alpha}$ to infinity diverges, so these tail expectations are infinite rather than given by the finite formula.

For a capped value $\min(X,u)$, where $u\geq\theta$, the expectation is finite for every $\alpha>0$, even if $E[X]$ is infinite. The cap restricts the value being averaged. One useful identity is

E[\min(X,u)]=\int_0^u S_X(x)\,dx =\theta+\int_\theta^u\left(\frac{\theta}{x}\right)^\alpha dx.

The identity follows because, for a nonnegative random variable $Z$, $E[Z]=\int_0^\infty P(Z>x)\,dx$; for $Z=\min(X,u)$, its survival probability is zero above $u$ and equals $S_X(x)$ below $u$. This integral is bounded by $u$, so it remains finite regardless of whether the uncapped mean exists. For $\alpha\neq1$, it evaluates to

E[\min(X,u)] =\theta+\theta^\alpha\frac{u^{1-\alpha}-\theta^{1-\alpha}}{1-\alpha}.

At $\alpha=1$, the integral instead gives $E[\min(X,u)]=\theta+\theta\ln(u/\theta)$. The formula for $\alpha\neq1$ must not be used by substituting $\alpha=1$; the logarithmic case is the appropriate evaluation.

Which Pareto Moments Exist?

For a positive order $r$, the raw moment is the integral of $x^r f_X(x)$ over the support. Evaluating it gives a finite result only when the resulting improper integral converges:

E[X^r]= \begin{cases} \displaystyle\frac{\alpha\theta^r}{\alpha-r}, & \alpha>r,\\[6pt] \infty, & 0<\alpha\leq r. \end{cases}

The condition $\alpha>r$ is part of the formula, not an optional technical detail. At the boundary $\alpha=r$, the integral contains $x^{-1}$ and diverges logarithmically; below that boundary it diverges as a power. In particular, the mean exists exactly when $\alpha>1$. The variance exists exactly when $\alpha>2$, and when it exists it is

E[X]=\frac{\alpha\theta}{\alpha-1},\qquad \operatorname{Var}(X)=\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)}.

When $\alpha\leq1$, it is incorrect to calculate a finite variance by subtracting squared moments: the mean itself is infinite. When $1<\alpha\leq2$, the mean is finite but the second raw moment, and hence the variance, is infinite. A finite probability or a finite capped expectation does not make the full mean or variance finite.

Applying the Threshold and Moment Checks

Worked Example: A High Threshold With Infinite Variance

A hypothetical portfolio has a loss $X$, measured in dollars, with a Pareto Type I distribution, scale $\theta=\$1{,}000$, and shape $\alpha=2$. Calculate the probability that the loss exceeds $\$4{,}000$, the expected contribution from losses above that threshold, and the conditional mean given that the threshold is exceeded. Determine whether the variance exists.

Method and model. The loss is continuous and follows the stated Pareto Type I model. The threshold $t=\$4{,}000$ is at least the scale. First calculate the tail probability:

P(X>4{,}000)=\left(\frac{1{,}000}{4{,}000}\right)^2 =\left(\frac14\right)^2 =\frac1{16} =0.0625.

Since $\alpha=2>1$, the tail expectation formula applies:

E[X\mathbf{1}_{\{X>4{,}000\}}] =\frac{2(1{,}000)^2}{2-1}(4{,}000)^{-1} =\frac{2{,}000{,}000}{4{,}000} =\$500.

The conditional mean is the tail contribution divided by the tail probability:

E[X\mid X>4{,}000] =\frac{\$500}{1/16} =\$8{,}000.

As a check, the direct conditional formula gives $\frac{2}{2-1}\times\$4{,}000=\$8{,}000$, matching the quotient. The mean exists because $\alpha>1$, and it equals $2(1{,}000)/(2-1)=\$2{,}000$. The variance does not exist as a finite number because $\alpha=2$ does not satisfy the required strict condition $\alpha>2$. The common mistake avoided here is treating the boundary $\alpha=2$ as sufficient for finite variance.

Worked Example: Tail Contribution and a Capped Loss

A hypothetical loss $X$ is measured in thousands of dollars and has a Pareto Type I distribution with scale $\theta=2$ and shape $\alpha=1.5$. At a threshold $t=8$, calculate $P(X>8)$, $E[X\mathbf{1}_{\{X>8\}}]$, $E[X\mid X>8]$, and $E[\min(X,8)]$. State which full moments exist.

Method and model. All values are in thousands of dollars. The mean exists because $1.5>1$; the variance does not because $1.5\leq2$. The threshold ratio is $2/8=1/4$, so

P(X>8)=\left(\frac{2}{8}\right)^{1.5} =\left(\frac14\right)^{3/2} =\frac18 =0.1250.

The unconditional tail expectation is

E[X\mathbf{1}_{\{X>8\}}] =\frac{1.5(2)^{1.5}}{1.5-1}(8)^{-0.5} =\frac{3}{0.5}\frac{\sqrt{2}}{\sqrt{8}} =6\left(\frac12\right) =3.

Thus the tail contributes an expected 3 thousand dollars. Conditioning on exceeding 8 gives

E[X\mid X>8]=\frac{3}{1/8}=24,

or 24 thousand dollars. To calculate the capped expectation, use the finite full mean $E[X]=1.5(2)/(1.5-1)=6$ and split it at 8:

E[\min(X,8)] =E[X\mathbf{1}_{\{X\leq8\}}]+8P(X>8) =(6-3)+8\left(\frac18\right) =3+1=4.

The result is 4 thousand dollars, which is between the minimum possible loss of 2 and the cap of 8. The calculation checks because the below-threshold contribution is $6-3=3$, and the capped tail adds 1. A common mistake is to report 3 as the capped expectation; that is only the expected loss contribution strictly above the threshold, not the contribution below the threshold plus the capped amount above it.

Worked Example: A Finite Cap When the Mean Is Infinite

A hypothetical claim severity $X$, measured in thousands of dollars, has a Pareto Type I distribution with scale $\theta=4$ and shape $\alpha=0.5$. Calculate $P(X>16)$, $E[X\mathbf{1}_{\{X\leq16\}}]$, and $E[\min(X,16)]$. Explain why the uncapped mean cannot be used to find the capped expectation.

Method and model. Here $\alpha=0.5\leq1$, so the full mean and the expected contribution from the unbounded upper tail are infinite. The threshold probability is nevertheless well defined:

P(X>16)=\left(\frac4{16}\right)^{0.5} =\sqrt{\frac14} =\frac12 =0.5000.

The truncated expectation below the cap is a bounded integral:

\begin{aligned} E[X\mathbf{1}_{\{X\leq16\}}] &=\int_4^{16}x\frac{0.5(4)^{0.5}}{x^{1.5}}\,dx\\ &=\int_4^{16}x^{-0.5}\,dx\\ &=2\left(\sqrt{16}-\sqrt4\right)\\ &=2(4-2)=4. \end{aligned}

The values are in thousands of dollars, so this truncated expectation is 4 thousand dollars. Above the cap, the capped value is 16, and the probability of reaching that region is $1/2$. Therefore,

E[\min(X,16)] =E[X\mathbf{1}_{\{X\leq16\}}]+16P(X>16) =4+16\left(\frac12\right) =12.

The check $4\leq12\leq16$ is consistent with a nonnegative capped loss that never exceeds 16. The uncapped mean is infinite, but the cap makes the averaged quantity bounded and finite. The common mistake avoided is substituting an infinite mean into a capped-loss calculation instead of using the truncated part and the probability above the cap.

Common Exam Trap: A Formula Without Its Condition

Suppose a Pareto Type I variable has $\theta=3$ and $\alpha=0.8$. A tempting but invalid calculation for its mean is

\frac{\alpha\theta}{\alpha-1} =\frac{0.8(3)}{0.8-1} =-12.

A negative mean is impossible here because $X\geq3$. The problem is not a rounding error: the formula was derived only for $\alpha>1$, and that condition fails. The correct conclusion is $E[X]=\infty$, because the integral for the mean diverges. By contrast, for any finite cap $u\geq3$, $E[\min(X,u)]$ is finite. Always record the parameter condition before evaluating a moment formula.

Key takeaway: For a Pareto Type I variable, threshold probabilities are determined by $(\theta/t)^\alpha$, but moment existence is determined by comparing the shape $\alpha$ with the moment order. A finite tail probability or capped expectation does not imply a finite uncapped moment.

Check Your Understanding

For each problem, identify the relevant parameter condition before applying a formula. Unless stated otherwise, use the Pareto Type I convention defined above.

A hypothetical severity has scale $\theta=3$, shape $\alpha=3$, and is measured in dollars. Find $P(X>9)$ and $E[X\mid X>9]$. (Answer: $0.0370$; $\$13.50$)
A hypothetical loss in thousands of dollars has scale 5 and shape 2.5. Find the probability it exceeds 20 and the conditional mean given that it does. (Answer: $0.0313$; $33.333$ thousand dollars, rounded to 3 decimals)
A hypothetical variable has scale 1 and shape 0.5. Find $P(X>9)$ and $E[\min(X,9)]$. (Answer: $0.3333$; 5)
A hypothetical loss has scale 2 and shape 3. Find $E[X^2]$, $\operatorname{Var}(X)$, and $P(X>4)$. (Answer: $12$; $3$; $0.1250$)
A hypothetical severity has scale 4 and shape 1. Find $P(X>16)$, $E[X\mathbf{1}_{\{X>16\}}]$, and $E[\min(X,16)]$. (Answer: $0.2500$; infinite; $4+4\ln(4)$, approximately $9.545$)

What You'll Learn