Delta-Method Uncertainty for Fitted Quantiles

Why a Fitted Quantile Has Estimation Uncertainty

A fitted severity quantile is often used to describe a high-loss threshold, set a hypothetical coverage feature, or compare candidate models. It is not known exactly when the model parameters have been estimated from data: changing those estimates changes the fitted quantile. The delta method translates the estimated uncertainty in the parameters into an approximate standard error for that quantile.

This tutorial concerns parameter-estimation uncertainty in a model-based quantile. It does not include the randomness of a future individual loss, uncertainty from model choice, or uncertainty caused by an incorrectly specified severity distribution. The severity models and covariance estimates below are hypothetical.

Definition: Let $X$ have a continuous severity distribution $F(x;\boldsymbol{\theta})$, with density $f(x;\boldsymbol{\theta})$ and parameter vector $\boldsymbol{\theta}$. For $0<p<1$, the $p$-quantile $q_p(\boldsymbol{\theta})$ satisfies $F(q_p(\boldsymbol{\theta});\boldsymbol{\theta})=p$. Given an estimator $\widehat{\boldsymbol{\theta}}$, the fitted quantile is $\widehat q_p=q_p(\widehat{\boldsymbol{\theta}})$.

For the usual smooth delta-method approximation, assume that the true parameter lies in a regular part of the model, that $F$ is differentiable in its arguments near the quantile, and that $f(q_p;\boldsymbol{\theta})>0$. Assume also that $\widehat{\boldsymbol{\theta}}$ is approximately unbiased at the relevant order and has covariance matrix $\boldsymbol{\Sigma}$, estimated from the data. If the density is zero at the quantile or the parameter estimate lies near a boundary, this approximation may be unreliable.

The Quantile Gradient and Its Approximate Variance

The quantile is defined implicitly rather than, in general, by a simple explicit formula. Differentiating its defining equation gives a direct way to find how it changes as each parameter changes.

Start with the quantile equation.
Write $F(q_p(\boldsymbol{\theta});\boldsymbol{\theta})=p$, holding $p$ fixed.

Differentiate with respect to a parameter.
For component $\theta_j$, the chain rule gives $f(q_p;\boldsymbol{\theta})\,\partial q_p/\partial\theta_j+\partial F(q_p;\boldsymbol{\theta})/\partial\theta_j=0$. The second derivative holds the loss value fixed when differentiating $F$ with respect to $\theta_j$.

Calculate the gradient and variance.
Evaluate the resulting gradient at the fitted parameters, then use the full covariance matrix to obtain the first-order variance approximation.

Theorem: If $f(q_p;\boldsymbol{\theta})>0$, implicit differentiation gives $$ \frac{\partial q_p}{\partial\theta_j} = -\frac{\partial F(q_p;\boldsymbol{\theta})/\partial\theta_j} {f(q_p;\boldsymbol{\theta})}. $$ Writing the gradient as $\nabla q_p$, the fitted-quantile variance and standard error are approximated by $$ \operatorname{Var}(\widehat q_p) \approx \nabla q_p(\widehat{\boldsymbol{\theta}})^{\mathsf T} \boldsymbol{\Sigma} \nabla q_p(\widehat{\boldsymbol{\theta}}), \qquad \operatorname{SE}(\widehat q_p) \approx \sqrt{\operatorname{Var}(\widehat q_p)}. $$

The reason for the variance formula is the first-order Taylor approximation $q_p(\widehat{\boldsymbol{\theta}})\approx q_p(\boldsymbol{\theta})+ \nabla q_p(\boldsymbol{\theta})^{\mathsf T}(\widehat{\boldsymbol{\theta}}-\boldsymbol{\theta})$. The variance of this linear combination is the gradient multiplied on both sides of the parameter covariance matrix. Thus the covariance between parameter estimates matters whenever the corresponding gradient components are nonzero. If the parameter-estimator covariance is instead supplied in asymptotic form $\mathbf V/n$, use $\mathbf V/n$ for $\boldsymbol{\Sigma}$.

Worked Applications

Worked Example: An Exponential Severity Quantile

A hypothetical equipment-repair portfolio is modeled with an exponential severity distribution with scale parameter $\theta$, so $F(x;\theta)=1-\exp(-x/\theta)$ for $x\geq 0$. The fitted scale is $\widehat\theta=\$120$, and the estimated variance of $\widehat\theta$ is $9$ dollars squared. Find the fitted 80th percentile and its approximate standard error. There is no interest or mortality assumption in this severity-only calculation.

Method. Solve $1-\exp(-q_{0.80}/\theta)=0.80$. Rearranging gives $\exp(-q_{0.80}/\theta)=0.20$, and hence $q_{0.80}=-\theta\ln(0.20)=\theta\ln(5)$.

The fitted quantile is $120\ln(5)=120(1.6094379)=\$193.1325$ before final rounding. Its derivative with respect to the scale is $\partial q_{0.80}/\partial\theta=\ln(5)=1.6094379$. Therefore, $\operatorname{Var}(\widehat q_{0.80})\approx(1.6094379)^2(9)=23.3126$ dollars squared, and $\operatorname{SE}(\widehat q_{0.80})\approx\sqrt{23.3126}=\$4.8283$.

Check and trap avoided. The quantile exceeds the scale, as expected for an 80th percentile. Using $-\ln(0.80)$ in place of $-\ln(1-0.80)$ would incorrectly produce the 20th-percentile multiplier. Rounded to the nearest cent, the fitted quantile is $\$193.13$ and its standard error is $\$4.83$.

Worked Example: A Lognormal 90th Percentile

A hypothetical specialty-materials severity model is lognormal: $\ln X\sim N(\mu,\sigma^2)$, where $\sigma$ is the standard deviation of log severity. The estimates are $\widehat\mu=\ln(50)$ and $\widehat\sigma=0.4$. The covariance matrix of $(\widehat\mu,\widehat\sigma)^{\mathsf T}$ is diagonal, with variances $0.0016$ and $0.0008$, respectively. Find the fitted 90th percentile and its approximate standard error. No interest or mortality assumptions apply.

Method. For a standard normal quantile $z_p$, the lognormal quantile is $q_p=\exp(\mu+\sigma z_p)$. At $p=0.90$, use $z_{0.90}=1.2815516$. Then $\widehat q_{0.90}=\exp(\ln(50)+0.4(1.2815516))=50\exp(0.5126206)\approx\$83.48$.

The gradient with respect to $(\mu,\sigma)$ is $(\partial q/\partial\mu,\partial q/\partial\sigma)^{\mathsf T} =(q_p,q_pz_p)^{\mathsf T}$. Evaluated at the fitted values, this is approximately $(83.483,\ 106.99)^{\mathsf T}$. Because the covariance matrix is diagonal, the variance is $83.483^2(0.0016)+106.99^2(0.0008)\approx11.15+9.16=20.31$ dollars squared. Thus the standard error is $\sqrt{20.31}\approx\$4.51$.

Check and trap avoided. Both gradient components are positive: increasing either $\mu$ or $\sigma$ increases this upper-tail quantile. The second component uses $z_{0.90}$, not $z_{0.90}^2$. Rounded to the nearest cent, the fitted quantile is $\$83.48$ and its standard error is $\$4.51$.

Worked Example: A Weibull Median With Shape Uncertainty

A hypothetical machinery-loss severity is modeled with survival function $S_X(x)=\exp[-(x/\lambda)^k]$, for $x\geq0$, with scale $\lambda>0$ and shape $k>0$. The fitted values are $\widehat\lambda=\$80$ and $\widehat k=1$. The estimator variances are $\operatorname{Var}(\widehat\lambda)=1$ dollar squared and $\operatorname{Var}(\widehat k)=0.0025$, with zero covariance. Find the fitted median and its approximate standard error. There is no interest or mortality assumption.

Method. Set $S_X(q_p)=1-p$, so $(q_p/\lambda)^k=-\ln(1-p)$. With $a=-\ln(1-p)$, the quantile is $q_p=\lambda a^{1/k}$. For the median, $p=0.50$ and $a=\ln(2)=0.6931472$. Therefore, $\widehat q_{0.50}=80(0.6931472)^{1}= \$55.4518$.

The derivatives are $\partial q/\partial\lambda=a^{1/k}/1=q/\lambda$, and $\partial q/\partial k=-\lambda a^{1/k}\ln(a)/k^2=-q\ln(a)/k^2$. At the fitted values, the gradient is approximately $(0.6931472,\ 20.33)^{\mathsf T}$, since $-55.4518\ln(0.6931472)\approx20.33$. The variance is approximately $0.6931472^2(1)+20.33^2(0.0025)=0.4805+1.033\approx1.514$ dollars squared. The standard error is therefore approximately $\sqrt{1.514}=\$1.23$.

Check and trap avoided. Although the fitted median is below the scale, uncertainty in the shape contributes materially to its variance. Omitting that contribution would leave only $0.4805$ dollars squared and understate the standard error. Rounded to the nearest cent, the fitted median is $\$55.45$ and its standard error is $\$1.23$.

Covariance, Interpretation, and Exam Traps

For a two-parameter estimate with gradient $(g_1,g_2)^{\mathsf T}$, the variance expansion is $g_1^2\operatorname{Var}(\widehat\theta_1)+g_2^2\operatorname{Var}(\widehat\theta_2)+2g_1g_2\operatorname{Cov}(\widehat\theta_1,\widehat\theta_2)$. The factor of two appears because both off-diagonal entries of the covariance matrix contribute to the matrix product.

For example, if the gradient is $(1.5,2)^{\mathsf T}$, the parameter variances are $0.09$ and $0.04$, and their covariance is $-0.015$, the wrong calculation that omits covariance gives $1.5^2(0.09)+2^2(0.04)=0.3625$. The correct covariance contribution is $2(1.5)(2)(-0.015)=-0.09$, so the variance is $0.3625-0.09=0.2725$ and the standard error is approximately $0.5220$. A negative covariance can reduce the variance; it does not justify deleting the covariance term.

A useful interpretation of the implicit derivative is that the density at the quantile appears in the denominator. When the density is small, a small change in the fitted cumulative probability can correspond to a relatively large movement in the loss value, so the quantile can be sensitive to parameter estimation. This is one reason upper-tail quantiles may have substantial uncertainty even when parameter estimates appear reasonably precise.

Differentiate the correct equation. Keep the target probability $p$ fixed when differentiating $F(q_p;\boldsymbol{\theta})=p$.
Use the full covariance matrix. Zero covariance may be used only when it is given or justified by the model or estimation result.
Check the parameterization. A lognormal $\sigma$ is the standard deviation of $\ln X$; the Weibull formula depends on whether the parameter is a scale or a rate.
Keep units consistent. A gradient with respect to a dollar-valued parameter and a dimensionless shape parameter has components with different units; the resulting quantile variance must be in dollars squared.
Do not confuse uncertainty with a prediction interval. The delta-method standard error describes estimation uncertainty in the fitted quantile, not the range of individual future losses.

If a symmetric normal approximation is appropriate, an approximate interval for the quantile can be formed as $\widehat q_p\pm z_{1-\alpha/2}\operatorname{SE}(\widehat q_p)$. Because severity quantiles are positive and their sampling distributions can be asymmetric, this interval can be unsuitable, particularly for a high quantile or a large standard error. The standard error remains a local, first-order approximation rather than a guarantee of accurate tail coverage.

Key takeaway: Differentiate the quantile equation to obtain its gradient, then propagate parameter uncertainty with $\nabla q_p^{\mathsf T}\boldsymbol{\Sigma}\nabla q_p$. Verify the model parameterization, retain covariance terms, and treat the result as an approximation whose reliability depends on smoothness and positive density at the fitted quantile.

Check Your Understanding

For each problem, calculate the requested fitted quantile uncertainty using the stated parameter covariance. Round dollar results to the nearest cent and other standard errors to four decimal places where appropriate.

A hypothetical exponential severity model has fitted scale $\widehat\theta=\$90$, with $\operatorname{Var}(\widehat\theta)=4$ dollars squared. Find the fitted 75th percentile and its approximate standard error. (Answer: fitted quantile $\$124.77$; standard error $\$2.77$.)
A hypothetical lognormal severity model has $\widehat\mu=\ln(40)$, $\widehat\sigma=0.3$, and $z_{0.95}=1.644854$. The estimator covariance matrix is diagonal with entries $0.0009$ and $0.0004$. Find the fitted 95th percentile and its approximate standard error. (Answer: fitted quantile $\$65.52$; standard error $\$2.92$.)
A fitted quantile has gradient $(2,-3)^{\mathsf T}$, and the parameter-estimator covariance matrix is $\begin{pmatrix}0.25&0.02\\0.02&0.04\end{pmatrix}$. Calculate its approximate variance and standard error, including the covariance contribution. (Answer: variance $1.12$; standard error $1.0583$.)
A hypothetical Weibull model has $\widehat\lambda=\$100$, $\widehat k=2$, and $p=0.75$. The estimator variances are $4$ dollars squared and $0.01$, respectively, with zero covariance. Using $a=-\ln(1-p)$, calculate the fitted quantile and approximate standard error. (Answer: fitted quantile $\$117.74$; standard error approximately $\$2.54$.)
For a quantile gradient $(1.5,2)^{\mathsf T}$, the parameter-estimator variances are $0.09$ and $0.04$, and their covariance is $-0.015$. Calculate the approximate quantile variance and standard error. (Answer: variance $0.2725$; standard error $0.5220$.)

What You'll Learn