Sensitivity of Capital to Tail Parameters

Why Sensitivity Matters

A capital estimate depends on both the risk measure and the model used to describe losses. Even when the confidence level stays fixed, changing a model parameter can move value-at-risk (VaR) or tail value-at-risk (TVaR), sometimes substantially. Sensitivity analysis quantifies that movement: it asks how much a risk measure changes for a small change in a specified parameter.

A derivative is a local measure, evaluated at a particular parameter value. It is not automatically the exact change for a large parameter revision. For a finite change, calculate the risk measure under both parameter values; a derivative can provide a first-order approximation or a check on that calculation.

Definition: Let $X_\theta$ have a loss distribution depending on a real parameter $\theta$, and fix a confidence level $p$, where $0<p<1$. Write $q_p(\theta)=\operatorname{VaR}_p(X_\theta)$. If the derivative exists, $\partial q_p/\partial\theta$ is the local change in VaR per unit change in $\theta$. Likewise, $\partial\operatorname{TVaR}_p(X_\theta)/\partial\theta$ is the local change in TVaR per unit change in $\theta$. The units of either derivative are loss units per unit of the parameter.

Quantile Sensitivity from the Distribution Function

Suppose $X_\theta$ has a continuous distribution with distribution function $F(x;\theta)$ and positive density $f(x;\theta)$ at its $p$-quantile. Assume $F$ is differentiable in $x$ and $\theta$, and that the quantile changes differentiably with $\theta$. The defining equation for the quantile is $F(q_p(\theta);\theta)=p$.

\frac{\partial q_p}{\partial\theta} =-\frac{\partial F(x;\theta)/\partial\theta\big|_{x=q_p}} {f(q_p;\theta)}.

The denominator matters: if the density near the quantile is small, a modest change in cumulative probability can correspond to a large movement in the quantile. The formula is not directly applicable at a point with zero density, a nondifferentiable distribution, or an atom where the quantile need not be differentiable. In those cases, use the distribution's quantile definition carefully and consider one-sided or finite changes instead.

Start with the quantile equation.
For fixed $p$, write $F(q_p(\theta);\theta)=p$.

Differentiate with respect to the parameter.
The chain rule gives $f(q_p;\theta)\,\partial q_p/\partial\theta+\partial F(q_p;\theta)/\partial\theta=0$.

Solve for the quantile derivative.
When $f(q_p;\theta)>0$, divide by that density to obtain the sensitivity formula.

For a continuous loss distribution with finite tail mean, TVaR can also be expressed as an average of quantiles above $p$:

\operatorname{TVaR}_p(X_\theta) =\frac{1}{1-p}\int_p^1 q_u(\theta)\,du.

This identity follows by representing the loss through its quantile function: a uniformly selected percentile above $p$ has average loss equal to the conditional mean in the upper $1-p$ fraction of the distribution. If differentiation can be interchanged with the integral, it follows that

\frac{\partial\operatorname{TVaR}_p(X_\theta)}{\partial\theta} =\frac{1}{1-p}\int_p^1\frac{\partial q_u(\theta)}{\partial\theta}\,du.

This formula explains why TVaR sensitivity reflects changes across the whole upper tail, rather than only the loss at the single VaR threshold. The required interchange is not automatic for every model; in the examples below, closed-form formulas and their differentiability justify the calculations directly.

Location and Scale: The Normal Model

Let $X\sim N(\mu,\sigma^2)$, where $\sigma>0$, and let $z_p$ be the $p$-quantile of the standard Normal distribution. Write $\phi$ for its density. The Normal-model results are

\operatorname{VaR}_p(X)=\mu+\sigma z_p,\qquad \operatorname{TVaR}_p(X)=\mu+\sigma\frac{\phi(z_p)}{1-p}.

The TVaR expression uses the conditional mean of a standard Normal variable above $z_p$, which is $\phi(z_p)/(1-p)$. Since $p$ is fixed, differentiating these expressions gives

\frac{\partial\operatorname{VaR}_p}{\partial\mu}=1,\qquad \frac{\partial\operatorname{VaR}_p}{\partial\sigma}=z_p,\qquad \frac{\partial\operatorname{TVaR}_p}{\partial\mu}=1,\qquad \frac{\partial\operatorname{TVaR}_p}{\partial\sigma} =\frac{\phi(z_p)}{1-p}.

Worked Example: Normal Capital Sensitivity to Location and Scale

A hypothetical portfolio loss is modeled as $X\sim N(1000,200^2)$, with losses measured in dollars. At $p=0.90$, use $z_{0.90}=1.2816$ and $\phi(z_{0.90})=0.1755$. Calculate VaR and TVaR, their local sensitivities to $\mu$ and $\sigma$, and the exact effect of increasing $\sigma$ from 200 to 210 while holding $\mu$ fixed.

Method. Use the Normal formulas and differentiate with respect to the parameter being changed. At this confidence level, $\phi(z_{0.90})/(1-p)=0.1755/0.10=1.7550$.

VaR. The quantile is $1000+200(1.2816)=1000+256.32=\$1,256.32$. Its sensitivities are $\partial\operatorname{VaR}_{0.90}/\partial\mu=1$ and $\partial\operatorname{VaR}_{0.90}/\partial\sigma=1.2816$. Thus a one-dollar increase in $\mu$ increases VaR by one dollar locally, and a one-unit increase in $\sigma$ increases it by about $\$1.2816$ locally.

TVaR. The value is $1000+200(1.7550)=1000+351.00=\$1,351.00$. Its sensitivities are $\partial\operatorname{TVaR}_{0.90}/\partial\mu=1$ and $\partial\operatorname{TVaR}_{0.90}/\partial\sigma=1.7550$.

Finite scale change. The increase in $\sigma$ is $210-200=10$. Because the Normal quantile and tail mean are linear in $\sigma$, the changes are exact: VaR increases by $10(1.2816)=\$12.816$, and TVaR increases by $10(1.7550)=\$17.550$. Rounded to cents, the changes are $\$12.82$ and $\$17.55$. The recalculated values are VaR $=\$1,269.14$ and TVaR $=\$1,368.55$, also rounded to cents. Subtracting the original values gives the same changes to the cent.

Check and trap avoided. Increasing $\sigma$ does not shift the Normal loss by the same amount as increasing $\mu$: the scale sensitivities are $z_p$ and $\phi(z_p)/(1-p)$, not 1. The recalculated differences confirm the exact scale changes for this model.

Scale Changes and Finite Differences

For a scale family $X_s=sZ$, with $s>0$ and a fixed nonnegative loss $Z$ having finite TVaR, both risk measures scale linearly:

\operatorname{VaR}_p(X_s)=s\operatorname{VaR}_p(Z),\qquad \operatorname{TVaR}_p(X_s)=s\operatorname{TVaR}_p(Z).

Consequently, their derivatives with respect to $s$ are the corresponding risk measures for $Z$. More generally, a finite difference between parameter values $\theta_1$ and $\theta_0$ is calculated as $R(\theta_1)-R(\theta_0)$, where $R$ is VaR or TVaR. The derivative at $\theta_0$, multiplied by $\theta_1-\theta_0$, is only a first-order approximation unless the risk measure is linear in that parameter.

Worked Example: Exponential Scale and an Exact Finite Change

A hypothetical loss $X$ has an exponential distribution with mean $\beta=500$ dollars, so $S_X(x)=e^{-x/\beta}$ for $x\geq0$. At $p=0.95$, calculate VaR and TVaR, their sensitivities to $\beta$, and the exact changes when the mean increases to 525 dollars.

Method. For an exponential loss, $F_X(x)=1-e^{-x/\beta}$. Setting this equal to $p$ gives $q_p=\beta[-\ln(1-p)]$. The exponential memoryless property gives the conditional mean above $q_p$ as $q_p+\beta$, so $\operatorname{TVaR}_p=\beta[-\ln(1-p)+1]$.

Here $-\ln(1-p)=-\ln(0.05)=\ln(20)=2.995732$. Therefore VaR is $500(2.995732)=\$1,497.866$, and TVaR is $500(2.995732+1)=500(3.995732)=\$1,997.866$.

Differentiating with respect to $\beta$ gives $\partial\operatorname{VaR}_{0.95}/\partial\beta=2.995732$ and $\partial\operatorname{TVaR}_{0.95}/\partial\beta=3.995732$. The parameter change is $525-500=25$ dollars. Since both risk measures are linear in $\beta$, the exact VaR increase is $25(2.995732)=\$74.8933$, and the exact TVaR increase is $25(3.995732)=\$99.8933$. Rounded to cents, these are $\$74.89$ and $\$99.89$.

The new values are VaR $=525(2.995732)=\$1,572.76$ and TVaR $=525(3.995732)=\$2,097.76$, rounded to cents. Subtracting the original values gives the same rounded changes. A common mistake is to use $\beta$ itself as the exponential VaR; the quantile also depends on the selected percentile through $-\ln(1-p)$.

Shape Sensitivity for a Pareto Type I Loss

Let $X$ have a Pareto Type I distribution with scale $\theta>0$ and shape $\alpha>1$, so $S_X(x)=(\theta/x)^\alpha$ for $x\geq\theta$. The restriction $\alpha>1$ ensures a finite mean and finite TVaR. For fixed $0<p<1$, its VaR and TVaR are

q_p=\theta(1-p)^{-1/\alpha},\qquad \operatorname{TVaR}_p=\frac{\alpha}{\alpha-1}q_p.

The VaR formula follows by solving $(\theta/q_p)^\alpha=1-p$. For the TVaR formula, integrate the survival function above $q_p$: for $t\geq q_p$, $P(X>t\mid X>q_p)=(q_p/t)^\alpha$. Thus

E[X\mid X>q_p] =q_p+\int_{q_p}^{\infty}\left(\frac{q_p}{t}\right)^\alpha dt =q_p+\frac{q_p}{\alpha-1} =\frac{\alpha}{\alpha-1}q_p.

Differentiating with respect to scale and shape gives

\frac{\partial q_p}{\partial\theta}=\frac{q_p}{\theta},\qquad \frac{\partial q_p}{\partial\alpha} =\frac{q_p\ln(1-p)}{\alpha^2}, $$ $$ \frac{\partial\operatorname{TVaR}_p}{\partial\theta} =\frac{\operatorname{TVaR}_p}{\theta},\qquad \frac{\partial\operatorname{TVaR}_p}{\partial\alpha} =q_p\left[-\frac{1}{(\alpha-1)^2} +\frac{\ln(1-p)}{\alpha(\alpha-1)}\right].

Since $\ln(1-p)<0$, increasing the shape parameter while holding scale and percentile fixed reduces Pareto VaR. It also reduces TVaR: both terms in the bracket of its shape derivative are negative for $\alpha>1$. These are local statements about this parameterization; changing the scale at the same time can alter the direction of the overall change.

Worked Example: Pareto Shape Sensitivity

A hypothetical individual loss follows a Pareto Type I distribution with $\theta=\$120$, $\alpha=3$, and $p=0.90$. Calculate VaR and TVaR and their local derivatives with respect to both $\theta$ and $\alpha$.

Method. Apply the Pareto formulas, keeping $p$ fixed. Since $1-p=0.10$, $(1-p)^{-1/\alpha}=10^{1/3}=2.1544347$. Thus $q_{0.90}=120(2.1544347)=\$258.5322$, or $\$258.53$ to cents. TVaR is $(3/2)(258.5322)=\$387.7982$, or $\$387.80$ to cents.

Scale sensitivities. The VaR derivative is $q_{0.90}/\theta=258.5322/120=2.1544347$ dollars per dollar of $\theta$. The TVaR derivative is $\operatorname{TVaR}_{0.90}/\theta=387.7982/120=3.2316515$ dollars per dollar of $\theta$.

Shape sensitivity of VaR. Using $\ln(1-p)=\ln(0.10)=-\ln(10)$, the derivative is $\displaystyle 258.5322\frac{-2.3025851}{9}=-\$66.14$ per unit of $\alpha$, rounded to cents. Equivalently, it is $\displaystyle -120(10^{1/3})\ln(10)/9$.

Shape sensitivity of TVaR. The derivative is $\displaystyle 258.5322\left[-\frac14-\frac{2.3025851}{6}\right]$. The bracket is approximately $-0.6337642$, giving $-\$163.85$ per unit of $\alpha$, rounded to cents. Equivalently, it is $\displaystyle -120(10^{1/3})[1/4+\ln(10)/6]$.

Check and trap avoided. Both shape derivatives are negative, as expected when increasing Pareto shape makes the tail lighter with scale held fixed. The numerical values are rounded only at the final step. A common mistake is to confuse the derivative with a dollar change for an arbitrary change in $\alpha$: the VaR derivative is dollars per unit of shape, not the exact VaR change for, say, a one-unit increase.

Interpreting and Using Sensitivities

A sensitivity is meaningful only with its parameter definition, units, confidence level, and model held clear. For example, the Pareto shape derivative above is conditional on the Pareto Type I survival function $S_X(x)=(\theta/x)^\alpha$; another parameterization may assign different meanings to its parameters. The derivative also does not measure uncertainty in the parameter estimate. It describes how the model-based risk measure responds if the parameter changes.

For a small change $\Delta\theta$, the first-order approximation is

R(\theta+\Delta\theta)\approx R(\theta) +\frac{\partial R}{\partial\theta}\Delta\theta,

where $R$ is VaR or TVaR. The approximation can be poor for a large change, especially when curvature is substantial. In practice, calculate the risk measure at the revised parameter as well, and compare that finite change with the derivative-based approximation. For multiple changing parameters, a first-order approximation adds the parameter-specific derivative times each parameter change, but it still omits higher-order interactions.

Check the direction. Confirm that the sign of the derivative is plausible for the model and parameterization.
Check the units. A derivative with respect to a dollar scale is not the same as one with respect to a dimensionless shape parameter.
Hold the percentile fixed. The formulas here measure parameter sensitivity at fixed $p$, not the effect of changing the confidence level.
Do not extrapolate blindly. A local derivative is not an exact finite change unless the risk measure is linear in the parameter, as it is for the scale changes shown above.

Key takeaway: Derive a risk measure as a function of its model parameters before differentiating. The resulting sensitivity has a direction and a unit, and it describes a local change at specified parameter values and a fixed confidence level.

Check Your Understanding

For each problem, compute the requested quantity at the stated confidence level. Answers are rounded to cents where loss amounts are requested.

A hypothetical exponential loss has mean $\beta=\$300$. At $p=0.80$, find VaR, TVaR, and both derivatives with respect to $\beta$. (Answer: VaR $=\$482.83$, TVaR $=\$782.83$; derivatives $1.6094$ and $2.6094$, respectively.)
A hypothetical Pareto Type I loss has $\theta=\$80$, $\alpha=4$, and $p=0.9375$. Find VaR, TVaR, and their derivatives with respect to $\theta$. (Answer: VaR $=\$160.00$, TVaR $=\$213.33$; derivatives $2.0000$ and $2.6667$ dollars per dollar of scale.)
A hypothetical loss is $N(400,50^2)$. At $p=0.95$, use $z_p=1.6449$ and $\phi(z_p)=0.1031$. Find VaR and the sensitivities of VaR to $\mu$ and $\sigma$. (Answer: VaR $=\$482.25$; sensitivities $1$ and $1.6449$.)
For the Normal loss in question 3, use $\phi(z_p)/(1-p)=0.1031/0.05=2.0620$. Find TVaR and its sensitivities to $\mu$ and $\sigma$. (Answer: TVaR $=\$503.10$; sensitivities $1$ and $2.0620$.)
A hypothetical exponential loss has mean $\$400$, which is revised to $\$440$. At $p=0.90$, find the exact finite increases in VaR and TVaR. Use $-\ln(0.10)=2.302585$. (Answer: VaR increases by $\$92.10$; TVaR increases by $\$132.10$.)

What You'll Learn