Hill Estimator and Extreme Quantile Confidence Intervals

/\alpha$). The tricky part is not the estimator itself but choosing how many top observations to use. Too few and you get huge variance; too many and you include data from the body of the distribution, introducing bias. This bias-variance tradeoff is the central practical challenge, and it is where interviewers will push you.

Quick Estimate: Suppose you have $n = 1000$ observations and you pick $k = 50$ (the top 5%). If the true $\alpha = 3$ (moderately heavy tail, like many financial return distributions), the Hill estimator has asymptotic standard deviation $\alpha / \sqrt{k} = 3/\sqrt{50} \approx 0.42$. So a 95% CI for $\alpha$ would be roughly $\hat{\alpha} \pm 0.83$, which is quite wide. This tells you that even with 1000 observations, tail index estimation is imprecise -- a fact that matters enormously for risk management.

Formal Solution:

Part (i): The Hill Estimator

The Hill estimator of the tail index $\alpha$ based on the top $k$ order statistics is:

$\hat{\alpha}_{\text{Hill}}(k) = \left( \frac{1}{k} \sum_{i=1}^{k} \ln X_{(i)} - \ln X_{(k+1)} \right)^{-1}$

Equivalently, define the reciprocal (the Hill estimator of $\gamma = 1/\alpha$):

$\hat{\gamma}(k) = \frac{1}{k} \sum_{i=1}^{k} \left( \ln X_{(i)} - \ln X_{(k+1)} \right)$

so that $\hat{\alpha}_{\text{Hill}}(k) = 1/\hat{\gamma}(k)$.

Why it works: For a Pareto-like tail, the log-excesses $\ln X_{(i)} - \ln X_{(k+1)}$ for $i = 1, \ldots, k$ behave approximately like the top order statistics from an $\text{Exp}(\alpha)$ distribution. The Hill estimator $\hat{\gamma}(k)$ is the sample mean of these log-spacings, which is the MLE for the exponential rate parameter's reciprocal. It is a conditional MLE for the tail index, conditioning on exceeding the threshold $X_{(k+1)}$.

Part (ii): Asymptotic Variance and Choosing $k$

Under regularity conditions (second-order regular variation), as $n \to \infty$ with $k = k(n) \to \infty$ and $k/n \to 0$:

$\sqrt{k} \left( \hat{\gamma}(k) - \gamma \right) \xrightarrow{d} N(0, \gamma^2)$

Equivalently, for $\hat{\alpha}(k)$ by the delta method:

$\sqrt{k} \left( \hat{\alpha}(k) - \alpha \right) \xrightarrow{d} N(0, \alpha^2)$

So the asymptotic variance of $\hat{\alpha}(k)$ is $\alpha^2 / k$.

The bias-variance tradeoff: - Variance $\approx \alpha^2 / k$: decreases as $k$ increases (more data points used). - Bias comes from including observations that are not truly in the tail. Under second-order regular variation with parameter $\rho < 0$, the bias is approximately $b \cdot (k/n)^{-\rho}$ for some constant $b$ depending on the specific distribution. Bias increases as $k$ increases. - The AMSE (Asymptotic Mean Squared Error) is:

$\text{AMSE}(k) = \text{Bias}^2(k) + \text{Var}(k) = b^2 \left(\frac{k}{n}\right)^{-2\rho} + \frac{\alpha^2}{k}$

Minimizing over $k$ gives the optimal $k^{*}$ that balances bias and variance.

Two practical methods for choosing $k$:

1. AMSE minimization (bootstrap or plug-in): Estimate the second-order parameters $(b, \rho)$ from the data (e.g., using the method of Beirlant et al. or a double-bootstrap), then minimize the estimated AMSE over $k$. This gives an explicit $\hat{k}^{*}$.

1. Hill plot diagnostic: Plot $\hat{\alpha}(k)$ against $k$ for $k = 1, 2, \ldots, n-1$. Look for a region where the plot is roughly stable (a plateau). Too small $k$: the plot is noisy. Too large $k$: the plot drifts (bias). Choose $k$ in the stable region. This is subjective but widely used as a sanity check.

Part (iii): Confidence Interval for $q_p$

For $p$ close to 1, the Pareto tail approximation gives:

$q_p \approx X_{(k+1)} \left( \frac{k+1}{n(1-p)} \right)^{\hat{\gamma}(k)}$

where $\hat{\gamma}(k) = 1/\hat{\alpha}(k)$. This is the Weissman extrapolation estimator: it uses the empirical threshold $X_{(k+1)}$ and extrapolates into the tail using the estimated Pareto exponent.

Derivation: Under the Pareto tail approximation, $\bar{F}(x) \approx (k+1)/n$ at $x = X_{(k+1)}$, and $\bar{F}(q_p) = 1 - p$. Taking ratios:

$\frac{1 - p}{(k+1)/n} \approx \left( \frac{q_p}{X_{(k+1)}} \right)^{-\alpha}$

Solving for $q_p$:

$q_p \approx X_{(k+1)} \left( \frac{k+1}{n(1-p)} \right)^{1/\alpha}$

Replace $\alpha$ with $\hat{\alpha}(k)$ to get the estimator $\hat{q}_p$.

For the confidence interval, the log of the quantile estimator is approximately normal. Applying the delta method to $\ln \hat{q}_p$:

$\ln \hat{q}_p \approx \ln X_{(k+1)} + \hat{\gamma}(k) \ln\left( \frac{k+1}{n(1-p)} \right)$

The dominant source of randomness is $\hat{\gamma}(k)$, which has asymptotic variance $\gamma^2/k$. So:

$\text{Var}(\ln \hat{q}_p) \approx \frac{\gamma^2}{k} \left[ \ln\left( \frac{k+1}{n(1-p)} \right) \right]^2$

A $(1 - \delta)$ confidence interval for $q_p$ is:

$\hat{q}_p \cdot \exp\left( \pm z_{\delta/2} \cdot \frac{\hat{\gamma}(k)}{\sqrt{k}} \cdot \ln\left( \frac{k+1}{n(1-p)} \right) \right)$

where $z_{\delta/2}$ is the standard normal critical value.

Answer:

• (i) The Hill estimator is $\hat{\alpha}(k) = \left[ \frac{1}{k} \sum_{i=1}^{k} (\ln X_{(i)} - \ln X_{(k+1)}) \right]^{-1}$, the reciprocal of the mean log-spacing of the top $k$ order statistics above the $(k+1)$-th.

• (ii) $\sqrt{k}(\hat{\alpha}(k) - \alpha) \xrightarrow{d} N(0, \alpha^2)$, so asymptotic variance is $\alpha^2/k$. Choose $k$ by minimizing estimated AMSE (bootstrap) or by identifying the stable plateau in the Hill plot.

• (iii) The quantile estimator is $\hat{q}_p = X_{(k+1)} \left( \frac{k+1}{n(1-p)} \right)^{1/\hat{\alpha}(k)}$, with CI given by exponentiating the normal CI for $\ln \hat{q}_p$ using the delta-method variance.