ETF Tracking Error: Tail Probability and VaR
The daily tracking error of an ETF (its return relative to its benchmark) is modeled as $X \sim N(0, \sigma^2)$ with $\sigma = 0.40\%$.
(i) Compute $P(|X| > 1.00\%)$ in terms of the standard normal CDF $\Phi$.
(ii) Compute the one-day 99% Value at Risk of $-X$ (i.e., the loss from the ETF underperforming its benchmark) and express the answer in basis points.
Hints
- Standardize: divide both the threshold and the random variable by $\sigma = 0.40\%$ to convert to a standard normal. The $z$-score for the $1\%$ threshold is $1.00/0.40 = 2.5$.
- For the two-tailed probability in part (i), use $P(|Z| > 2.5) = 2(1 - \Phi(2.5))$ by symmetry of the normal distribution.
- For VaR in part (ii), the 99% VaR of $-X$ equals $\sigma \cdot z_{0.99}$ where $z_{0.99} \approx 2.326$ is the 99th percentile of the standard normal.
Worked Solution
How to Think About It: Tracking error problems are pure normal distribution mechanics. You are standardizing by the volatility $\sigma$ to get a $z$-score, then reading off tail probabilities. For VaR, remember that VaR is a loss quantile -- you are asking how bad the underperformance can get in the worst 1% of days. Since $X$ is symmetric around zero, the 99% VaR of $-X$ is just the 99th percentile of $X$ flipped in sign, which equals $\sigma$ times the 99th standard normal quantile.
Quick Estimate: $\sigma = 40$ bps. The threshold $1.00\% = 100$ bps is $100/40 = 2.5$ standard deviations away. The two-tailed probability beyond $2.5\sigma$ is about $1.2\%$ (you can remember $2\sigma \approx 4.6\%$ and $3\sigma \approx 0.3\%$, so $2.5\sigma$ is roughly $1\%$). For VaR: the 99th percentile of a standard normal is about $2.326$, so VaR $\approx 2.326 \times 40 \approx 93$ bps.
Approach: Standardize $X$ to a standard normal, apply the symmetry of $\Phi$, and use $z_{0.99} \approx 2.326$.
Formal Solution:
Part (i): Let $Z = X / \sigma \sim N(0,1)$. Then: $$P(|X| > 1.00\%) = P\!\left(|Z| > \frac{1.00\%}{0.40\%}\right) = P(|Z| > 2.5)$$
Using the symmetry of the standard normal: $$P(|Z| > 2.5) = 2\,P(Z > 2.5) = 2(1 - \Phi(2.5)) = 2\Phi(-2.5)$$
$$\boxed{P(|X| > 1.00\%) = 2(1 - \Phi(2.5)) \approx 2(0.00621) \approx 1.24\%}$$
Part (ii): The one-day 99% VaR of $-X$ is the value $v$ such that $P(-X > v) = 0.01$, i.e., $P(X < -v) = 0.01$.
Since $X = \sigma Z$: $$P(X < -v) = P\!\left(Z < \frac{-v}{\sigma}\right) = 0.01$$ $$\frac{-v}{\sigma} = \Phi^{-1}(0.01) = -z_{0.99} \approx -2.326$$ $$v = \sigma \cdot z_{0.99} = 0.40\% \times 2.326 = 0.9304\%$$
$$\boxed{\text{VaR}_{99\%} = \sigma \cdot z_{0.99} \approx 0.9304\% = 93.04 \text{ bps}}$$
Answer: (i) $2(1 - \Phi(2.5)) \approx 1.24\%$. (ii) VaR $= \sigma \cdot z_{0.99} \approx 93$ bps.
Intuition
This problem is a clean application of normal distribution mechanics that every quant should be able to do in their head. The key ratio to internalize: $1\%$ threshold at $0.40\%$ vol is $2.5\sigma$, and the two-tailed probability beyond $2.5\sigma$ is about $1.2\%$ -- slightly more than the $1\%$ VaR threshold. That means the $1\%$ absolute deviation threshold is just barely above the 99th percentile level.
In practice, tracking error analysis is fundamental to ETF operations and index fund management. Risk teams monitor whether the fund is consistently within its tracking error budget. A 93 bps one-day VaR for a 40 bps daily vol fund means that on the worst 1% of days, the fund underperforms its benchmark by nearly a full percentage point -- that is significant and would trigger review. The relationship VaR $= \sigma \cdot z_{\alpha}$ is worth memorizing: it is the building block for all normal-model risk calculations.