University of Sydney · S1 2026 · FACULTY OF BUSINESS & ECONOMICS

FINC6023 · Financial Risk Management

Q: When is the √T rule wrong?

It assumes returns are independent and identically distributed. If returns trend (positive autocorrelation), √T underestimates long-horizon risk because losses compound; if returns mean-revert (negative autocorrelation), √T overestimates because losses partly cancel. Quote which direction the bias goes.

Q: Why is Expected Shortfall preferred to VaR?

VaR ignores how bad the tail is beyond the quantile and is not sub-additive, so a trader can split a position to lower reported VaR. ES averages the whole tail, is always at least as large as VaR, and is coherent — combining portfolios never makes ES look smaller than the sum, so it cannot be gamed the same way.

Q: Why is σ estimated more precisely than μ?

The standard error of the volatility estimate, se(σ̂) = σ̂·√(1/2n), shrinks faster with sample size than the mean's se(μ̂) = σ̂·√(1/n). This is why VaR models lean on volatility (which we can pin down) rather than the drift (which is noisy).

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Chapter 3 of 12 · FINC6023

VaR Accuracy, Backtesting, Scaling & Expected Shortfall

VaR Accuracy, Backtesting, Scaling & Expected Shortfall stress-tests the VaR number itself. You scale a VaR across confidence levels by the z-ratio and across horizons by the square-root-of-time rule (valid only for i.i.d. returns), backtest a model by counting exceptions against a binomial benchmark, and confront VaR's biggest weakness: it is not sub-additive, so it can be gamed. The fix is Expected Shortfall (ES), the mean loss given that the loss has already exceeded VaR — a coherent measure that is always at least as large as VaR.

In this chapter

What this chapter covers

01Scaling confidence: VaR_c2 = VaR_c1 · (z_c2 / z_c1)
02Scaling horizon: the √T rule and when it is valid (i.i.d. returns)
03Trend → √T underestimates; mean reversion → √T overestimates
04Precision of estimates: σ estimated more precisely than μ
05Binomial backtesting: exceptions, E[x] = pT, z = (x − pT)/√(p(1−p)T)
06Type I vs Type II error; Kupiec and Christoffersen tests
07Basel market-risk capital charge structure
08Expected Shortfall (ES/CVaR) = mean tail loss; coherence and sub-additivity

Worked example · free

Backtest a 99% VaR model with a binomial z-test

Q [7 marks]. A bank backtests its 1-day 99% VaR over T = 500 trading days and observes 9 exceptions (days where the loss exceeded the VaR). At a 99% model confidence (p = 0.01 exceptions expected), is the model producing too many exceptions? Use a normal approximation z = (x − pT)/√(p(1−p)T) and the critical value 1.96 for a two-sided 5% test.

1 markSet up the binomial benchmark: each day is a Bernoulli trial with exception probability p = 0.01, over T = 500 days.
1 markExpected exceptions E[x] = pT = 0.01 × 500 = 5.
2 marksVariance V[x] = p(1 − p)T = 0.01 × 0.99 × 500 = 4.95, so the standard deviation is √4.95 = 2.2249.
2 marksTest statistic z = (x − pT)/√(p(1−p)T) = (9 − 5)/2.2249 = 4/2.2249 = 1.798.
1 markCompare to the critical value: 1.798 < 1.96, so we do NOT reject the model at the 5% level.

z ≈ 1.80, which is below the 1.96 critical value, so 9 exceptions in 500 days is not statistically excessive at the 5% level — the model is not rejected, though it is on the high side.

Sia tip — Distinguish the two errors: rejecting a good model is Type I (set by your test size), accepting a bad model is Type II. Because exceptions are rare, you need many days before a backtest has the power to detect a poor model — quote that limitation if asked.

Glossary

Key terms

Square-root-of-time (√T) rule: Scale a 1-day VaR to T days by multiplying by √T. It is valid only for independent, identically distributed returns; a trend makes it underestimate, mean reversion makes it overestimate.
Backtesting: Counting how often the actual loss exceeds the predicted VaR (exceptions). Under the model, exceptions are binomial with E[x] = pT; a large-sample z-test or the Kupiec/Christoffersen tests judge whether there are too many or whether they cluster.
Type I vs Type II error: A Type I error rejects a correct VaR model (probability set by the test size); a Type II error fails to reject an incorrect model. Rare exceptions make backtests low-powered, raising Type II risk.
Expected Shortfall (ES / CVaR): The average loss given that the loss exceeds VaR — the mean of the tail beyond the VaR quantile. ES ≥ VaR and is coherent (sub-additive), so it cannot be reduced by splitting a portfolio.
Coherence / sub-additivity: A risk measure is coherent if it satisfies monotonicity, translation invariance, positive homogeneity and sub-additivity (the risk of a combined portfolio is no more than the sum of the parts). VaR can violate sub-additivity; ES does not.

FAQ

VaR Accuracy, Backtesting, Scaling & Expected Shortfall FAQ

When is the √T rule wrong?

It assumes returns are independent and identically distributed. If returns trend (positive autocorrelation), √T underestimates long-horizon risk because losses compound; if returns mean-revert (negative autocorrelation), √T overestimates because losses partly cancel. Quote which direction the bias goes.

Why is Expected Shortfall preferred to VaR?

VaR ignores how bad the tail is beyond the quantile and is not sub-additive, so a trader can split a position to lower reported VaR. ES averages the whole tail, is always at least as large as VaR, and is coherent — combining portfolios never makes ES look smaller than the sum, so it cannot be gamed the same way.

Why is σ estimated more precisely than μ?

The standard error of the volatility estimate, se(σ̂) = σ̂·√(1/2n), shrinks faster with sample size than the mean's se(μ̂) = σ̂·√(1/n). This is why VaR models lean on volatility (which we can pin down) rather than the drift (which is noisy).

Study strategy

Exam move

Practise the two scaling moves until automatic and always state the i.i.d. caveat with √T. For backtesting, learn the binomial setup (E[x] = pT, the z-statistic) and be ready to explain coherence as the one-line reason ES beats VaR.

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