Why the Time Horizon Matters
A VaR figure is incomplete unless its time horizon is stated. A one-day VaR describes the distribution of losses over one day; a ten-day VaR describes a different random variable, formed from losses over ten days. The relationship between the two depends on how losses accumulate and on the assumptions about their distributions and dependence.
A familiar approximation multiplies one-period VaR by the square root of the number of periods. This can be useful, but it is not a general rule. Mean losses, serial dependence, discrete outcomes, and changes in the loss process can all make direct horizon modeling necessary.
Direct Horizon Modeling
The direct approach is to specify or estimate the loss over the full horizon, then calculate the quantile of that horizon’s loss distribution. If the horizon contains period losses \(X_1,\ldots,X_T\), and these losses add in the same monetary units, the cumulative loss is
The distribution of this sum depends on the distribution of each \(X_t\) and their dependence. Independence is one possible assumption, not an automatic property of daily losses. If the period losses are independent and identically distributed, their distributions can be convolved to obtain the sum’s distribution. If they are dependent, the joint distribution is needed. For losses whose behavior changes over time, a single repeated-period model may not be suitable at all.
Worked Example: Ten-Day VaR With Nonzero Daily Mean
A hypothetical portfolio has independent daily losses that are normally distributed with mean \(\$200\) and standard deviation \(\$1{,}500\). Assume the daily losses are identically distributed, and use a 99% confidence level with \(z_{0.99}=2.3263\). Find the ten-day VaR. Round the final dollar amount to the nearest cent.
Method. Aggregate the daily losses first. A sum of independent normal losses is normal. Its mean is the sum of the daily means, and its variance is the sum of the daily variances.
For one day, the mean is \(\$200\) and the variance is \((\$1{,}500)^2\). Over ten days:
The ten-day standard deviation is \(\sqrt{22{,}500{,}000}=\$4{,}743.4165\ldots\). For a normal loss with mean \(\mu\) and standard deviation \(\sigma\), the \(\alpha\)-quantile is \(\mu+z_{\alpha}\sigma\). Therefore:
The ten-day 99% VaR is \(\$13{,}034.61\). As a check, the result exceeds the ten-day mean of \(\$2{,}000\), as expected for a 99th-percentile loss under this model.
Common mistake avoided: One-day 99% VaR is \(200+2.3263(1{,}500)=\$3{,}689.45\). Multiplying this by \(\sqrt{10}\) gives approximately \(\$11{,}667.06\), not the direct ten-day result. The one-day VaR includes a positive mean; that mean accumulates linearly over ten days, not by the square root of ten.
When Square-Root-of-Time Scaling Holds
The square-root result has a precise basis. Suppose each period loss \(X_t\) is independent and normally distributed with mean zero and variance \(\sigma^2\). The sum over \(T\) periods is then normal. Its mean is zero, and its variance is \(T\sigma^2\). Since standard deviation is the square root of variance, the horizon standard deviation is \(\sigma\sqrt{T}\).
Assume \(X_t\sim N(0,\sigma^2)\) for every period, with independent losses across periods.
The horizon loss \(L_T=\sum_{t=1}^{T}X_t\) is normal with mean zero and variance \(T\sigma^2\).
The \(\alpha\)-quantile of this normal loss is \(z_\alpha\sigma\sqrt{T}\).
Because the one-period VaR under this zero-mean model is \(z_\alpha\sigma\), the horizon VaR is \(\sqrt{T}\) times the one-period VaR. This is exact under the stated assumptions. In particular, the normality assumption ensures that a normal quantile fully describes the distribution at each horizon; independence ensures the variances add.
If independent daily losses instead have a normal distribution with mean \(\mu\) and variance \(\sigma^2\), then the exact horizon result is
The mean component scales with \(T\), whereas the standard-deviation component scales with \(\sqrt{T}\). Thus, even when losses are independent and normal, simply multiplying total one-period VaR by \(\sqrt{T}\) is generally incorrect when \(\mu\ne 0\).
Worked Example: A Discrete Loss Model Over Two Days
A hypothetical coverage has a daily loss of \(\$5{,}000\) with probability 0.20 and no loss with probability 0.80. Assume the two daily losses are independent and identically distributed. Find the two-day 95% VaR, and compare it with square-root-of-time scaling of the one-day 95% VaR.
Method. Enumerate the possible two-day sums and their probabilities. This preserves the discrete distribution and applies the quantile definition directly.
A two-day loss of \(\$0\) occurs when both days have no loss, with probability \(0.80^2=0.64\). A loss of \(\$5{,}000\) occurs when exactly one day has a loss, with probability \(2(0.20)(0.80)=0.32\). A loss of \(\$10{,}000\) occurs when both days have a loss, with probability \(0.20^2=0.04\). The probabilities sum to \(0.64+0.32+0.04=1\).
| Two-day loss | Probability | Cumulative probability |
|---|---|---|
| \(\$0\) | 0.64 | 0.64 |
| \(\$5{,}000\) | 0.32 | 0.96 |
| \(\$10{,}000\) | 0.04 | 1.00 |
The cumulative probability first reaches 0.95 at \(\$5{,}000\), so the direct two-day 95% VaR is \(\$5{,}000\). The one-day 95% VaR is also \(\$5{,}000\), since the one-day cumulative probability at zero is only 0.80. Scaling that one-day VaR by \(\sqrt{2}\) gives approximately \(\$7{,}071.07\), which is not the direct two-day VaR.
Common mistake avoided: Treating square-root scaling as a quantile rule for every distribution. Here the two-day loss distribution has only three possible values, and its 95th percentile is determined by the cumulative probabilities, not by a scaled one-day quantile.
Dependence Changes the Horizon Distribution
Independence is also important because it determines how period variances combine. For a weakly stationary sequence of losses with common variance \(\gamma_0\) and lag-\(k\) covariance \(\gamma_k=\operatorname{Cov}(X_t,X_{t+k})\), the variance of the horizon sum is
To see why, expand the variance of the sum into all variances and covariances. There are \(T\) individual variance terms, each equal to \(\gamma_0\). For each lag \(k\), there are \(T-k\) pairs separated by that lag; each covariance appears twice in the expansion, once in each order. If all lagged covariances are zero, the expression reduces to \(T\gamma_0\). Positive serial covariance increases horizon variance relative to independence; negative covariance decreases it. This variance result alone does not determine VaR unless the distributional assumptions are also sufficient to identify the relevant quantile.
Worked Example: Three-Day VaR With Serial Dependence
Suppose three daily losses have a joint normal distribution. Each daily loss has mean \(\$400\) and standard deviation \(\$2{,}000\). The correlation is 0.50 for each adjacent pair of days and 0.25 between days one and three. Use \(z_{0.95}=1.6449\) to find the three-day 95% VaR. Assume the stated joint normal model applies to the three days.
Method. Sum the means and calculate the variance of the total using each daily variance and each pairwise covariance. The total is normal because it is a linear combination of jointly normal losses.
Work in thousands of dollars. Each daily mean is 0.4 and each daily variance is \(2^2=4\), in squared thousands of dollars. The adjacent-day covariance is \(0.50(2)(2)=2\); the day-one/day-three covariance is \(0.25(2)(2)=1\). Therefore:
The standard deviation of the total is \(\sqrt{22}=4.6904\ldots\) thousand dollars. Its 95% VaR is
Rounded to the nearest cent, the three-day 95% VaR is \(\$8{,}915.26\). The variance calculation includes all three pairwise covariances, each twice in the variance expansion. Omitting those terms would incorrectly treat the daily losses as independent.
Common mistake avoided: Applying an independence-based scaling factor without checking the covariance assumptions. Here the daily losses are positively correlated, so the aggregate variance is 22 rather than the independent-loss variance \(3(4)=12\).
Practical Limits of the Approximation
For independent losses that are not normal, the square-root relationship is not generally exact, even if the losses have finite variance. A normal approximation to a sum may become useful for a sufficiently long horizon under suitable conditions, but that is an approximation to the sum’s distribution, not a guarantee that finite-horizon VaR equals \(\sqrt{T}\) times one-period VaR. Tail shape and discreteness can matter, especially at high confidence levels.
The assumptions should match the quantity being measured. For example, daily losses may be modeled as independent increments for a particular analysis, but real losses can exhibit serial dependence or changing volatility. Likewise, the time unit must be consistent: a daily distribution cannot be scaled to a different horizon merely by inserting a number of days if the underlying process or the definition of loss changes during that period.
| Situation | Implication for horizon VaR |
|---|---|
| Independent, zero-mean normal period losses | Square-root scaling is exact. |
| Independent normal losses with nonzero mean | Aggregate mean scales with \(T\); the standard-deviation term scales with \(\sqrt{T}\). |
| Serially dependent losses | Covariances affect the variance of the aggregate; full distributional effects may also matter. |
| Discrete or other nonnormal losses | Calculate or approximate the sum’s quantile; simple VaR scaling is not generally valid. |
| Changing loss process across the horizon | Model the periods and their conditions directly rather than assuming identical increments. |
A useful exam check is to write down what random variable the requested VaR describes before calculating. Then verify that the proposed scaling follows from that variable’s assumptions. This keeps the horizon, mean, variance, dependence, and confidence level aligned. Understanding these distinctions is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
Check Your Understanding
For each problem, identify the horizon model before calculating. Dollar answers are rounded to the nearest cent when needed.
- Independent daily losses are normal with mean \(\$100\) and variance \((\$900)^2\). Find the four-day 95% VaR using \(z_{0.95}=1.6449\). (Answer: \(\$3{,}360.82\))
- A hypothetical daily loss is \(\$10{,}000\) with probability 0.10 and zero otherwise. Daily losses are independent. Find the two-day 90% VaR by constructing the aggregate distribution. (Answer: \(\$10{,}000\))
- Independent, zero-mean normal daily losses have standard deviation \(\$3{,}000\). Find the five-day 99% VaR using \(z_{0.99}=2.3263\), and state the assumptions that permit square-root scaling. (Answer: \(\$15{,}605.29\); independent, identically distributed, zero-mean normal daily losses)
- Three jointly normal daily losses each have mean \(\$0\), variance 1 in squared thousands of dollars, and correlations 0.30 for each adjacent pair and 0.09 between days one and three. Find the three-day 95% VaR using \(z_{0.95}=1.6449\). (Answer: \(\$3{,}442.52\))