Why VaR Scales with Horizon and How Lognormal Loss Changes the Formula
Summary
The discussion compares the square-root-of-time VaR rule with a calculation that models portfolio value as the product of daily returns. Under independent, normally distributed daily returns, the question derives an approximate normal distribution for log portfolio value, then converts a lower-tail value quantile back to currency units. This produces an exponential expression that differs from the familiar linear VaR scaling rule.
The answer clarifies that VaR should represent a loss amount, not the portfolio's terminal value. It rewrites the probability in terms of the loss, obtains a loss expression from the lognormal model, and uses a first-order expansion of the exponential to recover the usual square-root-of-time approximation. The explanation is conditional on the stated distributional assumptions and small-return approximation; it does not establish that daily returns are independent or normal in practice. Care is also needed with tail probability conventions and signs when defining VaR.
Key ideas
- VaR is a loss quantile, whereas a portfolio value quantile is expressed in different units and has a different probability interpretation.
- Under independent normal daily returns, the variance of the sum of returns grows with the number of days, producing square-root-of-time scaling.
- Modeling compounded returns through log value leads to an exponential expression for the terminal value quantile.
- A first-order expansion of the exponential makes the lognormal loss expression approximately linear in volatility and horizon scale.
- The approximation depends on the return model and should not be treated as a general guarantee for real portfolios.
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# Understanding VaR rescaling
# Understanding VaR rescaling
Assume my portfolio has a current market value of $V_0$, that the daily returns are independent and identically distributed as a normal distribution $N(0, \sigma^2)$ and that there are $N$ trading days in a year. Also, let $\Phi$ denote the cdf of the standard normal distribution.
The Value at Risk rescaling formula says that $$\text{VaR}_{\alpha,N}=\text{VaR}_{\alpha,1}\times\sqrt{N}=V_0\times\Phi^{-1}(\alpha)\times\sigma\times\sqrt{N}$$
Now, let $V$ be the portfolio's value after the $N$-day period. It is clear that $$V=V_0\times\prod_{i=1}^{N}(1+R_i)$$ $$\log(V)=\log(V_0)+\sum_{i=1}^{N}\log(1+R_i)$$ and, by the Taylor expansion of $\log(1+x)$ we get that $$\log(V)=\log(V_0)+\sum_{i=1}^{N}R_i$$ plus smaller terms. Applying the central limit theorem we get that $$\log(V)\sim N\left(\log(V_0),N\sigma^2\right)$$ and thus $\text{VaR}_{\alpha,N}$ is the number $x$ such that $$\mathbf{P}(V\le x)=\alpha$$ But $$\mathbf{P}(V\le x)=\mathbf{P}(\log(V)\le \log(x))=\Phi\left(\frac{\log(x)-\log(V_0)}{\sqrt{N}\sigma}\right)$$ and therefore $$\text{VaR}_{\alpha,N}=x=\exp(\log(V_0)+\sqrt{N}\sigma\Phi^{-1}(\alpha))=V_0\exp(\sqrt{N}\sigma\Phi^{-1}(\alpha))$$ which is certainly different from the usual formula.
Question: why are the two formulas different? What am I missing? Is the difference caused solely by the approximation $\log(1+x)\approx x$?
Thanks in advance!
## Answer by KaiSqDist (score 2, accepted)
https://quant.stackexchange.com/a/78832
I looked through your proof and it looks mostly fine, but 2 things:
The scaling formula for VaR is given by (you wrote an extra $\sigma$) and I use $Z$ instead of $\Phi^{-1}(\alpha)$:
$$ VaR_{\alpha,T} = VaR_{\alpha,1} * \sqrt{T} = V_0 * \sigma * Z * \sqrt{T} $$
and your final formula was missing a $\sigma$ as well:
$$ VaR_{\alpha,T} = x = \exp(\ln(V_0) + \sqrt{T}\sigma\Phi^{-1}(\alpha)) = V_0\exp(\sqrt{T}\sigma\Phi^{-1}(\alpha)) $$
My feeling for where the problems started was your VaR formulation (the definition is wrong). The VaR is defined as the loss of a portfolio over a time horizon and not the value of the portfolio.
Therefore, the probability function should be written as (I use $X$ instead of $x$):
$$ P(V_0 - V \le X) = \alpha \rightarrow P(V \le V_0 - X) = 1 - \alpha $$
We can then get:
$$ 1 - \alpha = \Phi(\frac{\ln(V_0 - X) - \ln(V_0)}{\sqrt{T}\sigma}) \rightarrow -Z = \frac{\ln(V_0 - X) - \ln(V_0)}{\sqrt{T}\sigma} $$
If we make the VaR $X$ the subject of the equation:
$$ X = V_0 (1 - \exp(-Z\sigma\sqrt{T})) $$
Applying the Taylor series $\exp(-x) \approx 1 - x$ (and ignoring higher terms):
$$ X \approx V_0(1 - (1 - Z\sigma\sqrt{T})) = V_0Z\sigma\sqrt{T}\;(proven) $$
Please let me know if this helps you.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.