# 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 $$\log(1+x)\approx x$$?

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.