Problem in calculating a simple VaR

External question — Quantitative Finance Stack Exchange Author: CarLaTeX Original post: https://quant.stackexchange.com/questions/47499 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. In Alexander, Gordon J. and Alexandre M. Baptista (2006). Does the Basle Capital Accord reduce bank fragility? An assessment of the value-at-risk approach. Journal of Monetary Economics 53(7), 1631–1660. at page 1644 there is: Consider the following example that is based on a 10-trading day investment horizon and 99% confidence level, as required by the Basle Capital Accord. Suppose that the expected rate of return and standard deviation of efficient portfolios $S$ and $L$ are given by: $E[r_S] = 0.50\%$ , $\sigma[r_S] = 0.40\%$ ; $E[r_L] = 1.00\%$ , and $\sigma[r_L] = 0.60\%$ . It follows that $V[0.99; r_S] = 0.43\%$ and $V[0.99; r_L] = 0.40\%$ , ... and at page 1636 $V$ is defined as: For any $t\in(\frac{1}{2},1)$ , let $z_t \equiv -\Phi^{-1}(1-t)$ , where $\Phi(\cdot)$ is the standard normal cdf. Using the assumption of normality, portfolio $w$ ’s VaR at $100t\%$ confidence level is: $$V[t,r_w]\equiv z_t\sigma[r_w]-E[r_w].$$ I tried to calculate the VaRs in the example but I don't get the same results, even if I scale the $\sigma$ by a $\sqrt{10}$ factor. Could you please help me to understand how those VaRs were calculated?

External answer — Quantitative Finance Stack Exchange Author: Enrico Schumann Original post: https://quant.stackexchange.com/a/47500 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. The calculation assumes that returns are normally distributed. VaR is a percentile of the returns distribution, which in turn can be expressed as a multiple (here labelled $z$ ) of the standard deviation of returns. (This works as along as the standard deviation exists for the assumed distribution.) For the $99\,\%$ confidence under a normal distribution, the multiple is $2.33$ . So, in the first example, $2.33 \times 0.4 - 0.5 = 0.43$ .