Estimating One-Year VaR from Daily Returns and Overlapping Windows

# 1-year Var calculated from 1 year volatility


# 1-year Var calculated from 1 year volatility












I need to calculate Var with 1 year horizon.

I think the correct way to do it is to calculate standard deviation of daily log returns, then calculate daily Var and multiply it by sqrt(250).

But in our company there is another formula: 1. so at to calculate returns they divide today's level by level that was one year ago; 2. then take standard deviation of these returns 3. calculate Var using this standard deviation.

I believe that this approach is not correct, but I can't find how to prove my point of view.. could you please help)

## Answer by Ami44 (score 2, accepted)

https://quant.stackexchange.com/a/33848

You have Daily returns $x_{i}$ with $i = 1 \ldots N$ and lets call the corresponding yearly returns $a_{i} = \sum_{j=i}^{i+250}{x_{j}}$ with $i = 1 \ldots N-250$.

The $x_{i}$ are i.i.d. with standard derivation $\sigma_{x}$ and expected value $m_{x}$. For the $a_{i}$ we get than a standard derivation of $\sigma_{a} = \sqrt{250} * \sigma_{x}$ and a expected value of $m_{a} = 250 \cdot m_{x}$

If I understand you correctly you are comparing the following two approaches to estimate the 1 year standard derivation $s_{a}$:

- $s_{a} = \sqrt{\frac{250}{N-1} \cdot \sum_{i=1}^{N}{(x_{i}-\bar{x})^{2}}} = \sqrt{250} \cdot s_{x}$

- $s_{a} = \sqrt{\frac{1}{N-250-1} \cdot \sum_{i=1}^{N-250}{(a_{i}-\bar{a})^{2}}}$

Approach 1 is correct if the $x_{i}$ are truly independent. In the real world that is often only approximatly true. Nevertheless it is a very popular approach.

As you suspected, approach 2 is seriously flawed. The problem is, that the $a_{i}$ are not independent by design, which means the standard formula for estimating the standard deviation does not apply. $a_{i}$ and $a_{i+1}$ overlap in 249 $x_{i}$ and are therefore heavily dependent. The autocorrelation function $\rho_{k}$ which is the correlation between any $a_{i}$ and $a_{i+k}$ is for $k < 250$ given by $$\rho_{k} = corr(a_{i}, a_{i+k}) = (1- \frac{k}{250}) \cdot $$ For $k \geq 250$ it's $\rho_{k} = 0$.

As you can see in wikipedia in the presence of autocorrelation in the sample your result will be distorted like $$E(s^{2}) = s_{a} \cdot \left[ 1 - \frac{2}{N-250-1} \cdot \sum_{k=1}^{N-250}(1-\frac{k}{N-250}) \cdot \rho_{k} \right]$$

Which becomes in the case of the above auto correlation function: $$E(s^{2}) = s_{a} \cdot \left[ 1 - \frac{2}{N-250-1} \cdot \sum_{k=1}^{250}(1-\frac{k}{N-250}) \cdot (1-\frac{k}{250}) \right]$$

What you can see is that using approach 2 the estimate of the standard derivation will always be lower than the true value. The error will be smaller though the bigger your sample is compared to your aggregation interval. So if you have data from e.g. 50 years, the error of approach 2 might disappear.

Another approach might be to use only $a_{i}$ that don't overlap, i.e. use every 250 value. But than you loose a lot of information. In your example with a sample size of 500 you end up with just two values to estimate the standard deviation from.