Why is VaR proportional to square root of time? standard, should, sigma

meghanjackson

Why is VaR proportional to square root of time?
We know that VaR=R-z*sigma. R is proportional to time, standard deviation is proportional to square root of time. Therefore, VaR should NOT be proportional to square root of time. What am I missing?

iteracom

It’s because VaR is generally calculated from the standard deviation rather than the variance (which is SD^2).
If daily returns are stationary (mean and SD and other key statistical features over similar time periods do not change) and not autocorrelated (past changes do not inform the direction of future changes), then the total variance will rise linearly with time.  Basically, the longer you give a random process time to evolve, the farther you expect it to drift from where you initially were, and that’s what the total variance measures.
In practice, you might find momentum or mean reversion in these processes, and that will tend to show up as autocorrelation, which spoils the assumption (but can be corrected for as long as it’s stationary).
When you calculate the standard deviation by taking the square root of variance, you are also taking the square root of time.
From the standard devation, you can then calculate a bottom-most percentage return (and turn it into a $ figure if you desire).

brazilatz

var is proportional to sqrt of time when R=0