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Daily delta-normal VaR

Hi David,

If given an annual drift and annual volatility, how should one go about finding the daily VaR?

I appologize, but this is from another source and I believe they did it incorrectly.

I think you divide the annual volatility by sqrt(250) and the annual drift by 250 and then use

VaR= (daily drift - 2.33 * daily vol)* size of position

The other source said to first find the annual VaR (using annual drift and annual vol) and then divide by sqrt(250). I was under the impression that you could only do this if there was no drift component.

Which way is correct?


David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi Mike,

you are exactly correct, you cannot do what the other source suggests. you can't apply the S.R.R. to VaR if it contains a non-zero drift. For example, if the absolute VaR is $100, as there is no way of knowing how much is due to drift, you cannot scale the number directly in any way, it must be parsed into the two components before scaling.
(this is the same as saying, you can S.R.R. scale the relative VaR but not the absolute VaR as the absolute VaR contains drift).
Similar, i suppose, to the idea we discussed on another thread, the restriction to relative VaR in the mean-variance formula (http://www.bionicturtle.com/forum/threads/portfolio-var.4846/)
... I think what happens is that that many folks implicitly are referring to relative VaR, without saying it. But this is imprecise, the "default assumption" should be absolute VaR and then it can be relaxed with an explicit statement that drift is zero. The reason this matters is that, nowadays, VaR is not just daily market VaR, it's credit and operational and longer-term market VaR.

but here is just a case that volatility and drift don't scale the same, so they can't be scaled together. Drift scales with time, but volatility scales with square root of time (b/c variance scales with time). So, I am aware of no way to reduce an absolute VaR (itself the sum of drift and volatility) to treatment, in one swoop, with the square root rule.

thanks, David