VaR非弱增性的證明 [推广有奖]

VAR is not sub-additive

There are different ways to measure risk. Harry Markowitz measured risk as a standard deviation, so that is the norm in portfolio theory. Today, in VAR and credit work, risk is often measured as a percentile (or an absolute value of a percentile). There is no "right" approach. Each has its own advantages and disadvantages. One difference between standard deviations and percentiles is that the former satisfies the sub-additive property. The latter does not.

Suppose you have two random variables X1 and X2 with standard deviations S1 and S2 and percentiles (5% level) of P1 and P2. Let Y = X1 + X2. Without knowing anything else, we can say that the standard deviation S of Y is between 0 and S1 + S2. All we can say about the percentile P of Y is that it is between 0 and infinity.

I’ll give you a concrete example. In this example, the two random variables X1 and X2 will each have a (5% ) percentile of P1 = P2 = -10. Their sum Y, however, will have a (5%) percentile P = -1,000,000.

Suppose X1 and X2 are independent random variables. Assume they are identically distributed with the discrete probability distribution:

Pr(-1,000,000) = 0.03

Pr(-10) = 0.03

Pr(0) = 0.94.

In this case, the (5%) percentiles satisfy P1 = P2 = -10. However, based upon the independence of X1 and X2, it is easy to calculate the probability distribution of Y. It is:

Pr(-2,000,000) = .0009

Pr(-1,000,010) = .0018

Pr(-1,000,000) = .0564

Pr(-20) = .0009

Pr(-10) = .0564

Pr(0) = .8836

From this, we see that the (5%) percentile P of Y is -1,000,000.

Obviously, this is a contrived example to demonstrate a mathematical property of percentiles. In practice, VAR results are usually much better behaved. So, while it is possible in theory for VAR results to violate the sub-additive property. In practice they generally do not. If you are experiencing such violations, you should probably investigate your analysis to understand why.

[此贴子已经被作者于2007-5-18 19:55:14编辑过]