Math Is Fun Forum

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Computing Expectation of Negative returns

Hi boys and girls

I recently encounter an interesting task on studying fund rating methodologies. And I fund Lipper's Preservation Measure is

Sum(Min(0,ri))/T or Sum(Min(0,ri))/N*(T/N)

What it actually does is turn all the positive return r's to 0 and compute the average.

If we assume r normally distributed as N(u,s²)

The negative expectation can be modeled as  E- =∫r*pdf dr   over (-∞,0)

I came up with the answer

E-= u*N(-u/s)-(s/√2π)*exp(-u²/2s²)

But Michael Stutzer in his paper Mutual Fund Ratings: What is the Risk in Risk-Adjusted Fund Returns? derived an approximation as

Could you check this out and tell me why the difference? Thanks!

Last edited by George,Y (2010-06-23 14:52:16)

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Computing Expectation of Negative returns

Here is the paper.  I can't find the computation you claim.

Also, be careful George.  You need to assume an infinite amount of real numbers in order to calculate that integral

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Computing Expectation of Negative returns

at page 19

"You need to assume an infinite amount "
If infinite independent factors were real, normal curve would have explained everything. Unfortunately, there is always "fat tail" phenomenon, which indicates only finite factors in the real world.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Computing Expectation of Negative returns

page 39 appendix

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X'(y-Xβ)=0