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IC reduction in Grinold

Kashif Khalid

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Hi @David Harper CFA FRM CIPM

This is probably a trivial question.

In the notes to Grinold (and in the reading itself) about Alpha scaling it states:

...the original alphas have a standard deviation of 2.00 percent and the modified alphas have a standard deviation of 0.57 percent. This implies that the constraints in this example effectively shrank the IC by 62 percent, a significant reduction.

using Alpha = Vol * IC * Score how do we get to IC being reduced by 62% in the above? Score we assume is 0 so I tried to scale the average alpha of the original and modified and i got 72% reduction but I have a feeling I'm reading this incorrectly!.


Thanks
 

David Harper CFA FRM

David Harper CFA FRM
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Hi @Kashkha This has me stumped :eek: But here is something interesting if I work backwards from his conclusion:
  • Let alpha = sigma * IC, such that IC = alpha/sigma
  • IC[original] = .02/sigma
  • Solving for IC[modified]/IC[original] = 0.380; i.e., "shrinking the IC by 62%"
  • IC[modified] = 0.380*IC[original] = (0.380*0.02)/sigma, and if IC[modified] = 0.570/(sigma*X) then:
  • 0.570/(sigma*X) = (0.380*0.02)/sigma, and
  • X = 0.570/(0.380*0.02) = 0.750
In other words, under the formula (alpha = sigma * IC), if alpha drops from 2.0% to 0.575 and if volatility is multiplied by exactly 0.750 (i.e., i think we could call this a 25 percentage reduction but it's not -250 basis points but a ratio; it's not 25 percentage points), then the IC shrinks by 62% (i.e., the new IC is 0.38 times the original). That's my guess, is that the residual risk shrinks to make this true ...
 
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