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FRM Part II · FRM Exam Part II · Alpha (and the Low-Risk Anomaly)

A long-only manager with a benchmark tracking error of 5% has an IR of 0.6. The manager can instead run a long-short version with the same IC but constraints removed, which raises the transfer coefficient from 0.5 to 0.8, leaving breadth and IC unchanged. Using IR = TC × IC × √BR, and assuming the original IR of 0.6 reflects TC = 0.5, what is the new IR and the alpha at 5% active risk?

The new information ratio is 0.96 and alpha is 4.8%. Dividing the original 0.6 by a transfer coefficient of 0.5 gives an unconstrained IR of 1.2, which times the new coefficient of 0.8 equals 0.96. Multiplying by 5% active risk gives 4.8%.

  1. AIR 0.96; alpha 4.8%Correct
  2. BIR 0.75; alpha 3.75%
  3. CIR 0.96; alpha 3.0%
  4. DIR 1.20; alpha 6.0%

Explanation

IC×√BR = 0.6/0.5 = 1.2. New IR = 0.8 × 1.2 = 0.96. Alpha = 0.96 × 5% = 4.8%. Option 1.20 ignores the transfer coefficient of 0.8; 3.0% uses the old IR.

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