This question appeared in one of the university exams for portfolio management. I’ve provided the question along with the detailed answer.

Consider the two (excess return) index-model regression results for Portfolio A and B. The risk-free rate over the period was 6%, and the market’s average return was 14%. Performance is measured using an index model regression on excess returns.

**Results for Portfolio A**

r_{A} – r_{f} = 1% + 1.2(r_{M} – r_{f})

R-square = 0.57;

Residual standard deviation, s(e_{A}) =10.3%;

Standard deviation of (r_{A} -r_{f}) = 26.1%.

**Results for Portfolio B **

r_{B} – r_{f} = 2% + 0.8(r_{M} – r_{f})

R-square = 0.43;

Residual standard deviation, s(e_{B}) =19.1%;

Standard deviation of (r_{B} -r_{f}) = 24.9%.

**Solution**

To compute the Sharpe measure, note that for each portfolio, (r_{p} – r_{f}) can be computed from the right-hand side of the regression equation using the assumed parameters r_{M} = 14% and r_{f} = 6%.

The standard deviation of each stock’s returns is given in the problem.

The beta to use for the Treynor measure is the slope coefficient of the regression equation presented in the problem.

A | B | |

Alpha is the intercept of the regression (a) | 1% | 2% |

Information ratio or Appraisal ratio = a/s(e) | =1%/10.3%) = 0.097 | =2%/ 19.1% = 0.1047 |

Sharpe measure = (r_{p}– r_{f})/ s
| =(1% + 1.2(14% – 6%))/26.1%
0.4061 | =(2% + 0.8(14% – 6%))/24.9%
0.3373 |

Treynor measure = (r_{p} – r_{f})/ b | =(1% + 1.2(14% – 6%))/1.2
0.833 | =(2% + 0.8(14% – 6%))/0.8
10.5 |