In Robert Merton's derivation of the efficient frontier of a portfolio, he minimizes $\frac{1}{2}\sigma^2 $ over the investment weights in each asset, where $\sigma^2$ represents portfolio variance. I am confused why the function he minimizes is half the variance, instead of just the variance. It doesn't make a difference in calculations, but I cannot figure out why he (and all other derivations) do this.


I assume you're talking about this formula:

$$U(w) = w'\mu - \frac{1}{2} \lambda w' \Sigma w = w'\mu - \frac{1}{2} \lambda \sigma_\omega^2$$

where $\sigma_\omega^2$ denotes the portfolio variance for a portfolio with weights $\omega$.

Dividing by two is purely done for convenience, optimizing this formula requires taking the derivative with respect to $\omega$ and setting it to $0$. When the derivative is taken the factor $\frac{1}{2}$ is canceled by the square.

See this question on more information on setting $\lambda$.

| improve this answer | |
  • $\begingroup$ Clear answer - there is nothing more to it. $\endgroup$ – Ric Oct 1 '14 at 7:02
  • $\begingroup$ does having 1/2 in the front of the objective function give a different solution for $w$ than when there is no 1/2? If so, how to account for the discrepancy between the two approaches $\endgroup$ – develarist Jun 28 '19 at 14:34
  • $\begingroup$ @develarist This might be material for a different question. The answer is that it depends: you can change $\lambda$ to compensate. With fixed $\lambda$, yes the result will change and I expect the change to be significant in all cases encountered in practice. $\endgroup$ – Bob Jansen Jun 28 '19 at 14:37
  • $\begingroup$ so if we forget about the coefficient $\lambda$ and look at the more practical non-utility portfolio optimization,then $\min w'\Sigma w \neq \min (1/2) w'\Sigma w$ $\endgroup$ – develarist Jun 28 '19 at 14:40
  • $\begingroup$ I don't follow the argument, what happened to $\omega' \mu$ in your equation? $\endgroup$ – Bob Jansen Jun 28 '19 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.