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.
1 Answer
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 $\lambda$.
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$\begingroup$ 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 $\endgroup$ Jun 28, 2019 at 14:34
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$\begingroup$ @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, 2019 at 14:37
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$\begingroup$ so if we forget about the coefficient $\lambda$ and look at the more practical non-utility portfolio optimization,then $\min w'\Sigma w \neq \min (1/2) w'\Sigma w$ $\endgroup$ Jun 28, 2019 at 14:40
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$\begingroup$ I don't follow the argument, what happened to $\omega' \mu$ in your equation? $\endgroup$– Bob Jansen ♦Jun 28, 2019 at 14:42