Why Markowitz Variance Objectives Use a Negative Sign and One-Half

Summary

The document explains two features of a rewritten global minimum variance portfolio objective: the negative sign and the factor of one-half. A variance-minimization problem can be written as a maximization by negating the objective, which accounts for the minus sign. Multiplying the objective by a nonzero constant does not change the location of its optimum, so the one-half does not alter the portfolio weights that solve the problem.

The factor of one-half is included to simplify differentiation. In a quadratic expression, differentiating introduces a factor of two; the one-half cancels it, making first-order conditions cleaner. The explanation is conceptual and uses a two-dimensional illustration. It does not derive the full constrained Markowitz solution or discuss how constraints, expected returns, or covariance estimation affect portfolio construction.

Key ideas

• Negating variance converts a minimization objective into an equivalent maximization objective.
• Multiplying the objective by one-half changes its value but not the location of its optimum.
• The one-half factor cancels the factor of two that appears when differentiating a quadratic term.
• The explanation addresses objective notation and derivatives, not the full constrained portfolio solution.

Tags

# Formula in Markowitz Optimization Problem (without riskless asset)


# Formula in Markowitz Optimization Problem (without riskless asset)












(hope this is not too basic, I'm new to this forum) Im struggling to understand the optimization problem (global minimum variance portfolio) formula in Markowitz Theory:

$$\arg\ \min\ Var(Return\ x) = [\max_x (-\frac{1}{2} x^{\mathrm{T}}Vx)]$$

The only thing I dont understand is where the -1/2 is coming from, in all the sources I could find it wasn't explained and just taken as given...

Thanks in advance

## Answer by AdB (score 2, accepted)

https://quant.stackexchange.com/a/44984

Note that the solution to the problem is the same with or without the $\frac{1}{2}$, since multiplying it only changes the value of the objective function, but not where its extrema are located. The "$-$" comes from the fact that you switched it from a minimization problem to a maximization problem.

The reason behind the choice of $\frac{1}{2}$ simply makes the derivative nicer. Take the 2-dimensional case, where $x$ is squared. Then the $\frac{1}{2}$ cancels out the $2$ from taking the derivative with respect to $x$, and the first order conditions look neater.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.