When Mean-Variance Utility and Maximum Sharpe Optimization Differ
Summary
The document examines whether maximizing a mean-variance objective is equivalent to maximizing a portfolio’s Sharpe ratio. It contrasts a quadratic objective that trades expected return against variance through a risk-aversion parameter with a ratio objective that divides expected return by volatility. One response says the objectives are not generally equivalent in practice, while explaining why the quadratic form is easier to optimize under common portfolio constraints.
A second response emphasizes that the comparison depends on how the problem is posed, including the risk-free asset and the budget constraint. Mean-variance optimization across different risk-aversion settings traces the efficient frontier, whereas Sharpe maximization selects a tangency portfolio under suitable assumptions. The discussion is conceptual and supplies no full proof or numerical example; equivalence depends on portfolio constraints and model setup.
Key ideas
- Mean-variance utility and Sharpe-ratio maximization have different objective functions and are not generally identical.
- Changing risk aversion in a mean-variance problem can trace different points on the efficient frontier.
- Sharpe maximization selects a tangency portfolio under assumptions that include an appropriate risk-free asset and constraints.
- Quadratic mean-variance objectives are often easier to optimize with practical portfolio constraints than the ratio objective.
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Full text
# mean variance optimization vs max sharpe ratio
# mean variance optimization vs max sharpe ratio
I keep reading/hearing that the results from mean-var optimization is max Sharpe ratio. It seems making sense if you fix either target return or target risk, but in general, it doesn't seems right, for example, $J1$ and $J2$ are target function:
$J1 = \mu\prime w - \lambda w\prime\Sigma w.$
$J2 = (\mu\prime w)/\sqrt{w\prime\sigma w}$
The optimal solution of $J1$ and $J2$ should be very different, because $J1$ depends on lambda, $J2$ does not, not to mention the derivatives respect to w are very different.
what am I missing here?
## Answer by NegativeJo (score 2)
https://quant.stackexchange.com/a/36603
In theory in the case of a constrained optimisation and in practice they are not.
However... A lot of practitioner wants to achieve the best Sharpe Ratio for their portfolio. But as you describe it in J2 the term is not linear nor quadratic and is much harder to optimise especially in the context of the multitude of constraints that would occur in a typical portfolio optimisation framework
J1 is nicely quadratic so it is a lot easier to optimise. And it has this nice property that you would want to maximise u'w and minimise wSw which aligns in terms of conceptual goals with getting the best possible Sharpe Ratio
But in reality they are not equivalent and J2 is highly unpractical and rarely used. Also with J2 a passive portfolio with 0 tracking error would be always the best solution in the absence of other constraints... So the vast majority of practitioner would use a variant of J1
## Answer by markowitz (score 0)
https://quant.stackexchange.com/a/42714
You do not find a correct solution because, mathematically speaking, the problem are not well posed.
Firstly, in $J1$ seems that you have in mind the pure risky ptf while in the $J2$ case this is not possible. For simplification you can assume that $r_f =0$ but it exist, otherwise Sharpe ratio don’t have any sense.
Moreover, probably you have in mind the unconstrained version but also in this case you have to note that in $J1$ like in $J2$ case the the minimal constraint $w’1=1$ holds.
Third and, and maybe most important, in $J1$ the optimization strategy return you entirely the efficient frontier (through lambda) while $J2$ return you only one point. In this point $w$= tangent ptf.
The problem that, probably, you have in mind is well posed if in $J1$ you add the riskless asset. In this case $w$ is interpretable as the weight of risky ptf and $(1-w)$ as riskless asset weight. Then $w$ become unique and equal to tangent ptf as in $J2$ case. I got the proof but it is not short and now it is in some notebook.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.