Why Different Expected Returns Can Produce the Same Portfolio Weights
Summary
The document addresses whether a positive-weight, minimum-variance portfolio optimization maps each expected-return vector to a distinct set of weights. Its answer is no: distinct return inputs can lead to the same portfolio, so the mapping from expected returns to optimal weights need not be one-to-one. A two-asset example uses two different expected-return vectors and reports the same equal-weight solution for each, with the other optimization inputs held fixed.
It also distinguishes this question from uniqueness in optimization. A unique solution means that a particular problem has exactly one optimizer; it does not mean that different inputs must produce different optimizers. The document offers a counterexample rather than a general proof and does not fully develop the assumptions behind its dimensionality explanation. Its central lesson is about interpreting uniqueness, not about guaranteeing portfolio stability or the uniqueness of every Markowitz solution.
Key ideas
- The mapping from expected returns to optimal portfolio weights need not be injective.
- Different expected-return vectors can produce the same weights in the stated example.
- A unique optimizer for one problem does not imply a one-to-one mapping across different problems.
- The document gives a counterexample and does not establish conditions for uniqueness in every portfolio optimization.
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# Markowitz optimization - can two sets of returns produce the same set of weights?
# Markowitz optimization - can two sets of returns produce the same set of weights?
The portfolio optimization problem I have in mind is a minimum variance optimization with positive weights, formulated as below:
I am trying to show that the solution is unique, specifically in the sense that you could not input two different sets of returns and get the same portfolio (holding all other inputs and higher order moments constant). Is there a concise way to demonstrate that this holds?
Furthermore - does this property hold for all possible portfolios, or only those that lie on the efficient frontier?
## Answer by Matthew Gunn (score 4, accepted)
https://quant.stackexchange.com/a/42489
Yes, two different set of returns can lead to the same weights (so you won't be able to prove the opposite). Also, the term "unique solution" means something different than how you used it.
#### Taking $p$ and $Q$ as given, the mapping from $\boldsymbol{\mu}$ to solutions $\mathbf{x}^*$ is not injective
I'll give a simple counterexample that shows the mapping isn't injective. Let's assume $p=2$. $Q$ can be anything. Now consider two scenarios:
- Scenario 1: $\boldsymbol{\mu} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$
- Scenario 2: $\boldsymbol{\mu} = \begin{bmatrix} 0 \\ 4 \end{bmatrix}$.
The solution to your above optimization problem in both cases is $ x= \begin{bmatrix}.5 \\ .5 \end{bmatrix}$. Two different inputs map to the same output, hence the mapping isn't injective.
What's going on?
If you have $k$ assets, vector $\boldsymbol{\mu}$ lives in a $k$ dimensional space. On the other hand, the constraints on $\mathbf{x}$ imply that solution $\mathbf{x}^*$ can only take values in a $k-2$ dimensional space. You're not going to have a one to one correspondance between spaces of different dimension.
#### What people mean by unique solution
When people say an optimization problem has a unique solution, they mean it has a solution and the solution is unique.
They do not mean that the mapping from optimization problem inputs to optimal choice variables is an injective function.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.