Why the Long-Only Mean-Variance Opportunity Set Is Closed
Summary
The document explains why the mean and volatility pairs attainable from long-only portfolios in a finite asset universe form a closed set. It frames portfolio construction as a continuous map from asset weights in the simplex to the portfolio’s mean and standard deviation. The weights are nonnegative and sum to one, so the feasible weight set is closed and bounded, hence compact.
For any convergent sequence of attainable mean-volatility pairs, choose a portfolio producing each pair. Compactness guarantees a convergent subsequence of those portfolios’ weights; its limit remains feasible. Continuity of the map then shows that the limiting weights produce the original sequence’s limiting mean-volatility pair, establishing closure. The argument does not rely on the weights alone being compact: it also needs the continuous mapping from weights to portfolio statistics. The result is scoped to finitely many assets and long-only weights; the note offers no extension to short-selling or other portfolio constraints.
Key ideas
- Portfolio weights for long-only portfolios with weights summing to one form a compact simplex.
- Portfolio mean and standard deviation vary continuously with the asset weights.
- A convergent sequence of attainable portfolio statistics has a subsequence of weights converging to a feasible portfolio.
- Continuity ensures that the limiting feasible portfolio attains the limit of the statistics sequence.
- The closure argument depends on both compactness of the weight set and continuity of the portfolio-statistics map.
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# Proof that mean-variance opportunity set is closed
# Proof that mean-variance opportunity set is closed
In the book Financial Economics (2010) by Hens and Rieger, on page 101 we find the following Lemma 3.1: If we have finitely many assets, the minimum-variance opportunity set is closed and connected.
We have: K number of assets, a sequence of points $x_n = (\mu_n, \sigma_n) (n=1,2,...)$ in the opportunity set with $x_n\rightarrow x= (\mu,\sigma)$, each $x_n$ corrsponds to a portfolio characterized by asset weights $\lambda_1^n,...,\lambda_K^n$ with $\lambda_k^n\geq 0$ for all $k = 1,...,K$ and $\sum_{k=1}^{K}\lambda_k^n=1$
The proof that the mean-variance-opportunity set is closed, it is stated that the vector of asset weights $\lambda = (\lambda_1^n, ... ,\lambda_k^n)$ is for all $n\in \mathbb{N}$ in a compact set. Does the proof infer that since $\lambda$ is compact (ie closed and bounded), the opportunity set also must be closed? Why does an arbitrary point $x_n\rightarrow x= (\mu,\sigma)$?
## Answer by algebruh (score 1, accepted)
https://quant.stackexchange.com/a/67996
So the setting is as follows. We have a map $F: A \to \mathbb{R^2}, \lambda \to (\mu_\lambda, \sigma_\lambda)$ where $\mu_\lambda, \sigma_\lambda$ are the mean and variance of the portfolio with weights $\lambda$ and $A=\{\lambda_1, \ldots, \lambda_K|\lambda_i \geq 0 \text{ and } \sum_i \lambda_i=1\} $. Consider a sequence $(\mu_n, \sigma_n)$ converging to $(\mu,\sigma)$. Since they are result of a portfolio they are in the range of $F$. Take a point $\lambda^n \in F^{-1}((\mu_n, \sigma_n))$. Then $(\lambda^n)_{n} $ is a sequence in $A$. Since $A$ is bounded there exists a converging subsequence $(\lambda^{n_i})_i$. Since $A$ is closed the limit of this sequence $\tilde{\lambda}$ is in $A$.Since $F$ is continuous we have
$$ F(\tilde{\lambda})= F(\lim\limits_{i\to \infty} \lambda^{n_i}) =\lim\limits_{i\to \infty} F(\lambda^{n_i}) = \lim\limits_{i\to \infty} (\mu_{n_i}, \sigma_{n_i}) =\lim\limits_{n\to \infty} (\mu_{n}, \sigma_{n}) = (\mu,\sigma)$$
So we have found a portfolio with weights $\tilde{\lambda}$ that has the mean and variance of the limit. Thus the mean-variance opportunity set is closedShown 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.