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 $x_n\rightarrow x= (\mu,\sigma)$?


1 Answer 1


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 closed


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