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In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$ and that linear combination has no variance (because $\mathbf{w}$ lies in the null space of $C$).

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{1}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that estimating $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.

In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$ and that linear combination has no variance (because $\mathbf{w}$ lies in the null space of $C$).

(As @kjetil b halvorsen points out and explores more deeply in his answer, negative weights aren't limited to degenerate cases like this.)

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{1}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that estimating $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.

In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$ and that linear combination has no variance (because $\mathbf{w}$ lies in the null space of $C$).

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{w}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that estimating $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.

In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$ and that linear combination has no variance (because $\mathbf{w}$ lies in the null space of $C$).

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{1}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that estimating $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.

In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$. The linear combination has no variance (because $C$ is rank deficient).

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{w}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that computing $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.

In abstract theory, yes, because of the correlation structure.

If you had uncorrelated measurements (with positive variance), then the weights could only be positive.

Example:

Let $\mu$ denote the true value. Let noisy measurement $X_1 \sim \mathcal{N}(\mu, 1)$. Let $X_2 = 2X_1 - \mu$, hence $\operatorname{E}[X_2] = \mu$ but $X_2$ is perfectly correlated with $X_1$. The covariance matrix is $ C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $.

The solution is $\mathbf{w} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} $. Observe $2X_1 - X_2 = \mu$ and that linear combination has no variance (because $\mathbf{w}$ lies in the null space of $C$).

An equivalent finance problem:

An almost equivalent problem is solving for the minimum variance portfolio in finance.

Let $R$ denote a vector of $n$ returns. Let $C = \operatorname{Cov}(R)$ denote the covariance matrix of $R$. Let $\mathbf{w}$ denote a vector of portfolio weights.

The minimum variance portfolio is found by solving:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}'\Sigma \mathbf{w} \\ \mbox{subject to} & \mathbf{w}'\mathbf{1} = 1 \end{array} \end{equation}

This is exactly the same problem and it has exactly the same solution. For invertible $C$:

$$ \mathbf{w}_{mvp} = \frac{C^{-1} \mathbf{w}}{\mathbf{1}'C^{-1}\mathbf{1}}$$

Perhaps in the portfolio context, it's more intuitive that the minimum variance portfolio may involve both going long and short assets? (Note: before you run off and try to start an investment fund realize that estimating $C$ has big time problems.)

Some linear algebra interpretation

Let $U'\Lambda U = C$ be the eigenvalue decomposition of $C$. (This is basically PCA). Then $Y = R U$ is a random vector of uncorrelated random variables whose variance is given by the diagonal matrix of eigenvalues $\Lambda$. The minimum variance portfolio will give you positive weights on these random variables $Y_1, \ldots, Y_n$ but since these components are themselves linear combinations of security returns, you may get positive and negative weights in security weight space.