Chapter 03

Hard

Mean-Variance Optimization

00 · Symbol Glossary

$\mathbf{w}$w — portfolio weights

Column vector in $\mathbb{R}^N$ . Coordinate $w_i$ is the share of the book in asset $i$ . A negative coordinate is a short sale. Fully invested means $\mathbf{1}^T\mathbf{w}=1$ .

$\boldsymbol{\mu}$mu — vector of expected returns

$\boldsymbol{\mu}\in\mathbb{R}^N$ with $\mu_i=\mathbb{E}[R_i]$ . The portfolio mean is the scalar $\mu_p=\boldsymbol{\mu}^T\mathbf{w}$ . Chapter 02 already has the one-line version $\mu_p=\sum_i w_i\mu_i$ .

$\mu_\star$mu star — target mean

A number the investor requires: $\boldsymbol{\mu}^T\mathbf{w}=\mu_\star$ . It is an input, not an optimum. Different targets produce different books.

$\Sigma$Sigma — covariance matrix

Symmetric $N\times N$ matrix, $\Sigma_{ij}=\mathrm{Cov}(R_i,R_j)$ . Portfolio variance is $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ . The optimizer below needs $\Sigma$ positive definite, which is Linear Algebra Chapter 14.

$\mathbf{1}$ones vector

The column of $N$ ones. $\mathbf{1}^T\mathbf{w}=\sum_i w_i$ is the budget, the sum of the weights.

$\lambda_1,\lambda_2$Lagrange multipliers

$\lambda_1$ prices the mean constraint and $\lambda_2$ prices the budget. They are solved for, together with $\mathbf{w}$ . They are not the risk-aversion parameter.

$\lambda$lambda — risk aversion

Positive scalar in the objective $\boldsymbol{\mu}^T\mathbf{w}-\frac{\lambda}{2}\mathbf{w}^T\Sigma\mathbf{w}$ . Large $\lambda$ punishes variance more heavily. A different object from $\lambda_1$ and $\lambda_2$ .

$A,B,C,\Delta$the four scalars in the closed form

$A=\mathbf{1}^T\Sigma^{-1}\mathbf{1}$ , $B=\mathbf{1}^T\Sigma^{-1}\boldsymbol{\mu}$ , $C=\boldsymbol{\mu}^T\Sigma^{-1}\boldsymbol{\mu}$ , and $\Delta=AC-B^2$ . All four are numbers. $\Delta\gt 0$ when $\Sigma$ is positive definite and the assets do not all share one mean.

$\mathbf{w}_{\mathrm{GMV}}$global minimum-variance portfolio

The fully invested book of smallest variance, with no target mean imposed. $\mathbf{w}_{\mathrm{GMV}}=\Sigma^{-1}\mathbf{1}/A$ , and its variance is $1/A$ .

01 · The Choice

Chapter 02 fixed a weight vector and computed $\mu_p=\boldsymbol{\mu}^T\mathbf{w}$ and $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ . This chapter turns that around. The means and the covariance are the inputs. The unknown is $\mathbf{w}$ .

An investor who cares only about mean and variance faces a quadratic program. The objective is quadratic because variance is the quadratic form $\mathbf{w}^T\Sigma\mathbf{w}$ . The constraints used here are linear equalities: spend the whole budget, and hit a chosen mean. A quadratic objective with linear equality constraints is the Markowitz problem in the form where a closed form still exists.

Two modeling choices are already settled, and both are easy to reopen by accident.

The budget is $\mathbf{1}^T\mathbf{w}=1$ . Drop it and the coordinates of $\mathbf{w}$ are no longer portfolio weights: they are an unscaled position, and $\boldsymbol{\mu}^T\mathbf{w}$ is no longer a return on wealth. The zero vector has variance zero and is the unconstrained minimizer of $\mathbf{w}^T\Sigma\mathbf{w}$ . It is also not a portfolio.

Short sales stay legal. A coordinate of $\mathbf{w}^\star$ may be negative. Forbidding that is an inequality constraint, $w_i\ge 0$ , and the problem becomes a quadratic program whose active set changes with the target. That version is Chapter 10. Nothing in the inverse formula below knows about a no-short rule.

A risk-free asset is not in the menu. Every name in $\boldsymbol{\mu}$ and $\Sigma$ is risky. Mixing cash with a risky book, and the capital market line that comes with it, is Chapter 05.

02 · Two Programs

The same preference can be written in either of two ways. They select the same books, for a reason derived in Section 08. The algebra is different enough that both are worth setting up.

Definition — Target-mean program

Choose $\mathbf{w}\in\mathbb{R}^N$ to

\min_{\mathbf{w}}\ \frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w} \qquad\text{subject to}\qquad \boldsymbol{\mu}^T\mathbf{w}=\mu_\star, \qquad \mathbf{1}^T\mathbf{w}=1

The factor $\frac{1}{2}$ does not move the minimizer. It makes the gradient of the objective equal to $\Sigma\mathbf{w}$ rather than $2\Sigma\mathbf{w}$ , which keeps the first-order condition free of a stray $2$ .

Definition — Risk-aversion program

For a fixed $\lambda\gt 0$ , choose $\mathbf{w}\in\mathbb{R}^N$ to

\max_{\mathbf{w}}\ \boldsymbol{\mu}^T\mathbf{w}-\frac{\lambda}{2}\mathbf{w}^T\Sigma\mathbf{w} \qquad\text{subject to}\qquad \mathbf{1}^T\mathbf{w}=1

The mean is rewarded one-for-one. Variance is charged at price $\lambda/2$ . There is no separate target $\mu_\star$ : the optimal mean is whatever this tradeoff produces.

The target-mean program is the one with an explicit closed form in $\Sigma^{-1}\boldsymbol{\mu}$ and $\Sigma^{-1}\mathbf{1}$ . The risk-aversion program is the one that says which of those targets a mean-loving investor will actually pick.

Two assets, two equalities, no tradeoff left

With $N=2$ the target-mean program has two unknowns and two linear equalities. If $\mu_1\neq\mu_2$ , the feasible set is a single point, and $\Sigma$ never votes on the weights.

Take $\mu_1=0.06$ , $\mu_2=0.12$ , and $\mu_\star=0.09$ . Then $w_1+w_2=1$ and $0.06w_1+0.12w_2=0.09$ give $w_1=(0.09-0.12)/(0.06-0.12)=1/2$ . The book is half and half for every covariance matrix. $\Sigma$ still sets the variance of that unique feasible book, and it still sets the multipliers, but it does not set $\mathbf{w}$ .

A genuine variance-versus-mean choice inside the target-mean program starts at $N\ge 3$ , where two equalities leave $N-2$ free directions. The risk-aversion program still has a real tradeoff at $N=2$ , because it imposes only the budget.

03 · The Lagrangian

Work with the target-mean program. Attach a multiplier to each equality and build

\mathcal{L}(\mathbf{w},\lambda_1,\lambda_2)=\frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}-\lambda_1(\boldsymbol{\mu}^T\mathbf{w}-\mu_\star)-\lambda_2(\mathbf{1}^T\mathbf{w}-1)

The sign in front of each multiplier is a convention. Flipping both signs flips the signs of $\lambda_1$ and $\lambda_2$ and leaves the combination $\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ unchanged, so the weight formula in the next section is the same either way. With the sign written here, $\lambda_1$ is the derivative of the minimized objective with respect to $\mu_\star$ , and $\lambda_2$ is the derivative with respect to the required sum of weights.

Differentiate in $\mathbf{w}$ . For a symmetric matrix the gradient of the quadratic form is

\nabla_{\mathbf{w}}\Big(\frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}\Big)=\Sigma\mathbf{w}

One line of coordinates is enough to see the factor. The form is $\frac{1}{2}\sum_i\sum_j w_i\Sigma_{ij}w_j$ . The derivative in $w_k$ hits the left factor and the right factor:

\frac{1}{2}\sum_j\Sigma_{kj}w_j+\frac{1}{2}\sum_i w_i\Sigma_{ik}=\sum_j\Sigma_{kj}w_j

Symmetry $\Sigma_{ik}=\Sigma_{ki}$ is what lets the two halves add into $(\Sigma\mathbf{w})_k$ rather than into a half-and-half of $\Sigma$ and $\Sigma^T$ . Feed $\Sigma$ in symmetric. A covariance matrix already is.

The linear terms contribute $-\lambda_1\boldsymbol{\mu}-\lambda_2\mathbf{1}$ . Stationarity, together with the two constraints that were built into $\mathcal{L}$ , is the first-order system.

Definition — First-order conditions

A triple $(\mathbf{w},\lambda_1,\lambda_2)$ is a critical point of the target-mean program when

\Sigma\mathbf{w}=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}, \qquad \boldsymbol{\mu}^T\mathbf{w}=\mu_\star, \qquad \mathbf{1}^T\mathbf{w}=1

If $\Sigma$ is invertible the first block is the weight structure

\mathbf{w}=\Sigma^{-1}(\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1})

The multipliers are still unknown. They are the numbers that make this $\mathbf{w}$ obey the two equalities.

Where the factor one half goes

Minimize $\mathbf{w}^T\Sigma\mathbf{w}$ instead of half of it, and the stationarity condition becomes $2\Sigma\mathbf{w}=\lambda_1'\boldsymbol{\mu}+\lambda_2'\mathbf{1}$ . The weights match the formula above with $\lambda_i=\lambda_i'/2$ . Quote multipliers from a paper only after checking which objective that paper differentiated. The identity $\sigma_p^2=\lambda_1\mu_\star+\lambda_2$ , coming up in Section 05, belongs to the half-objective convention, the one in which $\Sigma\mathbf{w}=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ .

04 · Positive Definiteness

Existence and uniqueness are the content of $\Sigma\succ 0$ , in the sense of Linear Algebra Chapter 14: $\mathbf{x}^T\Sigma\mathbf{x}\gt 0$ for every $\mathbf{x}\neq\mathbf{0}$ , equivalently every eigenvalue of $\Sigma$ strictly positive, equivalently a Cholesky factor with a positive diagonal.

Definition — What $\Sigma\succ 0$ buys

Assume $\Sigma$ is symmetric positive definite, and assume $\boldsymbol{\mu}$ is not a scalar multiple of $\mathbf{1}$ . Then for every target $\mu_\star$ the target-mean program has exactly one solution $\mathbf{w}^\star$ . The multipliers $\lambda_1,\lambda_2$ at that solution are unique as well.

Three separate facts are packed in there.

The objective is strictly convex. The Hessian of $\frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}$ is $\Sigma$ itself. Positive definite Hessian means strict convexity: along any line, the graph of the objective bends upward. A strictly convex function on a convex set has at most one minimizer. The feasible set is an affine subspace, cut out by two linear equalities, so it is convex.

A minimum is attained. Along the feasible set, $\mathbf{w}^T\Sigma\mathbf{w}\to\infty$ as $\|\mathbf{w}\|\to\infty$ . There is no feasible direction $\mathbf{d}\neq\mathbf{0}$ with $\mathbf{d}^T\Sigma\mathbf{d}=0$ , because there is no such direction anywhere. The objective cannot slide off to infinity and stay finite, so a minimizer exists. Combined with strict convexity, it is the only one.

The critical point is that minimizer. An affine constraint set needs no inequality-multiplier theory. The gradient condition $\Sigma\mathbf{w}=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ , plus feasibility, is necessary and sufficient. Sufficiency is the Hessian: the second derivative in $\mathbf{w}$ is $\Sigma\succ 0$ at every point, so the critical point cannot be a maximum or a saddle.

The multipliers are unique when the two constraint gradients are linearly independent. Those gradients are $\boldsymbol{\mu}$ and $\mathbf{1}$ . They are independent precisely when $\boldsymbol{\mu}$ is not a multiple of $\mathbf{1}$ , i.e. when at least two assets have different expected returns. That is the constraint qualification behind the "unique $\lambda_1,\lambda_2$ " half of the definition.

If instead $\boldsymbol{\mu}=k\mathbf{1}$ , every fully invested book has mean $k$ . A target $\mu_\star\neq k$ is infeasible: no weights satisfy both equalities. A target $\mu_\star=k$ makes the mean constraint a copy of the budget. The weight problem collapses to the global minimum-variance portfolio of Section 07, which is still unique under $\Sigma\succ 0$ . The multipliers are no longer unique, because a redundant constraint can have its shadow price split arbitrarily between $\lambda_1$ and $\lambda_2$ . The weight does not care how that split is made: the stationarity condition says $\Sigma\mathbf{w}$ is a multiple of $\mathbf{1}$ either way.

❌ Inverting a covariance that is only semidefinite

Let two assets satisfy $\Sigma=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ . The eigenvalues are $2$ and $0$ , so $\Sigma$ is positive semidefinite and singular. The formula asks for $\Sigma^{-1}$ and stops.

The kernel vector $\mathbf{d}=(1,-1)^T$ has $\Sigma\mathbf{d}=\mathbf{0}$ and $\mathbf{1}^T\mathbf{d}=0$ . Adding $t\mathbf{d}$ preserves the budget and preserves the variance: $(\mathbf{w}+t\mathbf{d})^T\Sigma(\mathbf{w}+t\mathbf{d})=\mathbf{w}^T\Sigma\mathbf{w}$ . Every fully invested book has

\mathbf{w}^T\Sigma\mathbf{w}=(w_1+w_2)^2=1

If the two means are equal, every fully invested book is optimal for the global minimum-variance problem, and the target-mean problem is either the same problem or infeasible. If the two means differ, $\boldsymbol{\mu}^T\mathbf{d}\neq 0$ , so moving along $\mathbf{d}$ hits every target mean at this same variance. The minimum is attained on a whole line, not at a point.

Consequence: a sample covariance from $T$ periods on $N$ assets has rank at most $T-1$ . When $T\le N$ it is singular for this exact reason, and Chapter 14's criterion says it is not positive definite. The closed form is not a formula for that matrix. A computer inverse of a matrix with a tiny eigenvalue is the ill-conditioned cousin of this failure, taken up in Section 09.

05 · The Closed Form

Impose $\Sigma\succ 0$ and substitute $\mathbf{w}=\Sigma^{-1}(\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1})$ into the two equalities. Positive definiteness also puts $\Sigma^{-1}\succ 0$ , because the eigenvalues of the inverse are the reciprocals of the eigenvalues of $\Sigma$ , and reciprocals of positive numbers are positive. The four scalars

A=\mathbf{1}^T\Sigma^{-1}\mathbf{1}, \qquad B=\mathbf{1}^T\Sigma^{-1}\boldsymbol{\mu}, \qquad C=\boldsymbol{\mu}^T\Sigma^{-1}\boldsymbol{\mu}, \qquad \Delta=AC-B^2

are therefore well defined, and $A\gt 0$ . Symmetry of $\Sigma^{-1}$ gives $B=\boldsymbol{\mu}^T\Sigma^{-1}\mathbf{1}$ as well. The inner product $\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{x}^T\Sigma^{-1}\mathbf{y}$ obeys Cauchy–Schwarz, so

(\boldsymbol{\mu}^T\Sigma^{-1}\boldsymbol{\mu})(\mathbf{1}^T\Sigma^{-1}\mathbf{1})\ \ge\ (\boldsymbol{\mu}^T\Sigma^{-1}\mathbf{1})^2

which is $\Delta\ge 0$ , with equality if and only if $\boldsymbol{\mu}$ and $\mathbf{1}$ are linearly dependent. Under the standing assumption that the means are not all equal, $\Delta\gt 0$ .

Step-by-step — Solving the $2\times 2$ system for the multipliers

Write the constraints in $A,B,C$ . Dot the weight formula with $\mathbf{1}$ and with $\boldsymbol{\mu}$ :

B\lambda_1+A\lambda_2=1, \qquad C\lambda_1+B\lambda_2=\mu_\star

Eliminate $\lambda_2$ . The budget row gives $\lambda_2=(1-B\lambda_1)/A$ . Substitute into the mean row and clear the denominator:

AC\lambda_1+B-B^2\lambda_1=A\mu_\star

Hence $\lambda_1(AC-B^2)=A\mu_\star-B$ .

Read off both multipliers. With $\Delta=AC-B^2$ ,

\lambda_1=\frac{A\mu_\star-B}{\Delta}, \qquad \lambda_2=\frac{C-B\mu_\star}{\Delta}

The second one is the same elimination run the other way. From $\lambda_2=(1-B\lambda_1)/A$ and $\Delta+B^2=AC$ ,

\lambda_2=\frac{1}{A}\cdot\frac{\Delta-AB\mu_\star+B^2}{\Delta}=\frac{C-B\mu_\star}{\Delta}

Assemble the weight. Substitute the multipliers back into the stationarity condition:

\mathbf{w}^\star=\Sigma^{-1}(\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1})=\frac{A\mu_\star-B}{\Delta}\,\Sigma^{-1}\boldsymbol{\mu}+\frac{C-B\mu_\star}{\Delta}\,\Sigma^{-1}\mathbf{1}

As $\mu_\star$ varies, $\mathbf{w}^\star$ moves on a straight line in weight space: it is an affine function of the target. The curve those books trace in the $(\sigma_p,\mu_p)$ plane is the subject of Chapter 04.

One identity falls out of stationarity and is worth keeping as a check. Because $\Sigma\mathbf{w}^\star=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ ,

\sigma_p^2=(\mathbf{w}^\star)^T\Sigma\mathbf{w}^\star=\lambda_1\mu_\star+\lambda_2=\frac{A\mu_\star^2-2B\mu_\star+C}{\Delta}

The right-hand side is a quadratic polynomial in the target. Its minimum over $\mu_\star$ sits at $\mu_\star=B/A$ , and the minimal value is $1/A$ . That minimizing target is the mean of the global minimum-variance portfolio, which is the next section. Chapter 04 reads the geometry of this quadratic; the algebra is already finished once the multipliers are known.

06 · Three Assets, Both Ways

The numbers are in percentage points: a mean of $2$ is two percent, and $\Sigma$ is in percent squared. Dividing every mean by $100$ and every entry of $\Sigma$ by $10000$ converts to decimal returns and does not change $\mathbf{w}^\star$ . The integers are chosen so the inverse stays integral.

\boldsymbol{\mu}=\begin{pmatrix}1\\3\\2\end{pmatrix}, \qquad \Sigma=\begin{pmatrix}3/4&1/2&1/4\\1/2&1&1/2\\1/4&1/2&3/4\end{pmatrix}, \qquad \mu_\star=2

The inverse is the integer matrix

\Sigma^{-1}=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}

Check the first column: $\Sigma^{-1}$ times $(3/4,\ 1/2,\ 1/4)^T$ equals $(1,\ 0,\ 0)^T$ . The other two columns match the same way. Leading principal minors of $\Sigma^{-1}$ are $2$ , $3$ , and $4$ , all positive, so Sylvester's criterion from Chapter 14 gives $\Sigma^{-1}\succ 0$ , and therefore $\Sigma\succ 0$ .

Step-by-step — Closed form at a target of $2$

The four scalars. $\Sigma^{-1}\mathbf{1}=(1,\ 0,\ 1)^T$ , so $A=\mathbf{1}^T\Sigma^{-1}\mathbf{1}=2$ . Then $B=\boldsymbol{\mu}^T(1,\ 0,\ 1)^T=1+2=3$ . And $\Sigma^{-1}\boldsymbol{\mu}=(-1,\ 3,\ 1)^T$ , so $C=\boldsymbol{\mu}^T(-1,\ 3,\ 1)^T=-1+9+2=10$ . Hence $\Delta=2\cdot 10-3^2=11$ .

Multipliers. $\lambda_1=(A\mu_\star-B)/\Delta=(4-3)/11=1/11$ and $\lambda_2=(C-B\mu_\star)/\Delta=(10-6)/11=4/11$ .

Weights. $\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}=(5,\ 7,\ 6)^T/11$ , and

\mathbf{w}^\star=\Sigma^{-1}\frac{1}{11}\begin{pmatrix}5\\7\\6\end{pmatrix}=\frac{1}{11}\begin{pmatrix}3\\3\\5\end{pmatrix}

The three coordinates sum to $1$ . The mean is $(1\cdot 3+3\cdot 3+2\cdot 5)/11=22/11=2$ .

Variance, two ways. The multiplier identity gives $\sigma_p^2=\lambda_1\mu_\star+\lambda_2=2/11+4/11=6/11$ . The quadratic formula gives $(A\mu_\star^2-2B\mu_\star+C)/\Delta=(8-12+10)/11=6/11$ . Stationarity itself: $\Sigma\mathbf{w}^\star=(5,\ 7,\ 6)^T/11$ , and dotting with $\mathbf{w}^\star$ recovers $6/11$ .

The same point should come out of elementary calculus, with the two equalities substituted in before any matrix inverse. That route is available because a three-asset problem with two equalities has one free variable.

Step-by-step — One free variable, same weights

Kill the budget. Set $w_3=1-w_1-w_2$ .

Kill the mean. $w_1+3w_2+2w_3=2$ becomes $w_1+3w_2+2(1-w_1-w_2)=2$ , which simplifies to $w_2=w_1$ . One free weight remains. Write it as $x$ , so $\mathbf{w}=(x,\ x,\ 1-2x)^T$ . This tie $w_1=w_2$ is special to the target $2$ ; a different $\mu_\star$ leaves $w_1$ and $w_2$ in a different linear relation.

Variance as a quadratic in $x$ . Applying $\Sigma$ produces $\Sigma\mathbf{w}=\frac{1}{4}(3x+1,\ 2x+2,\ 3-3x)^T$ . Dot with $\mathbf{w}$ :

\sigma_p^2=\frac{1}{4}\Big(x(3x+1)+x(2x+2)+(1-2x)\cdot 3(1-x)\Big)=\frac{1}{4}(11x^2-6x+3)

Minimize. The derivative $(22x-6)/4$ vanishes at $x=3/11$ , and the second derivative $22/4$ is positive. Then $w_3=1-6/11=5/11$ , so $\mathbf{w}^\star=(3/11,\ 3/11,\ 5/11)^T$ , in agreement with the matrix formula. The variance at this $x$ is $6/11$ again.

❌ Reporting a position that does not sum to one

Keep the running $\boldsymbol{\mu}$ and $\Sigma$ , and the target $\mu_\star=2$ , but drop the budget. Stationarity is $\Sigma\mathbf{w}=\lambda_1\boldsymbol{\mu}$ , so $\mathbf{w}=\lambda_1\Sigma^{-1}\boldsymbol{\mu}=\lambda_1(-1,\ 3,\ 1)^T$ . The mean constraint reads $\lambda_1 C=2$ . With $C=10$ this is $\lambda_1=1/5$ , and

\mathbf{w}=\begin{pmatrix}-1/5\\3/5\\1/5\end{pmatrix}

The coordinates sum to $3/5$ , not to $1$ . The variance is $\lambda_1\mu_\star=2/5$ , smaller than the fully invested $6/11$ .

Why it breaks: $2/5$ is the variance of a position with the right mean and the wrong gross exposure. Multiplying by $5/3$ restores the budget and scales the mean from $2$ to $10/3$ . The cheaper variance was never a variance at mean $2$ on a fully invested book.

Consequence: the zero vector is the same mistake taken to the end. It minimizes $\mathbf{w}^T\Sigma\mathbf{w}$ with no constraints, its mean is $0$ , and it is not a portfolio. Both equalities are part of the program.

07 · The Global Minimum-Variance Portfolio

Ignore the target. The investor who wants the smallest variance among fully invested books solves

\min_{\mathbf{w}}\ \frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w} \qquad\text{subject to}\qquad \mathbf{1}^T\mathbf{w}=1

The Lagrangian is $\frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}-\lambda(\mathbf{1}^T\mathbf{w}-1)$ . Stationarity is $\Sigma\mathbf{w}=\lambda\mathbf{1}$ . Under $\Sigma\succ 0$ ,

\mathbf{w}=\lambda\Sigma^{-1}\mathbf{1}, \qquad \lambda A=1, \qquad \mathbf{w}_{\mathrm{GMV}}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}=\frac{\Sigma^{-1}\mathbf{1}}{A}

The variance identity collapses to $\sigma_{\mathrm{GMV}}^2=\lambda=1/A$ , since $\Sigma\mathbf{w}_{\mathrm{GMV}}=(1/A)\mathbf{1}$ and $\mathbf{1}^T\mathbf{w}_{\mathrm{GMV}}=1$ . The mean is whatever that book happens to earn:

\mu_{\mathrm{GMV}}=\boldsymbol{\mu}^T\mathbf{w}_{\mathrm{GMV}}=\frac{B}{A}

This is the target-mean solution at $\mu_\star=B/A$ . The formula for $\lambda_1$ gives $(A\cdot B/A-B)/\Delta=0$ , so the mean constraint's multiplier vanishes exactly there: the target is the one the minimum-variance book already hits. In the running example, $B/A=3/2$ and

\mathbf{w}_{\mathrm{GMV}}=\frac{1}{2}\begin{pmatrix}1\\0\\1\end{pmatrix}, \qquad \sigma_{\mathrm{GMV}}^2=\frac{1}{2}, \qquad \mu_{\mathrm{GMV}}=\frac{3}{2}

Asset 2, the high-variance name, is held at zero weight. That is legal. Nothing in the equality-constrained problem forces an interior holding. Raise the target from $3/2$ to $2$ , as in Section 06, and asset 2's high mean pulls it back in, at weight $3/11$ .

A zero coordinate is still the equality-constrained solution

$\mathbf{w}_{\mathrm{GMV}}$ has a zero, and the Section 06 book is strictly positive, while other targets on this same $\Sigma$ produce a short. Sign patterns are an output. Imposing $w_i\ge 0$ in advance deletes some of these books and is a different program. Chapter 10 treats that quadratic program. This chapter's formula applies when the only restrictions are the two equalities.

08 · One Risk-Aversion Parameter

Now solve the risk-aversion program. The Lagrangian, with the budget multiplier written $\nu$ so it is not confused with $\lambda$ , is

\mathcal{L}=\boldsymbol{\mu}^T\mathbf{w}-\frac{\lambda}{2}\mathbf{w}^T\Sigma\mathbf{w}+\nu(1-\mathbf{1}^T\mathbf{w})

Stationarity is $\boldsymbol{\mu}-\lambda\Sigma\mathbf{w}-\nu\mathbf{1}=\mathbf{0}$ . For $\lambda\gt 0$ and $\Sigma\succ 0$ ,

\mathbf{w}=\frac{1}{\lambda}\Sigma^{-1}(\boldsymbol{\mu}-\nu\mathbf{1})

The budget $\mathbf{1}^T\mathbf{w}=1$ forces $(B-\nu A)/\lambda=1$ , so $\nu=(B-\lambda)/A$ . Rearrangement is the useful form: the optimal book is the global minimum-variance portfolio plus a tilt toward high mean, in the $\Sigma^{-1}$ geometry, scaled by $1/\lambda$ .

\mathbf{w}^\star=\mathbf{w}_{\mathrm{GMV}}+\frac{1}{\lambda}\Sigma^{-1}\Big(\boldsymbol{\mu}-\mu_{\mathrm{GMV}}\mathbf{1}\Big), \qquad \mu_{\mathrm{GMV}}=\frac{B}{A}

The tilt $\Sigma^{-1}(\boldsymbol{\mu}-\mu_{\mathrm{GMV}}\mathbf{1})$ is orthogonal to $\mathbf{1}$ in the ordinary dot product — its coordinates sum to zero — so every $\lambda$ preserves the budget. The mean it delivers is

\mu_p=\mu_{\mathrm{GMV}}+\frac{\Delta}{\lambda A}

For $\lambda\gt 0$ and $\Delta\gt 0$ this mean sits at or above $\mu_{\mathrm{GMV}}$ . Sending $\lambda\to\infty$ kills the tilt and returns $\mathbf{w}_{\mathrm{GMV}}$ . Sending $\lambda$ down toward $0$ inflates the tilt without bound: a vanishing penalty on variance is a license for arbitrarily large long-short positions in the direction $\Sigma^{-1}$ marks as high mean. Every target strictly below $\mu_{\mathrm{GMV}}$ solves the target-mean program for someone who asked for that mean, and it solves the risk-aversion program for no $\lambda\gt 0$ . Chapter 04 draws that distinction on the frontier. The algebra that forces it is the display above.

On the running example, take $\lambda=5$ . Then $\mu_{\mathrm{GMV}}=3/2$ , $\Delta/(\lambda A)=11/10$ , and the optimal mean is $3/2+11/10=13/5$ . The excess-mean vector is $(-1/2,\ 3/2,\ 1/2)^T$ , and $\Sigma^{-1}$ sends it to $(-5/2,\ 3,\ -1/2)^T$ . One fifth of that tilt, added to $\mathbf{w}_{\mathrm{GMV}}$ , is

\mathbf{w}^\star=\begin{pmatrix}1/2\\0\\1/2\end{pmatrix}+\begin{pmatrix}-1/2\\3/5\\-1/10\end{pmatrix}=\begin{pmatrix}0\\3/5\\2/5\end{pmatrix}

Check the first-order condition directly. $\Sigma\mathbf{w}^\star=(2/5,\ 4/5,\ 3/5)^T$ , so $5\Sigma\mathbf{w}^\star=(2,\ 4,\ 3)^T$ and $\boldsymbol{\mu}-5\Sigma\mathbf{w}^\star=(-1,\ -1,\ -1)^T$ . That is $\nu\mathbf{1}$ with $\nu=(B-\lambda)/A=(3-5)/2=-1$ . Budget and stationarity both hold.

The same book comes out of the target-mean formulas at $\mu_\star=13/5$ . There $\lambda_1=(2\cdot 13/5-3)/11=1/5$ and $\lambda_2=(10-3\cdot 13/5)/11=1/5$ , so

\mathbf{w}^\star=\frac{1}{5}\Sigma^{-1}\begin{pmatrix}2\\4\\3\end{pmatrix}=\frac{1}{5}\begin{pmatrix}0\\3\\2\end{pmatrix}

One risk aversion, one target mean, one weight vector. Varying $\lambda\gt 0$ and varying $\mu_\star\ge\mu_{\mathrm{GMV}}$ are two parameterizations of one set of books. The map between them is $\lambda=\Delta/(A(\mu_\star-\mu_{\mathrm{GMV}}))$ .

09 · Errors in the Inputs

The map $(\boldsymbol{\mu},\Sigma)\mapsto\mathbf{w}^\star$ is exact for the $\boldsymbol{\mu}$ and $\Sigma$ that are typed in. Those two objects are estimated. This section records how the closed form responds. It does not build a robust estimator.

Differentiate the risk-aversion stationarity condition in the mean, holding $\Sigma$ and $\lambda$ fixed. The Hessian in $\mathbf{w}$ is $\lambda\Sigma$ , so the weight moves by

\mathrm{d}\mathbf{w}=\frac{1}{\lambda}\Sigma^{-1}\,\mathrm{d}\boldsymbol{\mu}\quad\text{plus a budget correction in the direction }\Sigma^{-1}\mathbf{1}

The correction keeps $\mathbf{1}^T\mathrm{d}\mathbf{w}=0$ . The size of the move is governed by $\Sigma^{-1}/\lambda$ . A small risk aversion inflates every input error. A small eigenvalue of $\Sigma$ inflates the component of the error that lies in the corresponding eigenvector, because that eigenvalue is a large eigenvalue of $\Sigma^{-1}$ . Chapter 14's $\lambda_{\min}(\Sigma)$ is the quantitative version of "how positive definite": close to zero means a large inverse and unstable weights.

A two-asset family makes the factor visible. Normalize both variances to $1$ and set the correlation equal to $1-\varepsilon$ with $0\lt\varepsilon\le 1$ , so

\Sigma=\begin{pmatrix}1&1-\varepsilon\\1-\varepsilon&1\end{pmatrix}, \qquad \boldsymbol{\mu}=\begin{pmatrix}1\\1+\delta\end{pmatrix}

The eigenvalue check is immediate: $\Sigma(1,-1)^T=\varepsilon(1,-1)^T$ and $\Sigma(1,1)^T=(2-\varepsilon)(1,1)^T$ . The small eigenvalue is $\varepsilon$ , on the long-short direction, so $\Sigma^{-1}(-1,\ 1)^T=(1/\varepsilon)(-1,\ 1)^T$ . Equal variances force $\mathbf{w}_{\mathrm{GMV}}=(1/2,\ 1/2)^T$ , whose mean is $\mu_{\mathrm{GMV}}=1+\delta/2$ . The excess-mean vector is therefore $(\delta/2)(-1,\ 1)^T$ . Section 08's tilt at $\lambda=1$ multiplies that vector by $\Sigma^{-1}$ :

\mathbf{w}^\star=\begin{pmatrix}1/2\\1/2\end{pmatrix}+\frac{\delta}{2\varepsilon}\begin{pmatrix}-1\\1\end{pmatrix}

The long-short imbalance is the mean gap divided by twice the residual eigenvalue. At $\varepsilon=0.02$ and $\delta=0.02$ the book is $(0,\ 1)^T$ : a mean gap of $0.02$ standard deviations, on a pair with correlation $0.98$ , puts the entire budget in the higher-mean name. Misread the gap as $\delta=0.04$ and the same formula returns $(-1/2,\ 3/2)^T$ , a half-short of the other name. The first-order condition confirms the first of these: $\boldsymbol{\mu}-\Sigma(0,\ 1)^T=0.02\cdot\mathbf{1}$ , which is the budget multiplier $\nu=(\varepsilon+\delta)/2$ .

Nothing here says to shrink $\Sigma$ , to blend $\boldsymbol{\mu}$ toward a prior, or to cap the shorts. Each of those is a different estimator. The only claim is that the Markowitz map is steep exactly where $\lambda_{\min}(\Sigma)$ is small or $\lambda$ is small, and that a singular sample covariance is the cliff at the end of that slope. Adding a ridge $\varepsilon I$ is what the two-asset family just did by hand; choosing $\varepsilon$ , and imposing the short-sale inequalities of Chapter 10, are later repairs. They are not implied by the first-order condition.

❌ Treating an ill-conditioned inverse as an exact optimum

A covariance with eigenvalues $1.99$ and $0.01$ is positive definite, so Section 04's uniqueness theorem applies: there is exactly one $\mathbf{w}^\star$ for the matrix that was typed. Uniqueness is not stability. The inverse amplifies the low-eigenvalue direction by $100$ , and the weight formula multiplies $\boldsymbol{\mu}$ by that inverse.

Why it breaks: the theorem guarantees a unique solution of the problem you wrote down. It says nothing about the distance between that solution and the solution of the neighboring problem with a slightly different $\boldsymbol{\mu}$ or a slightly different small eigenvalue. Those two solutions can sit far apart in weight space while both being the exact optimum of their own inputs.

Consequence: report the condition number, or $\lambda_{\min}$ , next to any book that came out of $\Sigma^{-1}$ . A unique critical point of a badly conditioned quadratic is still a badly conditioned answer. If the inverse fails outright because a sample eigenvalue is zero, stop: that is the singular case in Section 04, and no weight vector is "the" solution.

10 · Exercises

EXERCISE 3.1

One constraint, the budget. Stationarity is $\Sigma\mathbf{w}=\lambda\mathbf{1}$ . Invert, then choose $\lambda$ so the weights sum to $1$ . The variance equals that same $\lambda$ under the half-objective convention. As a check, $\Sigma\mathbf{w}$ should be a constant vector.

$\det\Sigma=5$ and $\Sigma^{-1}=\frac{1}{5}\begin{pmatrix}3&-1\\-1&2\end{pmatrix}$ . Then $\Sigma^{-1}\mathbf{1}=\frac{1}{5}(2,\ 1)^T$ and $A=\mathbf{1}^T\Sigma^{-1}\mathbf{1}=3/5$ , so

\mathbf{w}_{\mathrm{GMV}}=\frac{5}{3}\cdot\frac{1}{5}\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}2/3\\1/3\end{pmatrix}

The variance is $1/A=5/3$ . The mean is $(2/3)(1)+(1/3)(2)=4/3$ . And $\Sigma\mathbf{w}_{\mathrm{GMV}}=(5/3,\ 5/3)^T=(5/3)\mathbf{1}$ , so the budget multiplier is $5/3$ , matching the variance. The lower-variance asset gets the larger weight. It does not get the whole book: the off-diagonal $1$ makes a mix cheaper than concentration, and equal marginal contributions $\Sigma\mathbf{w}\propto\mathbf{1}$ are the first-order condition for that mix.

Two assets have $\boldsymbol{\mu}=(1,\ 2)^T$ and $\Sigma=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ . Find the global minimum-variance weights, their variance, and their mean. Verify that $\Sigma\mathbf{w}_{\mathrm{GMV}}$ is a multiple of $\mathbf{1}$ , and name the multiple.

EXERCISE 3.2

With $\Sigma=I$ the inverse is free: $A=N=3$ , $B=\mathbf{1}^T\boldsymbol{\mu}$ , $C=\boldsymbol{\mu}^T\boldsymbol{\mu}$ . Use $\lambda_1=(A\mu_\star-B)/\Delta$ and $\lambda_2=(C-B\mu_\star)/\Delta$ , then $\mathbf{w}=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ . Check the variance both as $\mathbf{w}^T\mathbf{w}$ and as $\lambda_1\mu_\star+\lambda_2$ .

$A=3$ , $B=6$ , $C=1+4+9=14$ , and $\Delta=42-36=6$ . Then $\lambda_1=(9-6)/6=1/2$ and $\lambda_2=(14-18)/6=-2/3$ , so

\mathbf{w}^\star=\frac{1}{2}\begin{pmatrix}1\\2\\3\end{pmatrix}-\frac{2}{3}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1/6\\1/3\\5/6\end{pmatrix}

The coordinates sum to $1$ . The mean is $-1/6+2/3+15/6=3$ . The variance is $(-1/6)^2+(1/3)^2+(5/6)^2=1/36+4/36+25/36=30/36=5/6$ , and $\lambda_1\mu_\star+\lambda_2=3/2-2/3=5/6$ . The short is in asset $1$ , the lowest mean. Hitting a target mean of $3$ , above the equal-weight mean $2$ , takes a long position in the high-mean name funded in part by shorting the low-mean name. Because $\Sigma=I$ , covariance is not what creates the short. The means and the budget are.

Three uncorrelated assets have variance $1$ each, so $\Sigma=I_3$ , and $\boldsymbol{\mu}=(1,\ 2,\ 3)^T$ . Using the closed form, find $\mathbf{w}^\star$ at the target $\mu_\star=3$ . Confirm the budget, the mean, and the variance. Which asset is held short, and why does that short survive even though every covariance is zero?

EXERCISE 3.3

Keep $\Sigma=I_3$ and $\boldsymbol{\mu}=(1,\ 2,\ 3)^T$ . With only the mean constraint, stationarity is $\mathbf{w}=\lambda_1\boldsymbol{\mu}$ and $\boldsymbol{\mu}^T\mathbf{w}=\mu_\star$ , so $\lambda_1=\mu_\star/C$ . Compare the sum of the coordinates, and the variance, with Exercise 3.2. If you rescale to restore the budget, recompute the mean before comparing variances.

$C=14$ and $\lambda_1=3/14$ , so $\mathbf{w}=(3,\ 6,\ 9)^T/14$ . The coordinates sum to $18/14=9/7\neq 1$ . The variance is $\mathbf{w}^T\mathbf{w}=126/196=9/14$ , and $9/14\lt 5/6$ because $9/14=27/42$ while $5/6=35/42$ .

That comparison does not say the budget-free position is a better portfolio at the same mean. Divide by $9/7$ to force the weights to sum to $1$ : the rescaled book is $(1/6,\ 1/3,\ 1/2)^T$ , whose mean is $1/6+2/3+3/2=7/3$ , not $3$ . Leaving the target is what made the variance look cheap. The admissible variance at mean $3$ on a fully invested book is the $5/6$ from Exercise 3.2. The zero vector is the same error with both constraints dropped: variance $0$ , mean $0$ , budget $0$ .

Using the data of Exercise 3.2, minimize $\frac{1}{2}\mathbf{w}^T\mathbf{w}$ subject to $\boldsymbol{\mu}^T\mathbf{w}=3$ and with no budget constraint. Report the resulting vector, the sum of its coordinates, and its variance. Explain why that variance is not comparable to the variance $5/6$ of the fully invested solution, and what happens to the mean if the vector is rescaled to sum to $1$ .

EXERCISE 3.4

Section 09's two-asset family, at risk aversion $\lambda=1$ , has $\mathbf{w}^\star=(1/2,\ 1/2)^T+\frac{\delta}{2\varepsilon}(-1,\ 1)^T$ . The imbalance is the mean gap divided by twice the small eigenvalue. Compute that coefficient at each pair $(\varepsilon,\delta)$ . Correlation is $1-\varepsilon$ , so a larger $\varepsilon$ is a less collinear pair.

(a) $\delta/(2\varepsilon)=(1/10)/(2/5)=1/4$ , so $\mathbf{w}^\star=(1/2-1/4,\ 1/2+1/4)^T=(1/4,\ 3/4)^T$ .

(b) $\delta/(2\varepsilon)=(2/5)/(2/5)=1$ , so $\mathbf{w}^\star=(1/2-1,\ 1/2+1)^T=(-1/2,\ 3/2)^T$ . The same correlation, with the mean gap four times larger, flips asset $1$ from a long quarter to a half short.

(c) $\delta/(2\varepsilon)=(1/10)/1=1/10$ , so $\mathbf{w}^\star=(2/5,\ 3/5)^T$ . The mean gap is the original one from (a). The book stays near equal weight because $\varepsilon=1/2$ is two and a half times the eigenvalue $1/5$ of (a) and (b), and the long-short coefficient $\delta/(2\varepsilon)$ shrinks by that same factor. The change from (a) to (b) is the mean gap, four times larger at a fixed eigenvalue, and that is what opens the short. The change from (a) to (c) is the eigenvalue, and that is what keeps the book near $\mathbf{w}_{\mathrm{GMV}}$ .

In the two-asset family of Section 09, variances are normalized to $1$ , the correlation is $1-\varepsilon$ , the mean gap is $\delta$ , and $\lambda=1$ . (a) Compute $\mathbf{w}^\star$ at $\varepsilon=1/5$ and $\delta=1/10$ . (b) Keep $\varepsilon=1/5$ and replace the gap by the mismeasured value $\delta=2/5$ . (c) Compute $\mathbf{w}^\star$ at $\varepsilon=1/2$ and $\delta=1/10$ . Say which change, the mean gap or the small eigenvalue, turns the mild book in (a) into the short-sale book in (b), and why (c) stays close to equal weight.

11 · Chapter Summary

Object	Role
Target-mean program	$\min \frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}$ subject to $\boldsymbol{\mu}^T\mathbf{w}=\mu_\star$ and $\mathbf{1}^T\mathbf{w}=1$
Risk-aversion program	$\max \boldsymbol{\mu}^T\mathbf{w}-\frac{\lambda}{2}\mathbf{w}^T\Sigma\mathbf{w}$ subject to the budget alone
Stationarity	$\Sigma\mathbf{w}=\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1}$ , hence $\mathbf{w}=\Sigma^{-1}(\lambda_1\boldsymbol{\mu}+\lambda_2\mathbf{1})$ when $\Sigma$ is invertible
$\Sigma\succ 0$	Strictly convex objective, $\Sigma^{-1}$ exists, exactly one fully invested minimizer
$\Delta=AC-B^2$	Positive when means are not all equal; the denominator of both multipliers
$\sigma_p^2=\lambda_1\mu_\star+\lambda_2$	Variance check under the half-objective convention; a quadratic in $\mu_\star$
$\mathbf{w}_{\mathrm{GMV}}=\Sigma^{-1}\mathbf{1}/A$	Target-mean solution at $\mu_\star=B/A$ , where $\lambda_1=0$ ; variance $1/A$
Risk-aversion weights	$\mathbf{w}_{\mathrm{GMV}}$ plus a tilt $\Sigma^{-1}(\boldsymbol{\mu}-\mu_{\mathrm{GMV}}\mathbf{1})/\lambda$ ; mean $\mu_{\mathrm{GMV}}+\Delta/(\lambda A)$
$N=2$ target-mean	Two equalities pin $\mathbf{w}$ whenever $\mu_1\neq\mu_2$ ; $\Sigma$ prices the variance and does not choose the mix
Singular or tiny $\lambda_{\min}$	Inverse formula unavailable, or available and violently sensitive to $\boldsymbol{\mu}$

Varying $\mu_\star$ moves $\mathbf{w}^\star$ along a line in weight space and traces a curve of $(\sigma_p,\mu_p)$ . Drawing that curve, and separating the limb a risk-averse investor will hold from the rest of the minimum-variance set, is the next chapter. Cash, the capital market line, and the CAPM are not inputs here.

Next: Chapter 04 — The Minimum-Variance Frontier, which takes the weights $\mathbf{w}^\star(\mu_\star)$ computed here and reads off the curve they trace.