Chapter 04

Hard

The Minimum-Variance Frontier

00 · Symbol Glossary

$\mathbf{w},\ \mathbf{1}$weights and the ones vector

$\mathbf{w}\in\mathbb{R}^N$ holds portfolio weights. $\mathbf{1}$ is the vector of ones. Full investment means $\mathbf{w}^\top\mathbf{1}=1$ . A negative coordinate is a short sale. The budget identity still holds: the short finances part of the long side, and there is no separate cash asset in this chapter.

$\boldsymbol{\mu},\ \Sigma$mean vector and covariance

$\boldsymbol{\mu}$ is the vector of expected asset returns. $\Sigma$ is their covariance matrix, symmetric positive definite, with $\Sigma_{ij}=\mathrm{Cov}(r_i,r_j)$ . Portfolio mean and variance are $\mu_p=\mathbf{w}^\top\boldsymbol{\mu}$ and $\sigma_p^2=\mathbf{w}^\top\Sigma\mathbf{w}$ .

$\mu_\star,\ \mathbf{w}(\mu_\star)$target mean and its minimum-variance portfolio

$\mu_\star$ is a required expected return. $\mathbf{w}(\mu_\star)$ is the unique fully invested portfolio that attains $\mu_\star$ at the smallest possible variance. As $\mu_\star$ runs through the reals, these portfolios trace the minimum-variance frontier.

$\mathbf{w}_g,\ \mu_g,\ \sigma_g$global minimum-variance portfolio

The fully invested portfolio of smallest variance, with no mean target. It is the leftmost point of the frontier. $\mu_g=\mathbf{w}_g^\top\boldsymbol{\mu}$ and $\sigma_g^2=\mathbf{w}_g^\top\Sigma\mathbf{w}_g$ .

$A,\ B,\ C$frontier scalars

$A=\mathbf{1}^\top\Sigma^{-1}\mathbf{1}$ , $B=\mathbf{1}^\top\Sigma^{-1}\boldsymbol{\mu}$ , and $C=\boldsymbol{\mu}^\top\Sigma^{-1}\boldsymbol{\mu}$ . Three numbers built from $\Sigma^{-1}$ that determine every frontier mean and variance.

$\Delta$frontier discriminant

$\Delta=AC-B^2$ . Positive exactly when $\Sigma$ is positive definite and $\boldsymbol{\mu}$ is not a multiple of $\mathbf{1}$ . It is the denominator of the frontier equation.

$\mathbf{h}$dollar-neutral frontier overlay

The fixed vector $\mathbf{h}$ with $\mathbf{h}^\top\mathbf{1}=0$ and $\mathbf{h}^\top\boldsymbol{\mu}=1$ such that $\mathbf{w}(\mu_\star)=\mathbf{w}_g+(\mu_\star-\mu_g)\mathbf{h}$ . Adding a multiple of $\mathbf{h}$ changes the mean one-for-one and does not change the net investment.

$\alpha$two-fund mixing weight

The real number in $\mathbf{w}=\alpha\mathbf{w}_p+(1-\alpha)\mathbf{w}_q$ when a frontier portfolio is written from two distinct frontier portfolios $\mathbf{w}_p$ and $\mathbf{w}_q$ . $\alpha$ is not restricted to the unit interval. A value outside that interval is a short of one fund.

01 · One Target Becomes a Curve

Chapter 03 fixed one target mean and returned one weight vector. That vector is not a fund you choose once. Change the target and the same problem returns a different weight vector. This chapter is about the curve those solutions trace in the plane whose horizontal axis is risk and whose vertical axis is expected return, and about the fact that two solutions determine all the others.

The standing assumptions are these.

There are $N\geq 2$ risky assets and no risk-free asset.
$\Sigma$ is symmetric positive definite: $\mathbf{x}^\top\Sigma\mathbf{x}>0$ for every $\mathbf{x}\neq\mathbf{0}$ . Variance is then a strictly convex function of $\mathbf{w}$ , and each minimization below has a unique solution. For a symmetric matrix this is equivalent to every leading principal minor being positive, and it implies $\Sigma^{-1}$ exists.
Full investment is the single budget equation $\mathbf{w}^\top\mathbf{1}=1$ . Short sales are allowed.
$\boldsymbol{\mu}$ is not a scalar multiple of $\mathbf{1}$ . The assets do not all share one expected return.

If every mean is the same

If $\boldsymbol{\mu}=c\mathbf{1}$ , every fully invested portfolio has mean $c$ . The mean constraint is either redundant or impossible, and the only minimum-variance portfolio is $\mathbf{w}_g$ . The curve collapses to a point. Everything that follows uses the opposite case, so that every real target $\mu_\star$ is attainable.

Chapter 03's first-order condition, written so that the objective is half the portfolio variance, says that a variance-minimizing fully invested portfolio of mean $\mu_\star$ satisfies

\Sigma\mathbf{w}=\lambda\mathbf{1}+\gamma\boldsymbol{\mu}.

Here $\lambda$ is the multiplier on the budget and $\gamma$ is the multiplier on the mean target. If the Lagrangian was built from $\mathbf{w}^\top\Sigma\mathbf{w}$ without the factor one half, both multipliers are twice as large and the span is unchanged. The geometry uses only the span: $\Sigma\mathbf{w}$ lies in the plane spanned by $\mathbf{1}$ and $\boldsymbol{\mu}$ , so

\mathbf{w}\in\mathrm{span}\{\Sigma^{-1}\mathbf{1},\ \Sigma^{-1}\boldsymbol{\mu}\}.

The whole frontier lives in a two-dimensional subspace of weight space. The budget $\mathbf{w}^\top\mathbf{1}=1$ cuts that plane in a line. That line is the frontier, seen in weight space rather than in the $(\sigma,\mu)$ plane.

02 · Definition of the Frontier

Definition — Minimum-Variance Frontier

For each target $\mu_\star\in\mathbb{R}$ , let $\mathbf{w}(\mu_\star)$ solve

\min_{\mathbf{w}}\ \frac12\mathbf{w}^\top\Sigma\mathbf{w}\quad\text{subject to}\quad\mathbf{w}^\top\mathbf{1}=1,\ \mathbf{w}^\top\boldsymbol{\mu}=\mu_\star.

The minimum-variance frontier is the locus of points $\big(\sigma(\mu_\star),\ \mu_\star\big)$ as $\mu_\star$ varies, where $\sigma(\mu_\star)=\sqrt{\mathbf{w}(\mu_\star)^\top\Sigma\mathbf{w}(\mu_\star)}$ . Equivalently, it is the locus of $\big(\sigma^2(\mu_\star),\ \mu_\star\big)$ .

Plain language: for each required mean, keep only the fully invested portfolio that delivers that mean at the smallest attainable standard deviation, then plot those portfolios. Every other fully invested portfolio of the same mean sits strictly to the right.

The frontier is a property of the equality-constrained problem. It contains every solution of that problem, including solutions whose mean is poor. Section 03 splits the curve into the part an investor who prefers a higher mean will actually hold and the part that investor will not.

When $N=2$ , the two equality constraints determine $\mathbf{w}$ completely whenever the two means differ. Every fully invested portfolio is $\mathbf{w}(\mu_\star)$ for $\mu_\star=\mathbf{w}^\top\boldsymbol{\mu}$ . The opportunity set and the frontier are the same curve. There is nothing to the right of it.

When $N\geq 3$ , the two constraints leave $N-2$ free directions. Any $\mathbf{v}$ with $\mathbf{v}^\top\mathbf{1}=0$ and $\mathbf{v}^\top\boldsymbol{\mu}=0$ can be added to $\mathbf{w}(\mu_\star)$ without changing the budget or the mean. The cross term vanishes: $\Sigma\mathbf{w}(\mu_\star)$ lies in the span of $\mathbf{1}$ and $\boldsymbol{\mu}$ , so $\mathbf{w}(\mu_\star)^\top\Sigma\mathbf{v}=0$ . The variance therefore rises by exactly $\mathbf{v}^\top\Sigma\mathbf{v}$ , which is positive when $\mathbf{v}\neq\mathbf{0}$ . Scaling $\mathbf{v}$ sweeps out every variance above the frontier value. In the $(\sigma,\mu)$ plane the feasible fully invested portfolios fill the closed region on and to the right of the curve.

❌ Calling every fully invested portfolio minimum-variance

A report plots three assets and the curve through all long-only mixes, then labels the entire shaded region "the minimum-variance frontier."

Why it breaks: a minimum-variance portfolio is the left boundary at its own mean. An interior point has the same mean as exactly one frontier portfolio and a strictly larger variance. With three or more assets that interior is nonempty. With two assets the interior is empty, which is a special case, not the definition.

Consequence: an interior portfolio is treated as an optimizer output. Replacing it by $\mathbf{w}(\mu_p)$ keeps the mean and cuts the variance. The frontier is the boundary of the feasible region.

03 · The Two Limbs and the Global Minimum

Drop the mean constraint and minimize variance subject only to full investment. Stationarity loses the $\boldsymbol{\mu}$ term:

\Sigma\mathbf{w}_g=\lambda\mathbf{1}.

Hence $\mathbf{w}_g=\lambda\Sigma^{-1}\mathbf{1}$ . Write $A=\mathbf{1}^\top\Sigma^{-1}\mathbf{1}$ and $B=\mathbf{1}^\top\Sigma^{-1}\boldsymbol{\mu}$ . The budget $\mathbf{1}^\top\mathbf{w}_g=1$ forces $\lambda A=1$ , so $\lambda=1/A$ and

\mathbf{w}_g=\frac{\Sigma^{-1}\mathbf{1}}{A},\qquad \sigma_g^2=\mathbf{w}_g^\top\Sigma\mathbf{w}_g=\frac{1}{A},\qquad \mu_g=\mathbf{w}_g^\top\boldsymbol{\mu}=\frac{B}{A}.

The variance identity is $\mathbf{w}_g^\top(\lambda\mathbf{1})=\lambda$ . This portfolio is the global minimum-variance portfolio. It is the leftmost point of the locus: Section 04 shows that every other frontier variance equals $1/A$ plus a square, so nothing on the frontier has a smaller standard deviation, and nothing feasible does either.

Definition — Efficient and Inefficient Limbs

Fix a frontier variance $\sigma^2>\sigma_g^2$ . Exactly two frontier portfolios attain it. Their means are symmetric around $\mu_g$ : one mean is above $\mu_g$ and one is below.

The upper limb, the mean-variance efficient frontier, is the set of frontier portfolios with $\mu\geq\mu_g$ , including $\mathbf{w}_g$ itself. Each of them maximizes expected return among fully invested portfolios of variance at most its own.

The lower limb is the set of frontier portfolios with $\mu<\mu_g$ . Each is minimum-variance for its own mean target and is mean-variance inefficient: the upper portfolio of the same variance has a higher mean and the same risk.

$\mathbf{w}_g$ belongs to the efficient frontier. It is the only fully invested portfolio of variance $\sigma_g^2$ , so no portfolio dominates it in the mean-variance order. Every other frontier variance comes as a pair. The upper member of the pair dominates the lower member. An investor who prefers a higher mean at a given variance, which is the usual mean-variance preference, holds a portfolio on the upper limb. The lower limb solves a mathematics problem — hit this low mean as tightly as possible — and it is dominated.

The same split can be read off the inequality form of the optimizer. Minimize variance subject to $\mathbf{w}^\top\mathbf{1}=1$ and $\mathbf{w}^\top\boldsymbol{\mu}\geq\mu_\star$ . If $\mu_\star\leq\mu_g$ , the mean constraint does not bind and the solution is $\mathbf{w}_g$ . If $\mu_\star>\mu_g$ , the constraint binds and the solution is the upper-limb portfolio $\mathbf{w}(\mu_\star)$ . The lower-limb portfolio $\mathbf{w}(\mu_\star)$ for $\mu_\star<\mu_g$ solves the equality-constrained problem and does not solve the inequality-constrained problem.

One covariance fact fixes the vertex. Because $\Sigma\mathbf{w}_g=(1/A)\mathbf{1}$ , every fully invested portfolio $\mathbf{q}$ satisfies

\mathrm{Cov}(r_q,r_g)=\mathbf{q}^\top\Sigma\mathbf{w}_g=\frac{\mathbf{q}^\top\mathbf{1}}{A}=\sigma_g^2.

The covariance does not depend on which fully invested portfolio you chose. Cauchy–Schwarz for the two portfolio returns then gives $\sigma_g^2=\mathrm{Cov}(r_q,r_g)\leq\sigma_q\sigma_g$ , so $\sigma_q\geq\sigma_g$ , with equality only at $\mathbf{w}_g$ . That is a second proof that the vertex is a global minimum, and it shows there is no fully invested hedge of the global minimum-variance portfolio: every fully invested book has the same covariance with it.

In the $(\sigma,\mu)$ plane the tangent at the vertex is vertical. Near $\mu_g$ , excess variance grows like $(\mu-\mu_g)^2$ , so $d\sigma/d\mu=0$ at the vertex and $d\mu/d\sigma$ is infinite. Moving off the vertex changes the mean to first order and the standard deviation only to second order. Far from the vertex the curve bends toward a pair of slanted asymptotes, derived in Section 05.

❌ Treating the whole frontier as efficient

A plot of the curve is titled "efficient frontier," and a portfolio on the lower limb is recommended because "it lies on the frontier."

Why it breaks: minimum-variance and efficient are different sets. The lower-limb portfolio has a twin on the upper limb with the same standard deviation and a strictly higher mean. The twin dominates it. The global minimum-variance portfolio is the only frontier portfolio without a twin, and it is efficient.

Consequence: the book takes needless pain in expected return. Any mean target below $\mu_g$ , fed into the Chapter 03 equality-constrained solver, lands on the lower limb. The solver did its job. The job was not mean-variance efficiency. Efficiency requires the target to sit at or above $\mu_g$ , or the mean constraint to be imposed as an inequality.

❌ Claiming every mean-variance solution is efficient

An optimizer is run at a target below the global minimum-variance mean. The output is labeled efficient because the first-order condition $\Sigma\mathbf{w}=\lambda\mathbf{1}+\gamma\boldsymbol{\mu}$ holds.

Why it breaks: that first-order condition is the stationarity condition for every point on both limbs. It characterizes the minimum-variance frontier. The sign of $\mu_\star-\mu_g$ , or equivalently the sign of the mean multiplier once the budget multiplier is fixed by the formulas in Section 04, separates the limbs. Stationarity alone does not.

Consequence: a dominated portfolio is booked as the risk-efficient choice at its volatility. The efficient portfolio at that same volatility is the upper twin.

04 · Scalars A, B, C and the Frontier Equation

Apply $\Sigma^{-1}$ to the first-order condition:

\mathbf{w}=\Sigma^{-1}(\lambda\mathbf{1}+\gamma\boldsymbol{\mu})=\lambda\Sigma^{-1}\mathbf{1}+\gamma\Sigma^{-1}\boldsymbol{\mu}.

Impose the two constraints. With $A$ and $B$ as in Section 03 and $C=\boldsymbol{\mu}^\top\Sigma^{-1}\boldsymbol{\mu}$ ,

\begin{aligned} \lambda A+\gamma B &= 1,\\ \lambda B+\gamma C &= \mu_\star. \end{aligned}

In matrix form the coefficient matrix is the Gram matrix of the columns $\mathbf{1}$ and $\boldsymbol{\mu}$ in the inner product $\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{x}^\top\Sigma^{-1}\mathbf{y}$ :

\begin{pmatrix} A & B \\ B & C \end{pmatrix} = \begin{pmatrix} \mathbf{1}^\top \\ \boldsymbol{\mu}^\top \end{pmatrix}\Sigma^{-1}\begin{pmatrix}\mathbf{1}&\boldsymbol{\mu}\end{pmatrix}.

Because $\Sigma^{-1}$ is positive definite, this $2\times 2$ matrix is positive definite if and only if $\mathbf{1}$ and $\boldsymbol{\mu}$ are linearly independent, which is the standing assumption that not all means agree. Its determinant is therefore positive:

\Delta=AC-B^2>0.

Solving the system,

\lambda=\frac{C-B\mu_\star}{\Delta},\qquad \gamma=\frac{A\mu_\star-B}{\Delta}.

Portfolio variance collapses along the first-order condition. Since $\Sigma\mathbf{w}=\lambda\mathbf{1}+\gamma\boldsymbol{\mu}$ ,

\sigma^2(\mu_\star)=\mathbf{w}^\top\Sigma\mathbf{w}=\lambda\,\mathbf{w}^\top\mathbf{1}+\gamma\,\mathbf{w}^\top\boldsymbol{\mu}=\lambda+\gamma\mu_\star.

Substitute $\lambda$ and $\gamma$ :

\sigma^2(\mu_\star)=\frac{A\mu_\star^2-2B\mu_\star+C}{\Delta}.

This is the equation of the frontier in the $(\sigma^2,\mu)$ plane. The right-hand side is a quadratic polynomial in the target mean, with leading coefficient $A/\Delta>0$ , so the parabola opens toward larger variance.

Complete the square. $A\mu^2-2B\mu+C=A(\mu-B/A)^2+(AC-B^2)/A$ , and $(AC-B^2)/A=\Delta/A$ , so

\sigma^2(\mu)=\frac{1}{A}+\frac{A}{\Delta}\left(\mu-\frac{B}{A}\right)^2=\sigma_g^2+\frac{A}{\Delta}(\mu-\mu_g)^2.

The first term is the variance already computed for $\mathbf{w}_g$ . The second term is a square, hence nonnegative, and it is zero only at $\mu=\mu_g$ . The vertex of the parabola is exactly the global minimum-variance portfolio, and the two means that share a common variance $\sigma^2>\sigma_g^2$ are

\mu=\mu_g\pm\sqrt{\frac{\Delta}{A}\big(\sigma^2-\sigma_g^2\big)}.

The positive branch is the upper limb. The negative branch is the lower limb. Differentiating the parabola gives $d\sigma^2/d\mu=(2A/\Delta)(\mu-\mu_g)$ , which is zero at the vertex, positive on the upper limb, and negative on the lower limb. On the lower limb, raising the mean target lowers frontier variance, until the vertex is reached. That derivative is the algebraic form of domination.

The mean multiplier has the same sign as $\mu_\star-\mu_g$ , because $\gamma=(A/\Delta)(\mu_\star-\mu_g)$ . On the upper limb $\gamma>0$ . At the vertex $\gamma=0$ , which recovers $\Sigma\mathbf{w}_g=\lambda\mathbf{1}$ . On the lower limb $\gamma<0$ .

Banning short sales changes the curve

The equation above uses an unconstrained weight vector apart from the two equalities. A ban on negative weights, or any other inequality constraint, replaces the single hyperbola with a chain of curve pieces, one piece for each set of assets that are held long. That critical-line geometry is Chapter 10. This chapter keeps short sales available so the frontier is one smooth curve and the spanning theorem has a single statement.

05 · Parabola in Variance, Hyperbola in Volatility

Read the completed square in two coordinate systems. Both use expected return on the vertical axis.

In the plane of variance and mean, $\sigma^2$ horizontal and $\mu$ vertical, the equation $\sigma^2=\sigma_g^2+(A/\Delta)(\mu-\mu_g)^2$ is a parabola opening to the right, with vertex $(\sigma_g^2,\mu_g)$ .

In the plane of standard deviation and mean, the same portfolios form a hyperbola. Rearrange the completed square:

\frac{\sigma^2}{1/A}-\frac{(\mu-\mu_g)^2}{\Delta/A^2}=1.

This is the right-hand branch of a hyperbola centered at $(\sigma,\mu)=(0,\mu_g)$ , with vertex $(\sigma_g,\mu_g)$ . The left-hand branch has $\sigma<0$ and is discarded, because a standard deviation is nonnegative. The asymptotes pass through $(0,\mu_g)$ with slopes $\pm\sqrt{\Delta/A}$ , measured as rise in mean per unit of standard deviation. Along the upper limb the ratio $(\mu-\mu_g)/\sigma$ approaches $\sqrt{\Delta/A}$ . That limiting slope is finite. The vertical tangent lives only at the vertex.

Calling the curve in the $(\sigma,\mu)$ plane a parabola is a slip of coordinates. The parabola is the picture in variance. Volatility is the square root of a quadratic function of $\mu$ , which is the hyperbola just written. Software that plots $\sigma$ and a textbook that plots $\sigma^2$ are drawing the same portfolios in different horizontal units.

Example — Two assets, so the frontier is the whole opportunity set

Two assets, returns in percent, variances in percent squared, correlation zero so the inverse stays diagonal:

\boldsymbol{\mu}=\begin{pmatrix}5\\10\end{pmatrix},\qquad \Sigma=\begin{pmatrix}100&0\\0&400\end{pmatrix}.

$\Sigma^{-1}$ is diagonal with entries $1/100$ and $1/400$ , so $A=1/100+1/400=1/80$ , $B=5/100+10/400=3/40$ , $C=25/100+100/400=1/2$ , and $\Delta=(1/80)(1/2)-(3/40)^2=1/1600$ . Then $\mu_g=6$ , $\sigma_g^2=80$ , and $A/\Delta=20$ , so every fully invested portfolio satisfies

\sigma^2=80+20(\mu-6)^2,\qquad \frac{\sigma^2}{80}-\frac{(\mu-6)^2}{4}=1.

Let $w$ be the weight on the first asset. Then $\mu=10-5w$ and $\sigma^2=100w^2+400(1-w)^2$ .

$w$ on asset 1	Mean	Variance	Location
$4/5$	$6$	$80$	vertex, efficient
$1$	$5$	$100$	lower limb (asset 1 itself)
$2/5$	$8$	$160$	upper limb
$0$	$10$	$400$	upper limb (asset 2 itself)
$8/5$	$2$	$400$	lower limb, twin of asset 2

Asset 1 sits on the frontier and is inefficient: its mean $5$ is below $\mu_g=6$ . Asset 2 has the same variance as the portfolio $w=8/5$ , which shorts asset 2, and asset 2 is the efficient twin of that portfolio. The global minimum puts weight $4/5$ on the quieter asset and $1/5$ on the noisier one, and its standard deviation $\sqrt{80}\approx 8.94$ is smaller than both assets' standard deviations $10$ and $20$ . With two assets there is no portfolio of mean $8$ and variance $200$ . The frontier variance at that mean is $160$ , and the constraints leave no free direction in which to add risk.

Zero correlation was only a convenience for the inverse. For any correlation strictly inside $(-1,1)$ , two assets with different means still trace a hyperbola, and both assets still lie on it.

06 · A Three-Asset Frontier, Built by Hand

With a third asset the curve stops being the whole opportunity set. The same scalars still give the boundary. The market below is scaled in percent and percent squared so that every fraction stays exact. The shape does not depend on that choice of unit.

Step-by-step — Frontier of a three-asset market

Write the inputs and check positive definiteness.

\boldsymbol{\mu}=\begin{pmatrix}3\\1\\2\end{pmatrix},\qquad \Sigma=\begin{pmatrix}2&1&0\\1&2&0\\0&0&1\end{pmatrix}.

Leading principal minors: $2>0$ , $\det\begin{pmatrix}2&1\\1&2\end{pmatrix}=3>0$ , and $\det\Sigma=3>0$ . Sylvester's criterion says $\Sigma$ is positive definite, so the Chapter 03 minimizer at each target is unique.

Invert $\Sigma$ . The third asset is uncorrelated with the first two, and the upper $2\times 2$ block has determinant $3$ , so

\Sigma^{-1}=\begin{pmatrix}2/3&-1/3&0\\-1/3&2/3&0\\0&0&1\end{pmatrix}.

Multiplying $\Sigma\Sigma^{-1}$ recovers the identity.

Compute the three scalars. $\Sigma^{-1}\mathbf{1}=(1/3,\ 1/3,\ 1)^\top$ , so

A=\frac13+\frac13+1=\frac53.

$\Sigma^{-1}\boldsymbol{\mu}=(5/3,\ -1/3,\ 2)^\top$ , so $B=5/3-1/3+2=10/3$ . Then $C=3\cdot(5/3)+1\cdot(-1/3)+2\cdot 2=26/3$ . The discriminant is

\Delta=\frac53\cdot\frac{26}{3}-\left(\frac{10}{3}\right)^2=\frac{130-100}{9}=\frac{10}{3}>0.

Read off the vertex. $\mu_g=B/A=2$ and $\sigma_g^2=1/A=3/5$ . The weights are $\Sigma^{-1}\mathbf{1}$ divided by $A$ :

\mathbf{w}_g=\frac{3}{5}\begin{pmatrix}1/3\\1/3\\1\end{pmatrix}=\begin{pmatrix}1/5\\1/5\\3/5\end{pmatrix}.

Mean check: $(1/5)\cdot 3+(1/5)\cdot 1+(3/5)\cdot 2=2$ . Asset 3 alone has the same mean and variance $1>3/5$ , so asset 3 is feasible and strictly inside the frontier. None of the three unit portfolios is minimum-variance. That is the geometric difference from the two-asset example.

Hit the target $\mu_\star=5/2$ , on the upper limb. The multipliers are

\lambda=\frac{26/3-(10/3)\cdot(5/2)}{10/3}=\frac{1}{10},\qquad \gamma=\frac{(5/3)\cdot(5/2)-10/3}{10/3}=\frac14.

$\gamma>0$ agrees with $\mu_\star>\mu_g$ . Then $\lambda\mathbf{1}+\gamma\boldsymbol{\mu}=(17/20,\ 7/20,\ 3/5)^\top$ , and $\mathbf{w}=\Sigma^{-1}(\lambda\mathbf{1}+\gamma\boldsymbol{\mu})$ is

\mathbf{w}\!\left(\frac52\right)=\begin{pmatrix}9/20\\-1/20\\3/5\end{pmatrix}.

The short of the low-mean second asset is allowed. Variance is $\lambda+\gamma\mu_\star=1/10+(1/4)\cdot(5/2)=29/40$ . The completed square agrees: $3/5+(1/2)(5/2-2)^2=3/5+1/8=29/40$ , since $A/\Delta=1/2$ .

Pair it with the lower twin, then with an interior portfolio of the same mean. Set $29/40=3/5+(1/2)(\mu-2)^2$ . Then $(\mu-2)^2=1/4$ , so $\mu=5/2$ or $\mu=3/2$ . The inefficient twin has mean $3/2$ and, by the same multiplier formulas, weights $(-1/20,\ 9/20,\ 3/5)^\top$ . Same variance $29/40$ , lower mean.

The fully invested portfolio $(7/10,\ 1/5,\ 1/10)^\top$ also has mean $5/2$ , but its variance is $27/20$ . Since $27/20=54/40>29/40$ , it sits to the right of $\mathbf{w}(5/2)$ . Given the target $5/2$ , the Chapter 03 problem returns the boundary portfolio.

Example — Reading the three points in the $(\sigma,\mu)$ plane

Plot standard deviation horizontally and mean vertically. The vertex is at $\big(\sqrt{3/5},\ 2\big)\approx(0.775,\ 2)$ . The upper and lower portfolios of variance $29/40$ sit on the vertical line $\sigma=\sqrt{29/40}\approx 0.851$ , at heights $2.5$ and $1.5$ . Asset 3 sits at $(\sigma,\mu)=(1,\ 2)$ , directly to the right of the vertex. The portfolio of variance $27/20$ sits near $\sigma\approx 1.16$ at height $2.5$ , directly to the right of the efficient portfolio. The asymptotes have slopes $\pm\sqrt{\Delta/A}=\pm\sqrt{2}$ and pass through $(0,\ 2)$ on the mean axis. The curve leaves the vertex vertically and bends toward those lines. Everything a fully invested investor can hold is on that curve or to its right. Everything to its left is unreachable at any weights that sum to one.

07 · Two-Fund Spanning among Risky Assets

The first-order condition put every frontier weight vector in one plane, and the budget cut that plane down to a line. A line is determined by any two of its points. That is the spanning theorem. The explicit parametrization makes the coefficient obvious.

From $\gamma=(A/\Delta)(\mu_\star-\mu_g)$ and $\mathbf{w}=\lambda\Sigma^{-1}\mathbf{1}+\gamma\Sigma^{-1}\boldsymbol{\mu}$ , the map $\mu_\star\mapsto\mathbf{w}(\mu_\star)$ is affine. Matching the vertex and the slope in $\mu_\star$ gives

\mathbf{w}(\mu_\star)=\mathbf{w}_g+(\mu_\star-\mu_g)\mathbf{h},

where the overlay

\mathbf{h}=\frac{A\Sigma^{-1}\boldsymbol{\mu}-B\Sigma^{-1}\mathbf{1}}{\Delta}

satisfies $\mathbf{h}^\top\mathbf{1}=0$ and $\mathbf{h}^\top\boldsymbol{\mu}=1$ . In the three-asset market, $\mathbf{h}=(1/2,\ -1/2,\ 0)^\top$ . Check the target $5/2$ : $\mathbf{w}_g+(1/2)\mathbf{h}=(1/5,\ 1/5,\ 3/5)^\top+(1/4,\ -1/4,\ 0)^\top=(9/20,\ -1/20,\ 3/5)^\top$ , which matches the portfolio built in Section 06. The cross term $2\mathbf{w}_g^\top\Sigma\mathbf{h}$ vanishes because $\Sigma\mathbf{w}_g$ is a multiple of $\mathbf{1}$ and $\mathbf{h}$ is dollar-neutral, so

\sigma^2\big(\mathbf{w}_g+(\mu_\star-\mu_g)\mathbf{h}\big)=\sigma_g^2+(\mu_\star-\mu_g)^2\,\mathbf{h}^\top\Sigma\mathbf{h},

and $\mathbf{h}^\top\Sigma\mathbf{h}=A/\Delta$ . Excess frontier variance is exactly the variance of a zero-net-investment overlay, uncorrelated with the global minimum-variance portfolio.

Definition — Two-Fund Spanning (Risky Assets)

Let $\mathbf{w}_p=\mathbf{w}(\mu_p)$ and $\mathbf{w}_q=\mathbf{w}(\mu_q)$ be two distinct minimum-variance frontier portfolios, so $\mu_p\neq\mu_q$ . A fully invested portfolio lies on the minimum-variance frontier if and only if its weights equal

\mathbf{w}=\alpha\mathbf{w}_p+(1-\alpha)\mathbf{w}_q

for the real number $\alpha=(\mu-\mu_q)/(\mu_p-\mu_q)$ , where $\mu$ is the portfolio's mean. Both limbs are included. $\alpha$ runs through all of $\mathbb{R}$ as $\mu$ does.

The identity is substitution. Because $\mathbf{w}(\mu)=\mathbf{w}_g+(\mu-\mu_g)\mathbf{h}$ ,

\alpha\mathbf{w}(\mu_p)+(1-\alpha)\mathbf{w}(\mu_q)=\mathbf{w}_g+\big(\alpha\mu_p+(1-\alpha)\mu_q-\mu_g\big)\mathbf{h}=\mathbf{w}(\alpha\mu_p+(1-\alpha)\mu_q).

An affine combination of frontier portfolios is the frontier portfolio whose mean is the same affine combination of the means. Conversely every frontier portfolio arises this way: choose $\alpha$ to match the desired mean. The two funds are not unique. Any distinct pair spans the same line. The pair $\mathbf{w}_g$ and $\mathbf{w}(\mu_g+1)=\mathbf{w}_g+\mathbf{h}$ is a convenient basis because the coefficient of the second fund equals the excess mean over $\mu_g$ . The vector $\mathbf{h}$ itself is not a fund: its weights sum to zero. It is the difference of two funds.

When $\alpha$ lies between $0$ and $1$ inclusive, the mean $\mu$ lies between $\mu_p$ and $\mu_q$ , and both funds are held long. That segment is only an arc of the frontier. Reaching a mean outside the interval requires $\alpha$ outside the unit interval, which is a short of one fund and a leveraged long position in the other. In particular, two upper-limb funds span the lower limb once short sales of a fund are allowed. Restricting $\alpha$ to the unit interval cuts the frontier down to the arc between the two funds and hides the twin portfolios.

Investors who share $\boldsymbol{\mu}$ and $\Sigma$ , and who differ only in the mean they require on the upper limb, can all be served by two mutual funds that are themselves frontier portfolios. Each investor's risky portfolio is one number $\alpha$ times those two funds. Investors who want a mean between the two funds hold both long. Investors who want a mean beyond the higher-mean fund short the lower-mean fund. This is spanning inside the risky universe. It is the reason a later chapter can replace a whole menu of risky portfolios by fund holdings.

What Chapter 05 adds, and what it does not reuse unchanged

A risk-free asset is still outside this chapter. Chapter 05 introduces one. The frontier of the enlarged universe is a pair of straight half-lines leaving the risk-free point and tangent to the risky hyperbola of this chapter. The upper half-line is the capital market line. Every portfolio on it is a mix of the risk-free asset and a single risky portfolio, the tangency portfolio. That is two-fund separation in the sense of cash plus one fund of risky assets. The theorem proved here uses two risky frontier portfolios and no risk-free asset. The tangency portfolio can stand in for the whole risky frontier only after cash exists, because cash selects one point of tangency on the hyperbola this chapter drew. Equilibrium identification of a frontier portfolio with the market portfolio is Chapter 06, not a consequence of the geometry alone.

❌ Reading this theorem as the capital market line

A slide says: "by two-fund separation, every efficient investor holds the risk-free asset and the tangency portfolio, so the frontier is a straight line."

Why it breaks: the straight line appears after a risk-free asset is added, which is Chapter 05. In the risky universe the frontier is the hyperbola of Section 05. Spanning here means every frontier portfolio, efficient or not, is an affine mix of two risky frontier portfolios. There is no tangency portfolio until there is a risk-free return to be tangent to.

Consequence: the risky hyperbola is replaced by a ray too early, and the lower limb drops out of the picture for the wrong reason. The tangency portfolio is one upper-limb portfolio, selected by the risk-free rate. It is not $\mathbf{w}_g$ . The tangent at the vertex is vertical, so no line drawn from a finite risk-free point on the mean axis is tangent there.

08 · Exercises

EXERCISE 4.1

On the two-asset frontier of Section 05, a point $(\sigma^2,\mu)$ is attainable if and only if $\sigma^2=80+20(\mu-6)^2$ . If equality holds, the point is efficient when $\mu\geq 6$ and inefficient when $\mu<6$ . If the stated variance is larger, the point would sit to the right of the curve, and two assets have no such portfolio. If it is smaller, the point is to the left of the curve and no fully invested portfolio reaches it.

(a) $80+20(0)^2=80$ . Attainable. Mean equals $\mu_g$ , so this is the global minimum-variance portfolio, and it is efficient. (b) Frontier variance at mean $5$ is $80+20(1)^2=100$ . Attainable, and $5<6$ , so it is the lower limb. It is asset 1. (c) Frontier variance at mean $8$ is $80+20(4)^2=160$ . Attainable and efficient. (d) Frontier variance at mean $4$ is also $160$ . Attainable and inefficient: the twin of (c). (e) Frontier variance at mean $8$ is $160$ , but the stated variance is $200$ . Unattainable in this two-asset market. The point lies to the right of the hyperbola. (f) Variance $40$ is below $\sigma_g^2=80$ . Unattainable. It lies to the left of the vertex.

The two-asset market of Section 05 has frontier $\sigma^2=80+20(\mu-6)^2$ , with variance in percent squared and mean in percent. For each point $(\sigma^2,\mu)$ below, say whether a fully invested portfolio attains it. If it does, say whether that portfolio is efficient or inefficient. If it does not, say whether the point lies to the left of the frontier or to the right.

(a) $(80,\ 6)$ . (b) $(100,\ 5)$ . (c) $(160,\ 8)$ . (d) $(160,\ 4)$ . (e) $(200,\ 8)$ . (f) $(40,\ 6)$ .

EXERCISE 4.2

Invert the $2\times 2$ matrix first: the determinant is $3$ . Then $A=\mathbf{1}^\top\Sigma^{-1}\mathbf{1}$ , $B=\mathbf{1}^\top\Sigma^{-1}\boldsymbol{\mu}$ , $C=\boldsymbol{\mu}^\top\Sigma^{-1}\boldsymbol{\mu}$ , and $\Delta=AC-B^2$ . The vertex formulas are $\mathbf{w}_g=\Sigma^{-1}\mathbf{1}/A$ , $\mu_g=B/A$ , and $\sigma_g^2=1/A$ . At a target, $\sigma^2=(A\mu_\star^2-2B\mu_\star+C)/\Delta$ , and the twin mean is the other root of the completed square $\sigma^2=\sigma_g^2+(A/\Delta)(\mu-\mu_g)^2$ .

$\det\Sigma=3$ and $\Sigma^{-1}=(1/3)\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ . Then $\Sigma^{-1}\mathbf{1}=(1/3,\ 1/3)^\top$ , so $A=2/3$ . And $\Sigma^{-1}\boldsymbol{\mu}=(-1/3,\ 5/3)^\top$ , so $B=4/3$ and $C=1\cdot(-1/3)+3\cdot(5/3)=14/3$ . Hence $\Delta=(2/3)(14/3)-(4/3)^2=12/9=4/3$ .

The global minimum-variance portfolio has $\mu_g=(4/3)/(2/3)=2$ , $\sigma_g^2=3/2$ , and $\mathbf{w}_g=(1/2,\ 1/2)^\top$ . Mean check: half of $1$ plus half of $3$ is $2$ . Variance check: $\mathbf{w}_g^\top\Sigma\mathbf{w}_g=3/2$ .

At target mean $4$ , $\sigma^2=\big((2/3)\cdot 16-2\cdot(4/3)\cdot 4+14/3\big)/(4/3)=(14/3)/(4/3)=7/2$ . Since $N=2$ , the weights are the unique solution of $w_1+w_2=1$ and $1\cdot w_1+3\cdot w_2=4$ , namely $\mathbf{w}=(-1/2,\ 3/2)^\top$ . Here $A/\Delta=1/2$ , so $7/2=3/2+(1/2)(\mu-2)^2$ forces $(\mu-2)^2=4$ . The two means are $4$ and $0$ . The portfolio of mean $4$ is efficient. Its twin has mean $0$ , weights $(3/2,\ -1/2)^\top$ , and the same variance $7/2$ , and it is inefficient.

Two assets have $\Sigma=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $\boldsymbol{\mu}=\begin{pmatrix}1\\3\end{pmatrix}$ , in the same percent units as Section 06. (a) Compute $\Sigma^{-1}$ , $A$ , $B$ , $C$ , and $\Delta$ . (b) Compute the global minimum-variance weights, mean, and variance. (c) Compute the frontier variance at target mean $4$ , and the weights. (d) Find the mean and the weights of the other frontier portfolio with that same variance, and say which of the two is efficient.

EXERCISE 4.3

Mean is linear in the weights, and an affine mix of two frontier portfolios is the frontier portfolio at the mixed mean. Solve $\alpha\cdot 2+(1-\alpha)\cdot(5/2)=3/2$ for $\alpha$ , then apply that same $\alpha$ to the weight vectors. The value of $\alpha$ tells you whether a fund is held short.

The mean equation is $2\alpha+(5/2)(1-\alpha)=3/2$ , so $5/2-(1/2)\alpha=3/2$ , hence $(1/2)\alpha=1$ and $\alpha=2$ . The weights are

2\mathbf{w}_g-\mathbf{w}\!\left(\frac52\right)=2\begin{pmatrix}1/5\\1/5\\3/5\end{pmatrix}-\begin{pmatrix}9/20\\-1/20\\3/5\end{pmatrix}=\begin{pmatrix}-1/20\\9/20\\3/5\end{pmatrix},

which is $\mathbf{w}(3/2)$ from Section 06. Since $\alpha=2$ lies outside the unit interval, the position in $\mathbf{w}(5/2)$ is a short of that fund, and the position in the global minimum-variance fund is leveraged. That short is what reaches the lower limb from two portfolios on the upper limb (the vertex counts as upper). Equal weight $(1/3,\ 1/3,\ 1/3)^\top$ has the same mean $2$ , but $\Sigma$ times that weight is $(1,\ 1,\ 1/3)^\top$ and the variance is $1/3+1/3+1/9=7/9>3/5$ , so equal weight is not on the frontier. The same $\alpha=2$ still produces mean $3/2$ and full investment, and the resulting weight vector equals $\mathbf{w}(3/2)$ only if equal weight equals $\mathbf{w}_g$ . It does not. Uniqueness of the minimizer at mean $3/2$ then says the mix is not minimum-variance. Both funds in the theorem have to lie on the frontier.

In the three-asset market of Section 06, $\mathbf{w}_g=(1/5,\ 1/5,\ 3/5)^\top$ has mean $2$ , and $\mathbf{w}(5/2)=(9/20,\ -1/20,\ 3/5)^\top$ has mean $5/2$ . Find $\alpha$ such that $\alpha\mathbf{w}_g+(1-\alpha)\mathbf{w}(5/2)$ equals the frontier portfolio of mean $3/2$ , and write that weight vector. Does the investor hold both funds long? What fails if $\mathbf{w}_g$ is replaced by the equal-weight portfolio, which also has mean $2$ ?

EXERCISE 4.4

Use $\Sigma\mathbf{w}_g=(1/A)\mathbf{1}$ and $\sigma_g^2=1/A$ . Correlation divides covariance by the product of the two standard deviations. For the overlay, expand $(\mathbf{w}_g+\mathbf{v})^\top\Sigma(\mathbf{w}_g+\mathbf{v})$ and use $\mathbf{v}^\top\mathbf{1}=0$ to kill the cross term. The vector $(1,\ 1,\ -2)^\top$ is dollar-neutral and has mean contribution zero against $(3,\ 1,\ 2)^\top$ , and its quadratic form on $\Sigma$ equals $10$ .

(a) $\mathrm{Cov}(r_3,r_g)=\mathbf{e}_3^\top\Sigma\mathbf{w}_g=\mathbf{e}_3^\top(\sigma_g^2\mathbf{1})=\sigma_g^2=3/5$ . Asset 3 has variance $1$ , so $\sigma_3=1$ and $\rho(r_3,r_g)=(3/5)/\sqrt{3/5}=\sqrt{3/5}$ . The same ratio $\sigma_g/\sigma_3$ is what the general identity $\mathrm{Cov}=\sigma_g^2$ produces for every fully invested portfolio. (b) $\mathbf{v}^\top\Sigma\mathbf{v}=10$ and $\mathbf{w}_g^\top\Sigma\mathbf{v}=(1/A)\mathbf{v}^\top\mathbf{1}=0$ , so the variance of $\mathbf{w}_g+0.1\mathbf{v}$ is $3/5+(0.1)^2\cdot 10=3/5+1/10=7/10$ . The mean is still $2$ because $\mathbf{v}^\top\boldsymbol{\mu}=0$ . Frontier variance at mean $2$ is $3/5<7/10$ , so the portfolio is inside the frontier. (c) The cross term $2\mathbf{w}_g^\top\Sigma\mathbf{v}$ really is zero, so the overlay return is uncorrelated with $r_g$ . The colleague dropped $\mathbf{v}^\top\Sigma\mathbf{v}=10$ , which is why the variance is $7/10$ rather than $3/5$ . To undercut the vertex, a fully invested portfolio would also have to escape the Section 03 argument: covariance with $\mathbf{w}_g$ is stuck at $\sigma_g^2$ , so Cauchy–Schwarz forces $\sigma_q\geq\sigma_g$ . An uncorrelated overlay does not create a portfolio on the wrong side of that bound. It moves you to the right of the vertex.

Keep the three-asset market of Section 06, where $\sigma_g^2=3/5$ and $\Sigma\mathbf{w}_g=(3/5)\mathbf{1}$ . (a) Compute $\mathrm{Cov}(r_3,r_g)$ and the correlation of asset 3 with the global minimum-variance portfolio. (b) Let $\mathbf{v}=(1,\ 1,\ -2)^\top$ . Compute the variance and the mean of $\mathbf{w}_g+0.1\mathbf{v}$ , and say where that portfolio sits relative to the frontier. (c) A colleague notes that $\mathbf{v}$ is uncorrelated with the global minimum-variance portfolio and concludes that $\mathbf{w}_g+0.1\mathbf{v}$ still has variance $3/5$ . Which term in the expansion did the colleague drop? Separately, the covariance identity in Section 03 already forbids every fully invested portfolio from having variance below $3/5$ . Which step of that argument is the one the colleague's conclusion would have to escape in order to undercut the vertex?

09 · Chapter Summary

Object	What it is
Minimum-variance frontier	For each mean $\mu_\star$ , the fully invested portfolio of smallest variance, plotted in $(\sigma,\mu)$
Efficient frontier	The upper limb, $\mu\geq\mu_g$ , including the vertex. Maximizes mean at a given variance
Lower limb	Frontier portfolios with $\mu<\mu_g$ . Minimum-variance for that target, and dominated by the upper twin
Global minimum $\mathbf{w}_g$	$\Sigma^{-1}\mathbf{1}/A$ , variance $1/A$ , mean $B/A$ . Leftmost point. Covariance $\sigma_g^2$ with every fully invested portfolio
Frontier equation	$\sigma^2=(A\mu^2-2B\mu+C)/\Delta=\sigma_g^2+(A/\Delta)(\mu-\mu_g)^2$ , a parabola in variance and a hyperbola in volatility
Discriminant	$\Delta=AC-B^2>0$ when $\Sigma$ is positive definite and the assets do not all have the same mean
Two-fund spanning	Any frontier portfolio is an affine mix $\alpha\mathbf{w}_p+(1-\alpha)\mathbf{w}_q$ of two distinct frontier portfolios. $\alpha\in\mathbb{R}$ , so one fund may be short
Interior, $N\geq 3$	Same mean as some frontier portfolio, strictly larger variance. Empty when $N=2$

The capital market line is not this hyperbola. It is the efficient straight half-line obtained after a risk-free asset is introduced.

Next: Chapter 05 — Two-Fund Separation and the Capital Market Line (coming soon). It adds a risk-free asset, draws the tangency from that asset to the risky frontier of this chapter, and shows that cash plus one tangency portfolio spans every efficient portfolio of the enlarged universe.