Multivariable Calculus and Lagrange Multipliers

1Introduction

Every portfolio decision turns more than one dial: how much 0050, how much bond, how much leverage. Single-variable calculus (LM2, LM4) turns one dial at a time; this module turns them together, adds constraints and prices the constraints themselves. Applied to two assets, it yields the minimum-variance portfolio, the tangency portfolio and the Kelly fractions, and it shows why an answer fitted to the past is not the answer you will get.

Sections 2–5 extend derivatives, Taylor expansions and optimization to several variables and add equality constraints. Section 6 applies them to two assets and resolves the case, Section 7 adds inequality constraints and Section 8 solves everything in Python. Matrix notation for $N$ assets waits for LM10 and LM32.

2Functions of Several Variables and Partial Derivatives

A portfolio's volatility depends on two weights, two volatilities and a correlation at once, and a TWD investor's return on a US bond depends on the bond and on the currency.

Functions, surfaces and level curves

Level curves are the map view of the surface. The central example is the variance of a two-asset portfolio. With weights $w_1,w_2$ and daily returns $R_{1t},R_{2t}$, the portfolio return $R_{pt}=w_1{R}_{1t}+w_2{R}_{2t}$ deviates from its mean by $w_1(R_{1t}-{\bar{R}}_1)+w_2(R_{2t}-{\bar{R}}_2)$. Squaring and averaging gives

The middle average is the covariance of the two returns, positive when they tend to fall on the same side of their means (LM16). Divided by the two standard deviations it gives the correlation $\rho$. It lies between −1 and 1 by the Cauchy–Schwarz inequality $\left|{\sum }_t{x}_t{y}_t\right|\le \sqrt{{\sum }_t{x}_t^2}\sqrt{{\sum }_t{y}_t^2}$, applied to the deviations (stated without proof; LM8). Annualizing each average with $A$,

The same expansion gives two facts used later. Averaging $(R_{1t}-{\bar{R}}_1)(R_{pt}-{\bar{R}}_p)$ instead shows that the covariance of $R_1$ with the portfolio is $w_1{\sigma }_1^2+w_2{\sigma }_{12}$, and weights $(1,-1)$ give $\mathrm{Var}(R_1-R_2)={\sigma }_1^2+{\sigma }_2^2-2{\sigma }_{12}$.

When $|\rho |<1$, the level curves of $V$ are ellipses centred on the all-cash point $(0,0)$, one for each variance level. The budget rule $w_1+w_2=1$ is a straight line across them. Section 5 shows that the minimum-variance portfolio sits where that line just touches an ellipse, and Exhibit 5 draws the picture.

Partial derivatives

To compute a partial derivative, treat every other variable as a constant and apply the rules of LM2. For the portfolio variance,

The first bracket is the covariance of asset 1 with the portfolio, found above. So the marginal variance of a position is twice its covariance with the portfolio it joins, a fact LM36 builds into risk budgeting.

Higher-order partial derivatives. Differentiating again gives $h_{xx}$, $h_{yy}$ and the mixed partials $h_{xy}=(h_x{)}_y$ and $h_{yx}=(h_y{)}_x$. For every function in this book the order does not matter.

The total differential and the linear approximation

Move both inputs at once, by $\Delta x$ and $\Delta y$. Going first along $x$ and then along $y$, and using the one-variable linear approximation of LM2 on each leg, gives

When $h$ has continuous partial derivatives, the error is small relative to the step; when its second partial derivatives are continuous too, the error is of second order in $(\Delta x,\Delta y)$. The linear part, $dh=h_{x}dx+h_{y}dy$, is the total differential. Dividing by a time step gives the multivariable chain rule: if $x=x(t)$ and $y=y(t)$, then $\frac{d}{dt}h(x(t),y(t))=h_x{x}^{\prime }(t)+h_y{y}^{\prime }(t)$.

3Gradient, Hessian and the Multivariate Taylor Expansion

Directional derivatives and the gradient

Pick a direction, given by a unit vector $\mathbf{u}=(u_1,u_2)$, and walk from $\mathbf{a}$ along it; the height of the surface along the walk is $z(t):=h(\mathbf{a}+t\mathbf{u})$. By the chain rule, its slope at the start, the directional derivative, is

Two geometric facts follow. First, by the Cauchy–Schwarz inequality $\nabla h\cdot \mathbf{u}\le \Vert \nabla h\Vert \Vert \mathbf{u}\Vert =\Vert \nabla h\Vert$, with equality only when $\mathbf{u}$ points along $\nabla h$. So the gradient points in the direction of steepest ascent, and its length is the steepest slope.

Second, if a point $\mathbf{x}(t)$ moves along a level curve, then $h(\mathbf{x}(t))=b$, and differentiating gives $\nabla h\cdot {\mathbf{x}}^{\prime }(t)=0$. So the gradient is perpendicular to the level curves.

For a step $\mathbf{\Delta }$ that is not a unit vector, $\nabla h\cdot \mathbf{\Delta }$ is the first-order change in $h$; case (b) of Example 2 used exactly this, with $\mathbf{\Delta }=(0.01,-0.01)$.

The Hessian

The Hessian measures curvature in every direction at once. Differentiating $z(t)=h(\mathbf{a}+t\mathbf{u})$ twice with the chain rule gives

the curvature of $h$ along $\mathbf{u}$. For the portfolio variance the Hessian is constant, $H=2\left(\begin{array}{cc}{\sigma }_1^2& {\sigma }_{12}\\ {\sigma }_{12}& {\sigma }_2^2\end{array}\right)=2\Sigma$: twice the covariance matrix $\Sigma$, the object LM10 studies.

The second-order Taylor expansion

The one-variable Taylor polynomial of LM3, applied to $z(t)=h(\mathbf{a}+t\mathbf{\Delta })$ at $t=1$, gives $z(1)\approx z(0)+z^{\prime }(0)+\frac{1}{2}{z}^{\prime \prime }(0)$. Substituting the two derivatives just computed:

with all derivatives evaluated at $\mathbf{a}$. If $h$ has continuous third partial derivatives, the Lagrange remainder of LM3 (3.5), applied to $z$, puts the error at $z^{\prime \prime \prime }(\xi )/6$ for some $\xi \in (0,1)$: third order in $\mathbf{\Delta }$. With three variables the pattern is the same, with three first derivatives and a $3\times 3$ Hessian. A polynomial of degree two is reproduced exactly by (7.2), and one of degree three once the third-order terms are added.

A Taylor attribution of the 42/58 mix's extra risk

Hold the weights of the case fixed at $w_1=42.317\%$ in 0050 and $w_2=57.683\%$ in 00679B. Their variance is then a function of three inputs,

Between the estimation and test windows the inputs moved from $(16.37\%,13.55\%,-0.221)$ to $(23.40\%,13.43\%,-0.022)$. The partial derivatives at the starting point, from the unrounded estimates, are $V_{{\sigma }_1}=2w_1^2{\sigma }_1+2w_1{w}_2\rho {\sigma }_2=0.043994$, $V_{{\sigma }_2}=2w_2^2{\sigma }_2+2w_1{w}_2\rho {\sigma }_1=0.072465$ and $V_{\rho }=2w_1{w}_2{\sigma }_1{\sigma }_2=0.010827$. Multiplying each by its input's change gives the first-order terms of Exhibit 2.

The first-order approximation explains 51.70 of the 69.67 %² rise in variance, or 74%. The Hessian terms supply the rest, slightly overshooting. The two largest are $\frac{1}{2}{V}_{{\sigma }_1{\sigma }_1}(\Delta {\sigma }_1{)}^2=8.86$, with $V_{{\sigma }_1{\sigma }_1}=2w_1^2$, and the cross term $V_{{\sigma }_1\rho }\Delta {\sigma }_1\Delta \rho =9.28$, with $V_{{\sigma }_1\rho }=2w_1{w}_2{\sigma }_2$. A higher correlation hurts more when 0050 is more volatile, and only a second-order model can see that. Because $V$ is a polynomial of degree three in its inputs, the single third-order term $2w_1{w}_2\Delta {\sigma }_1\Delta {\sigma }_2\Delta \rho$, about −0.08, makes the expansion exact.

To first order, 0050's higher volatility accounts for 44% of the rise in variance and the higher correlation for 31%. The second-order terms add 26%, half of it from the two moving together.

A two-variable Taylor model of portfolio growth

The most important use of (7.2) in this book is the growth rate of a levered portfolio. Hold a fraction $f_1$ of capital in 0050 and $f_2$ in 00679B, and finance the rest at an illustrative risk-free rate $r_f=1.5\%$ a year, $r_f/A$ a day. The daily return is $R_p=r_f/A+f_1{X}_1+f_2{X}_2$, with excess returns $X_i=R_i-r_f/A$. The quantity a Kelly investor maximizes (Kelly 1956; LM4) is the average log growth, annualized:

Differentiate under the sum: $g_{f_i}=A\cdot \frac{1}{T}{\sum }_t{X}_{it}/(1+R_{pt})$ and $g_{f_i{f}_j}=-A\cdot \frac{1}{T}{\sum }_t{X}_{it}{X}_{jt}/(1+R_{pt}{)}^2$. At $\mathbf{f}=0$ these are the mean excess returns and minus the mean products, each divided by a power of $1+r_f/A$.

On the data, $\nabla g(0)=(0.2275,-0.0255)$, which is $(e_1,e_2)$ up to the factor $1/(1+r_f/A)$. The Hessian $H(0)$ has entries $(-0.0406,0.0029,-0.0182)$, against $-{\sigma }_1^2=-0.0404$, $-{\sigma }_{12}=0.0029$ and $-{\sigma }_2^2=-0.0182$. The small gap in the first entry is the squared daily mean, smaller than the variance by a factor of order $1/A$. Dropping it and the factors $1+r_f/A$, (7.2) becomes

The same model follows from expanding $\ln (1+R_p)\approx R_p-\frac{1}{2}{R}_p^2$ day by day and averaging, as in LM3 (3.8). It neglects the squared daily mean, $\frac{1}{2}A{\bar{R}}_p^2$, and the terms of third and higher order in the daily portfolio return. In matrix form (7.3) reads $g\approx r_f+{\mathbf{f}}^{\top }\mathbf{e}-\frac{1}{2}{\mathbf{f}}^{\top }\Sigma \mathbf{f}$, the starting point of LM43.

The model is excellent for unlevered mixes and degrades as leverage grows. For 0050 alone the 1.0 bp error is the squared mean's 1.2 bp less 0.2 bp of higher-order terms. At full Kelly the model overstates growth by 2.02 points a year: 0.35 from the dropped squared daily mean and 1.67 from the third- and higher-order terms of $\ln (1+R_p)$, which grow with the daily moves. The fourth-order term alone costs 1.87 points; the third adds back 0.49, and fifth and higher orders cost 0.29.

4Unconstrained Optimization in Two Variables

LM4 found optima of one-variable functions with two conditions: a zero derivative and the sign of the second derivative. Both carry over to two variables, the second with more care, because a surface can curve up in one direction and down in another.

First-order conditions

If $h$ is differentiable and has a local maximum or minimum at an interior point $\mathbf{a}$, then every one-variable slice through $\mathbf{a}$ has zero slope there. In particular $h_x(\mathbf{a})=h_y(\mathbf{a})=0$, that is, $\nabla h(\mathbf{a})=0$. Points where the gradient vanishes are critical (or stationary) points. As in one variable, the condition is necessary, not sufficient.

Second-order conditions: the 2×2 test

At a critical point the linear term of (7.2) vanishes, so for small steps $h(\mathbf{a}+\mathbf{\Delta })-h(\mathbf{a})\approx \frac{1}{2}{\mathbf{\Delta }}^{\top }H\mathbf{\Delta }$, and the sign of this quadratic form decides the shape. If $h_{xx}\ne 0$, completing the square gives

If $\det H>0$, both terms have the sign of $h_{xx}$, and the form is non-zero for every $\mathbf{\Delta }\ne 0$. If $\det H<0$, the two terms have opposite signs, and the form takes both signs. When $h$ has continuous second partial derivatives, the remainder of (7.2) is small next to $\Vert \mathbf{\Delta }{\Vert }^2$ near $\mathbf{a}$ and cannot overturn a definite form (stated without proof; Stewart, Clegg and Watson 2021, §14.7). When $\det H=0$ the test is silent and higher-order terms decide: $x^2+y^4$ has a minimum at the origin and $x^2+y^3$ does not, with the same Hessian there. Exhibit 4 collects the cases. In matrix language, $H$ is positive definite exactly when $h_{xx}>0$ and $\det H>0$, the $2\times 2$ case of a criterion LM10 extends to $N$ assets.

Global optima and convexity

A local test says nothing about the rest of the surface. One condition does: if ${\mathbf{\Delta }}^{\top }H\mathbf{\Delta }\ge 0$ for every $\mathbf{\Delta }$ at every point (the Hessian is positive semidefinite everywhere), $h$ is convex, and any critical point is a global minimum (stated without proof; LM34 develops convexity). Quadratic functions have a constant Hessian, so for them the local test is global.

Portfolio variance has $H=2\Sigma$ with $\det H=4{\sigma }_1^2{\sigma }_2^2(1-{\rho }^2)\ge 0$: it is convex, and strictly convex when $|\rho |<1$. The growth model (7.3) has $H=-\Sigma$ and is concave.

5Equality Constraints: Lagrange Multipliers

Most portfolio problems come with constraints: the weights must sum to one, the volatility must hit a target, an order must be completed. When a constraint can be solved for one variable, substitution reduces the problem to LM4. Lagrange's method works without solving the constraint, and it delivers a bonus: the price of the constraint.

The tangency condition

Consider optimizing $h(x,y)$ subject to $\psi (x,y)=b$, and walk along the constraint curve. At a constrained optimum the slope of $h$ along the curve must be zero, so $\nabla h$ is perpendicular to the curve there. The constraint curve is a level curve of $\psi$, so $\nabla \psi$ is perpendicular to it too (Section 3). Two vectors perpendicular to the same curve are parallel: at the optimum, the level curve of $h$ just touches the constraint curve. Exhibit 5 shows this for the minimum-variance portfolio.

Algebraically, suppose ${\psi }_y\ne 0$ at the optimum. Near it the constraint defines $y$ as a function of $x$ (the implicit function theorem, stated without proof), and differentiating $\psi (x,y(x))=b$ gives ${\psi }_x+{\psi }_y{y}^{\prime }(x)=0$, so $y^{\prime }=-{\psi }_x/{\psi }_y$. The one-variable function $F(x)=h(x,y(x))$ has a critical point at the optimum:

Writing $\lambda :=h_y/{\psi }_y$, this says $h_x=\lambda {\psi }_x$, and $h_y=\lambda {\psi }_y$ holds by definition of $\lambda$.

The three conditions ${\mathcal{L}}_x={\mathcal{L}}_y={\mathcal{L}}_{\lambda }=0$ give candidates; whether a candidate is a maximum, a minimum or neither needs a second-order or a global argument. A convex objective (for a minimum) or a concave one (for a maximum) with a linear constraint makes a candidate the global answer, as in the minimum-variance problems of Section 6; Section 7 states the general result. A curved constraint can produce several candidates: Example 5 has two.

The multiplier is a shadow price

Proof. By the chain rule and (7.5), $J^{\prime }(b)=h_x{x}^{\ast \prime }+h_y{y}^{\ast \prime }={\lambda }^{\ast }({\psi }_x{x}^{\ast \prime }+{\psi }_y{y}^{\ast \prime })$. Differentiating the identity $\psi (x^{\ast }(b),y^{\ast }(b))=b$ shows that the bracket equals 1. $■$

This is the envelope theorem of LM4 (4.16) applied to the Lagrangian, whose partial derivative in $b$ is $\lambda$. So $\Delta J\approx {\lambda }^{\ast }\Delta b$, with an error of second order in $\Delta b$. The units of ${\lambda }^{\ast }$ are units of objective per unit of constraint: NT$ per contract, return per unit of risk, variance per unit of capital.

6Two-Asset Portfolios by Calculus

This section applies Sections 2–5 to two assets, in the tradition of Markowitz (1952) and of Merton's (1972) closed forms: the minimum-variance weight, in and out of sample, then the tangency portfolio and the Kelly fractions.

Minimum variance, by substitution and by Lagrange

Substitution. With $w$ in asset 1 and $1-w$ in asset 2, the variance is a one-variable function,

Its derivative is $V^{\prime }(w)=2w{\sigma }_1^2-2(1-w){\sigma }_2^2+2(1-2w){\sigma }_{12}$. Setting it to zero:

The second derivative is $V^{\prime \prime }(w)=2{\sigma }_d^2$, where ${\sigma }_d^2:={\sigma }_1^2+{\sigma }_2^2-2{\sigma }_{12}=\mathrm{Var}(R_1-R_2)$ (Section 2). It is positive unless the two returns differ by a constant, so $w^{\ast }$ is the global minimum. Substituting $w^{\ast }$ back:

Because $V$ is a quadratic in $w$ with second derivative $2{\sigma }_d^2$, it can be written exactly around its minimum:

Lagrange. Minimize $V(w_1,w_2)$ subject to $w_1+w_2=1$. With $\mathcal{L}=V-\lambda (w_1+w_2-1)$, the conditions are

Subtracting gives $w_1({\sigma }_1^2-{\sigma }_{12})=w_2({\sigma }_2^2-{\sigma }_{12})$, which with $w_1+w_2=1$ is (7.7) again. The two conditions say more: at the minimum, each asset has the same covariance with the portfolio.

Multiply the first condition by $w_1$, the second by $w_2$ and add: $2V=\lambda (w_1+w_2)=\lambda$, so ${\lambda }^{\ast }=2{\sigma }_{\min }^2$. That is the shadow price (7.6) of the budget: investing $b$ units in the same proportions gives variance $b^2{\sigma }_{\min }^2$, whose derivative at $b=1$ is $2{\sigma }_{\min }^2$.

Correlation and diversification

Formula (7.8) shows how correlation sets the benefit of diversification.

• $\rho =+1$. Then ${\sigma }_{\min }^2=0$ only through a short position: $w^{\ast }=-{\sigma }_2/({\sigma }_1-{\sigma }_2)$ when ${\sigma }_1>{\sigma }_2$. Long-only, volatility is linear in $w$, and there is no diversification.
• $\rho =0$. Then $w^{\ast }={\sigma }_2^2/({\sigma }_1^2+{\sigma }_2^2)$: each asset is weighted by its inverse variance.
• $\rho =-1$. Then $w^{\ast }={\sigma }_2/({\sigma }_1+{\sigma }_2)$ and ${\sigma }_{\min }=0$: a perfect hedge without shorting.

The minimum-variance portfolio shorts the riskier asset when the numerator of (7.7) turns negative, ${\sigma }_2^2<{\sigma }_{12}$, that is, when $\rho >{\sigma }_2/{\sigma }_1$ (asset 1 being the riskier one). For 0050 and 00679B the ratio is $13.50/20.11=0.67$, far above any correlation in the sample, so the minimum is interior.

The envelope theorem of LM4 (4.16) prices a change in $\rho$ without differentiating (7.8): because $V^{\prime }(w^{\ast })=0$, $d{\sigma }_{\min }^2/d\rho =\partial V/\partial \rho =2w_1^{\ast }{w}_2^{\ast }{\sigma }_1{\sigma }_2$ at $w^{\ast }$. With the 2017–2021 estimates this is 0.0108, so a correlation 0.1 higher raises ${\sigma }_{\min }^2$ by 0.00108 (exact to that precision) and ${\sigma }_{\min }$ from 9.22% to 9.79%.

Back to the case: in sample and out of sample

The calculation was right; the inputs changed, as Exhibit 6 shows year by year. In 2022 both funds fell by about 21–22% while their daily correlation was −0.06: a correlation near zero does not prevent a joint loss in a year when both assets drift down.

From 2023 the trailing one-year correlation moved between −0.29 (April 2025) and +0.25 (December 2023) (Exhibit 7). At lower frequencies the correlation turned positive: between the two windows it rose from −0.29 to +0.21 for monthly returns and from −0.15 to +0.09 for weekly ones.

Identity (7.9) now splits the miss exactly, and Exhibit 8 draws it. With the test-window inputs, the lowest attainable variance was ${\sigma }_{\min }^2=133.15$ %² at $w^{\ast }=25.3\%$, and the coefficient of the quadratic term was ${\sigma }_d^2=0.07419$, half the curvature $V^{\prime \prime }=0.1484$. The in-sample weight was 0.17067 (17.07 points) too high, which adds $0.07419\times 0.17067^2=21.61$ %². Together, $133.15+21.61=154.76$ %², the 12.44% volatility the 42/58 mix actually realized.