Chapter 15

Covariance Matrices & Quadratic Forms

00 · Symbol Glossary

$\Sigma$Sigma — covariance matrix

The uppercase Greek letter sigma used for the covariance matrix of a random vector $\mathbf{r}\in\mathbb{R}^p$ . Entry $\Sigma_{ij}=\text{Cov}(r_i,r_j)$ . The diagonal entries are variances; off-diagonal entries are covariances. Not to be confused with the summation $\sum$ (lowercase indices) — context and font distinguish them.

$\mathbf{x}^T A \mathbf{x}$x transpose A x — quadratic form

A quadratic form in $\mathbf{x}$ defined by matrix $A$ . The result is a scalar — a degree-2 polynomial in the components of $\mathbf{x}$ . For symmetric $A$ , the quadratic form uniquely determines $A$ . Portfolio variance $\mathbf{w}^T\Sigma\mathbf{w}$ is the fundamental financial quadratic form.

$\bar{\mathbf{x}}$x bar — sample mean vector

The sample mean vector: $\bar{\mathbf{x}} = \frac{1}{n}\sum_{t=1}^n \mathbf{x}_t$ . Each component $\bar{x}_j$ is the sample mean of the $j$ -th variable. The mean-centered data matrix has rows $(\mathbf{x}_t-\bar{\mathbf{x}})^T$ — translating the data cloud to the origin before computing scatter.

$S$S — sample covariance matrix

The unbiased sample covariance matrix: $S=\frac{1}{n-1}\sum_{t=1}^n(\mathbf{x}_t-\bar{\mathbf{x}})(\mathbf{x}_t-\bar{\mathbf{x}})^T$ . Dividing by $n-1$ (Bessel's correction) rather than $n$ makes $S$ an unbiased estimator of the population covariance $\Sigma$ . Always symmetric PSD; PD when $n>p$ .

$\rho_{ij}$rho i j — correlation

The Pearson correlation between assets $i$ and $j$ : $\rho_{ij}=\frac{\Sigma_{ij}}{\sqrt{\Sigma_{ii}\Sigma_{jj}}}$ . Always in $[-1,1]$ . The correlation matrix $\mathbf{C}$ has $C_{ij}=\rho_{ij}$ and ones on the diagonal. $\Sigma = D\mathbf{C}D$ where $D=\text{diag}(\sqrt{\Sigma_{11}},\ldots,\sqrt{\Sigma_{pp}})$ .

01 · Quadratic Forms

A quadratic form is the generalisation of $ax^2$ to multiple variables. Instead of one variable, you have a vector $\mathbf{x}$ ; instead of the scalar coefficient $a$ , you have a matrix $A$ .

Definition — Quadratic Form

Given a symmetric matrix $A \in \mathbb{R}^{n\times n}$ , the quadratic form in $\mathbf{x}\in\mathbb{R}^n$ is:

Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = \sum_{i=1}^n\sum_{j=1}^n a_{ij} x_i x_j

Every quadratic polynomial with no linear or constant terms can be written as $\mathbf{x}^TA\mathbf{x}$ for a unique symmetric $A$ . The diagonal entry $a_{ii}$ is the coefficient of $x_i^2$ ; the off-diagonal entry $2a_{ij}$ (for $i\neq j$ ) is the coefficient of $x_ix_j$ .

Step-by-step — Expanding $\mathbf{x}^TA\mathbf{x}$ for $A=\begin{pmatrix}2&1\\1&3\end{pmatrix}$, $\mathbf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}$

Compute $A\mathbf{x}$ first:

A\mathbf{x} = \begin{pmatrix}2&1\\1&3\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix} = \begin{pmatrix}2x_1+x_2\\x_1+3x_2\end{pmatrix}

Multiply $\mathbf{x}^T(A\mathbf{x})$ :

$\mathbf{x}^T A\mathbf{x} = x_1(2x_1+x_2) + x_2(x_1+3x_2) = 2x_1^2 + x_1x_2 + x_1x_2 + 3x_2^2 = 2x_1^2 + 2x_1x_2 + 3x_2^2$ .

The two cross terms $x_1x_2$ add up because $a_{12}=a_{21}=1$ .

Read off the structure: coefficient of $x_1^2$ is $a_{11}=2$ ; coefficient of $x_2^2$ is $a_{22}=3$ ; coefficient of $x_1x_2$ is $a_{12}+a_{21}=2a_{12}=2$ . Always: for symmetric $A$ , the coefficient of $x_ix_j$ (with $i\neq j$ ) in $\mathbf{x}^TA\mathbf{x}$ is $2a_{ij}$ .

✓ Example — Evaluating a Quadratic Form at a Specific Point

$A=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ , $\mathbf{x}=\begin{pmatrix}2\\-1\end{pmatrix}$ .

$Q(\mathbf{x}) = 2(2)^2 + 2(2)(-1) + 3(-1)^2 = 8 - 4 + 3 = 7$ .

Check via matrix product: $A\mathbf{x}=\begin{pmatrix}3\\-1\end{pmatrix}$ ; $\mathbf{x}^T(A\mathbf{x})=2(3)+(-1)(-1)=6+1=7$ ✓.

❌ What Breaks — Asymmetric Matrix Changes the Quadratic Form

Writing a quadratic form using an asymmetric matrix is valid but non-unique. If $A$ is asymmetric, $\mathbf{x}^TA\mathbf{x} = \mathbf{x}^T(\frac{A+A^T}{2})\mathbf{x}$ — the quadratic form depends only on the symmetric part $\frac{A+A^T}{2}$ , not on the antisymmetric part $\frac{A-A^T}{2}$ . The antisymmetric contribution $\mathbf{x}^T\frac{A-A^T}{2}\mathbf{x}=0$ for all $\mathbf{x}$ . When working with quadratic forms, always symmetrise $A$ first.

02 · Classifying Quadratic Forms

The sign of $\mathbf{x}^TA\mathbf{x}$ classifies the matrix and the form.

Sign of $\mathbf{x}^TA\mathbf{x}$	Classification	Eigenvalues
$> 0$ for all $\mathbf{x}\neq\mathbf{0}$	Positive definite	All $\lambda_i > 0$
$\geq 0$ for all $\mathbf{x}$	Positive semi-definite	All $\lambda_i \geq 0$
$\leq 0$ for all $\mathbf{x}$	Negative semi-definite	All $\lambda_i \leq 0$
$< 0$ for all $\mathbf{x}\neq\mathbf{0}$	Negative definite	All $\lambda_i < 0$
Both signs possible	Indefinite	Mixed signs

Indefinite quadratic forms arise in saddle-point problems in optimisation — the Hessian is indefinite at a saddle point.

03 · The Sample Covariance Matrix

The sample covariance matrix is the key object connecting statistics to linear algebra.

Definition — Sample Covariance Matrix

Given $n$ observations of a $p$ -dimensional random vector, arranged as rows of a data matrix $X \in \mathbb{R}^{n\times p}$ , the sample covariance matrix is:

S = \frac{1}{n-1} X_c^T X_c

where $X_c$ is the mean-centered data matrix with rows $(\mathbf{x}_t - \bar{\mathbf{x}})^T$ and $\bar{\mathbf{x}} = \frac{1}{n}\sum_t \mathbf{x}_t$ .

Equivalently: $S_{ij} = \frac{1}{n-1}\sum_{t=1}^n (x_{ti}-\bar{x}_i)(x_{tj}-\bar{x}_j)$ .

Structure: $S$ is symmetric ( $S_{ij}=S_{ji}$ ), PSD ( $S=\frac{1}{n-1}X_c^TX_c\succeq0$ ), and PD when $\text{rank}(X_c)=p$ (i.e. $n>p$ and no column of $X$ is a linear combination of others).

Step-by-step — Computing the sample covariance matrix from 3 observations of 2 assets

Set up data: 3 daily returns for two assets.

X = \begin{pmatrix}0.01 & 0.02 \\ 0.03 & 0.01 \\ -0.01 & 0.03\end{pmatrix}

Rows = days. Columns = Asset 1, Asset 2.

Compute sample means: $\bar{x}_1 = (0.01+0.03-0.01)/3 = 0.03/3 = 0.01$ . $\bar{x}_2=(0.02+0.01+0.03)/3=0.06/3=0.02$ .

Mean-center the data: subtract $\bar{\mathbf{x}}^T=(0.01,0.02)$ from each row.

X_c = \begin{pmatrix}0.01-0.01 & 0.02-0.02\\0.03-0.01 & 0.01-0.02\\-0.01-0.01 & 0.03-0.02\end{pmatrix} = \begin{pmatrix}0 & 0\\0.02 & -0.01\\-0.02 & 0.01\end{pmatrix}

Check: each column of $X_c$ sums to zero — the centring worked. $(0+0.02-0.02=0)$ , $(0-0.01+0.01=0)$ ✓.

Compute $X_c^TX_c$ :

X_c^TX_c = \begin{pmatrix}0&0.02&-0.02\\0&-0.01&0.01\end{pmatrix}\begin{pmatrix}0&0\\0.02&-0.01\\-0.02&0.01\end{pmatrix}

$(1,1)$ : $0+0.0004+0.0004=0.0008$ . $(1,2)=(2,1)$ : $0-0.0002-0.0002=-0.0004$ . $(2,2)$ : $0+0.0001+0.0001=0.0002$ .

$X_c^TX_c = \begin{pmatrix}0.0008 & -0.0004\\-0.0004 & 0.0002\end{pmatrix}$ .

Divide by $n-1=2$ :

S = \frac{1}{2}\begin{pmatrix}0.0008 & -0.0004\\-0.0004 & 0.0002\end{pmatrix} = \begin{pmatrix}0.0004 & -0.0002\\-0.0002 & 0.0001\end{pmatrix}

$S_{11}=0.0004$ is the sample variance of Asset 1 (standard deviation $\approx2\%$ ). $S_{22}=0.0001$ is the sample variance of Asset 2 (standard deviation $1\%$ ). $S_{12}=-0.0002$ : assets move opposite each other (negative covariance).

Bessel's Correction: $n-1$ not $n$

Dividing by $n-1$ rather than $n$ corrects for the bias introduced by estimating the mean from the same data. With $n$ observations, once you fix the sample mean, only $n-1$ deviations are free — the last one is determined by the constraint that deviations sum to zero. Dividing by $n$ gives a biased estimate; $n-1$ gives the unbiased estimate.

04 · Portfolio Variance as a Quadratic Form

The most direct application of the covariance matrix in finance is portfolio variance.

Definition — Portfolio Variance

Given a portfolio with weight vector $\mathbf{w}\in\mathbb{R}^p$ (with $\sum_i w_i=1$ ) and asset covariance matrix $\Sigma$ , the portfolio variance is:

\sigma_p^2 = \mathbf{w}^T \Sigma \mathbf{w} = \sum_{i=1}^p\sum_{j=1}^p w_i \Sigma_{ij} w_j

This is a quadratic form in the weights. Since $\Sigma\succeq0$ , $\sigma_p^2\geq0$ for all $\mathbf{w}$ — portfolio variance cannot be negative. Since portfolio variance is a quadratic form, minimising variance over all weights $\mathbf{w}$ is a convex quadratic programme — it has a unique global minimum.

✓ Example — Two-Asset Portfolio Variance

Two assets with $\Sigma=\begin{pmatrix}0.04&0.02\\0.02&0.01\end{pmatrix}$ and weights $\mathbf{w}=\begin{pmatrix}0.6\\0.4\end{pmatrix}$ .

$\mathbf{w}^T\Sigma\mathbf{w}=0.6^2(0.04)+2(0.6)(0.4)(0.02)+0.4^2(0.01)$

$= 0.36(0.04)+0.48(0.02)+0.16(0.01) = 0.0144+0.0096+0.0016=0.0256$ .

Portfolio standard deviation: $\sigma_p=\sqrt{0.0256}=0.16=16\%$ . The $2\times$ cross term $2(0.6)(0.4)(0.02)$ comes from $\Sigma_{12}=\Sigma_{21}=0.02$ appearing in both $(i,j)=(1,2)$ and $(i,j)=(2,1)$ in the double sum.

05 · Spectral Decomposition of the Covariance Matrix

Since $\Sigma$ is symmetric PSD, the spectral theorem gives $\Sigma=Q\Lambda Q^T$ where $Q$ is orthogonal and $\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_p)$ with $\lambda_i\geq0$ .

Portfolio variance in the eigenvector basis: Let $\mathbf{f}=Q^T\mathbf{w}$ (coordinates of $\mathbf{w}$ in the eigenvector basis).

\sigma_p^2 = \mathbf{w}^T\Sigma\mathbf{w} = \mathbf{w}^TQ\Lambda Q^T\mathbf{w} = (Q^T\mathbf{w})^T\Lambda(Q^T\mathbf{w}) = \mathbf{f}^T\Lambda\mathbf{f} = \sum_{k=1}^p \lambda_k f_k^2

Each term $\lambda_k f_k^2$ is the variance contribution from principal component $k$ . The eigenvalue $\lambda_k$ is the variance of the $k$ -th PC; the coefficient $f_k$ is the portfolio's exposure to PC $k$ .

To minimise portfolio variance while maintaining expected return, you want to avoid loading heavily on large- $\lambda_k$ directions — the high-variance PCs.

06 · Correlation Matrix and Standardisation

The covariance matrix conflates two things: the scale of each asset (measured in $\text{return}^2$ ) and the linear dependence structure. The correlation matrix separates these.

Definition — Correlation Matrix

The correlation matrix $\mathbf{C}$ is obtained by standardising the covariance matrix:

C_{ij} = \rho_{ij} = \frac{\Sigma_{ij}}{\sqrt{\Sigma_{ii}}\sqrt{\Sigma_{jj}}}, \qquad \mathbf{C} = D^{-1}\Sigma D^{-1}

where $D = \text{diag}(\sqrt{\Sigma_{11}},\ldots,\sqrt{\Sigma_{pp}})$ is the diagonal matrix of standard deviations.

Equivalently: $\Sigma = D\mathbf{C}D$ .

$\mathbf{C}$ has ones on the diagonal and entries in $[-1,1]$ . $\mathbf{C}$ is PSD (since $\mathbf{x}^T\mathbf{C}\mathbf{x}=\|D^{-1}\mathbf{x}\|$ under the Mahalanobis metric... more precisely: $\mathbf{C}$ is PSD because $\Sigma=DCD$ and $D$ is invertible).

Step-by-step — Extracting correlations from $\Sigma=\begin{pmatrix}0.04&0.03\\0.03&0.09\end{pmatrix}$

Read off variances: $\Sigma_{11}=0.04$ (Asset 1 variance), $\Sigma_{22}=0.09$ (Asset 2 variance). Standard deviations: $\sigma_1=\sqrt{0.04}=0.2=20\%$ , $\sigma_2=\sqrt{0.09}=0.3=30\%$ .

Compute correlation: $\rho_{12}=\frac{\Sigma_{12}}{\sigma_1\sigma_2}=\frac{0.03}{0.2\times0.3}=\frac{0.03}{0.06}=0.5$ .

Write the correlation matrix:

\mathbf{C} = \begin{pmatrix}1 & 0.5 \\ 0.5 & 1\end{pmatrix}

$\rho=0.5$ means the two assets have moderate positive co-movement. The correlation does not depend on the units (percent, basis points, etc.) — it is dimensionless.

07 · Practice Exercises

EXERCISE 15.1

For diagonal entry $a_{ii}$ : the coefficient of $x_i^2$ in $\mathbf{x}^TA\mathbf{x}$ . For off-diagonal $a_{ij}$ : the coefficient of $x_ix_j$ in the expansion is $2a_{ij}$ (because $A$ is symmetric). Work backwards from the polynomial.

Polynomial: $Q(\mathbf{x})=5x_1^2+4x_1x_2-2x_1x_3+3x_2^2+6x_2x_3+x_3^2$ .

Reading coefficients: $a_{11}=5$ (coeff of $x_1^2$ ); $a_{22}=3$ (coeff of $x_2^2$ ); $a_{33}=1$ (coeff of $x_3^2$ ).

Cross terms: coeff of $x_1x_2$ is $4=2a_{12}\Rightarrow a_{12}=2$ . Coeff of $x_1x_3$ is $-2=2a_{13}\Rightarrow a_{13}=-1$ . Coeff of $x_2x_3$ is $6=2a_{23}\Rightarrow a_{23}=3$ .

A = \begin{pmatrix}5&2&-1\\2&3&3\\-1&3&1\end{pmatrix}

Verify: $\mathbf{e}_1^TA\mathbf{e}_1=5$ ✓; $\mathbf{e}_2^TA\mathbf{e}_2=3$ ✓; $(\mathbf{e}_1+\mathbf{e}_2)^TA(\mathbf{e}_1+\mathbf{e}_2)=5+2(2)+3=12$ , which matches $5(1)+4(1)(1)+3(1)=12$ ✓.

Find the symmetric matrix $A$ such that $\mathbf{x}^TA\mathbf{x} = 5x_1^2 + 4x_1x_2 - 2x_1x_3 + 3x_2^2 + 6x_2x_3 + x_3^2$ .

EXERCISE 15.2

Compute the sample mean vector, mean-center the data matrix, form $X_c^TX_c$ , and divide by $n-1$ . Then extract the diagonal for variances and use $\rho=S_{12}/\sqrt{S_{11}S_{22}}$ for correlation.

Data: $X=\begin{pmatrix}2&4\\4&2\\3&3\\5&1\\1&5\end{pmatrix}$ (5 observations, 2 assets).

$\bar{x}_1=(2+4+3+5+1)/5=15/5=3$ . $\bar{x}_2=(4+2+3+1+5)/5=15/5=3$ .

$X_c=\begin{pmatrix}-1&1\\1&-1\\0&0\\2&-2\\-2&2\end{pmatrix}$ .

$X_c^TX_c=\begin{pmatrix}(-1)^2+1^2+0^2+2^2+(-2)^2 & (-1)(1)+1(-1)+0+2(-2)+(-2)(2)\\(\text{same by symmetry}) & 1^2+(-1)^2+0^2+(-2)^2+2^2\end{pmatrix}$ .

$(1,1)$ : $1+1+0+4+4=10$ . $(1,2)$ : $-1-1+0-4-4=-10$ . $(2,2)$ : $1+1+0+4+4=10$ .

$S=\frac{1}{4}\begin{pmatrix}10&-10\\-10&10\end{pmatrix}=\begin{pmatrix}2.5&-2.5\\-2.5&2.5\end{pmatrix}$ .

$\sigma_1=\sigma_2=\sqrt{2.5}\approx1.58$ . Correlation: $\rho=\frac{-2.5}{2.5}=-1$ — perfect negative correlation. Asset 2's returns are exactly $6-$ Asset 1's returns — they always move in opposite directions by the same amount.

Compute the $2\times2$ sample covariance matrix $S$ from the data $X=\begin{pmatrix}2&4\\4&2\\3&3\\5&1\\1&5\end{pmatrix}$ (5 observations, 2 assets). Also compute the correlation $\rho_{12}$ and interpret the result.

EXERCISE 15.3

Expand $\mathbf{w}^T\Sigma\mathbf{w}=\sum_i\sum_j w_i\Sigma_{ij}w_j$ using $\Sigma_{11}$ , $\Sigma_{22}$ , $\Sigma_{12}=\Sigma_{21}$ . The cross term appears twice. Then minimise over $w_1=1-w_2$ .

$\Sigma=\begin{pmatrix}0.09&0.06\\0.06&0.16\end{pmatrix}$ , $\mathbf{w}=\begin{pmatrix}w\\1-w\end{pmatrix}$ .

$\sigma_p^2=w^2(0.09)+2w(1-w)(0.06)+(1-w)^2(0.16)$ .

$=0.09w^2+0.12w-0.12w^2+0.16-0.32w+0.16w^2$ .

$=(0.09-0.12+0.16)w^2+(0.12-0.32)w+0.16$ .

$=0.13w^2-0.20w+0.16$ .

Minimise: $\frac{d\sigma_p^2}{dw}=0.26w-0.20=0 \Rightarrow w^*=\frac{0.20}{0.26}=\frac{10}{13}\approx0.769$ .

$w_1^*\approx76.9\%$ , $w_2^*\approx23.1\%$ .

$\sigma_p^{*2}=0.13(10/13)^2-0.20(10/13)+0.16=0.13\cdot\frac{100}{169}-\frac{2}{13}+0.16=\frac{100}{1300}-\frac{200}{1300}+\frac{208}{1300}=\frac{108}{1300}\approx0.0831$ .

Minimum portfolio std dev $\approx28.8\%$ .

Two assets have covariance matrix $\Sigma=\begin{pmatrix}0.09&0.06\\0.06&0.16\end{pmatrix}$ . Write portfolio variance $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ as a function of $w_1=w$ (with $w_2=1-w$ ), find the minimum-variance portfolio weight, and compute the minimum portfolio standard deviation.

EXERCISE 15.4

$\Sigma=Q\Lambda Q^T$ where $Q$ has eigenvectors as columns. Portfolio variance becomes $\mathbf{f}^T\Lambda\mathbf{f}=\sum\lambda_kf_k^2$ where $\mathbf{f}=Q^T\mathbf{w}$ . The variance is dominated by the eigenvector direction with the largest $\lambda_k$ .

$\Sigma=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ , $\mathbf{w}=\begin{pmatrix}0.5\\0.5\end{pmatrix}$ .

Eigenvalues: $p(\lambda)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)$ . $\lambda_1=3$ , $\lambda_2=1$ .

Eigenvectors: $\lambda_1=3$ : $\mathbf{q}_1=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$ (equal-weight direction). $\lambda_2=1$ : $\mathbf{q}_2=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$ (long-short direction).

PC exposures: $\mathbf{f}=Q^T\mathbf{w}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}0.5\\0.5\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1/\sqrt{2}\\0\end{pmatrix}$ .

$\sigma_p^2=\mathbf{f}^T\Lambda\mathbf{f}=3(1/\sqrt{2})^2+1(0)^2=3/2$ .

Direct check: $\mathbf{w}^T\Sigma\mathbf{w}=0.25(2)+2(0.25)(1)+0.25(2)=0.5+0.5+0.5=1.5$ ✓.

Interpretation: the equal-weight portfolio has exposure only to PC1 ( $\lambda_1=3$ ). It has zero exposure to the long-short direction ( $\lambda_2=1$ ). All variance (1.5) comes from the market/PC1 direction.

For $\Sigma=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $\mathbf{w}=\begin{pmatrix}0.5\\0.5\end{pmatrix}$ , compute the portfolio variance using the spectral decomposition: (1) find eigenvalues and eigenvectors of $\Sigma$ , (2) compute PC exposures $\mathbf{f}=Q^T\mathbf{w}$ , (3) evaluate $\sigma_p^2=\mathbf{f}^T\Lambda\mathbf{f}$ . Verify against the direct formula.

EXERCISE 15.5

The correlation matrix $\mathbf{C}$ satisfies $\Sigma=D\mathbf{C}D$ where $D=\text{diag}(\sigma_1,\sigma_2,\sigma_3)$ . So $\mathbf{C}=D^{-1}\Sigma D^{-1}$ . Verify $\mathbf{C}$ has ones on the diagonal and check that $|\rho_{ij}|\leq1$ .

$\Sigma=\begin{pmatrix}4&2&-1\\2&9&3\\-1&3&1\end{pmatrix}$ .

Standard deviations: $\sigma_1=\sqrt{4}=2$ , $\sigma_2=\sqrt{9}=3$ , $\sigma_3=\sqrt{1}=1$ .

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

$\mathbf{C}_{ij}=\Sigma_{ij}/(\sigma_i\sigma_j)$ : $C_{11}=4/(2\cdot2)=1$ , $C_{22}=9/(3\cdot3)=1$ , $C_{33}=1/(1\cdot1)=1$ .

$C_{12}=2/(2\cdot3)=1/3\approx0.333$ . $C_{13}=-1/(2\cdot1)=-0.5$ . $C_{23}=3/(3\cdot1)=1$ .

\mathbf{C}=\begin{pmatrix}1&1/3&-1/2\\1/3&1&1\\-1/2&1&1\end{pmatrix}

$C_{23}=1$ means Assets 2 and 3 are perfectly positively correlated — they move in lockstep. $\det(\mathbf{C})=0$ — confirming the covariance matrix is PSD but not PD.

Extract the correlation matrix $\mathbf{C}$ from $\Sigma=\begin{pmatrix}4&2&-1\\2&9&3\\-1&3&1\end{pmatrix}$ . Compute all pairwise correlations, identify any perfectly correlated pair, and state what that implies about the rank of $\Sigma$ .

EXERCISE 15.6

The variance of a long-short portfolio $\mathbf{w}=(1,-1)^T/\sqrt{2}$ (normalised) is $\frac{1}{2}\mathbf{w}^T\Sigma\mathbf{w}$ with $\mathbf{w}=(1,-1)^T$ . Correlation $\rho$ appears in the off-diagonal; as $\rho\to1$ the variance goes to zero. Connect to the eigenvalue $\lambda_2=\sigma^2(1-\rho)$ .

$\Sigma=\sigma^2\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}$ for two assets with equal variance $\sigma^2$ and correlation $\rho$ .

Long-short portfolio $\mathbf{w}=(1,-1)^T$ :

$\sigma_{LS}^2=\mathbf{w}^T\Sigma\mathbf{w}=\sigma^2(1-2\rho+1)=2\sigma^2(1-\rho)$ .

Eigenvalues of $\Sigma$ : $\lambda_1=\sigma^2(1+\rho)$ , $\lambda_2=\sigma^2(1-\rho)$ .

The long-short portfolio is the eigenvector $\frac{1}{\sqrt{2}}(1,-1)^T$ associated with $\lambda_2=\sigma^2(1-\rho)$ . Portfolio variance $= \|(1,-1)^T\|^2\cdot\lambda_2/({\|(1,-1)^T\|^2})$ ... more directly: $\sigma_{LS}^2=2\sigma^2(1-\rho)=2\lambda_2$ .

As $\rho\to1$ : $\sigma_{LS}^2\to0$ — the long-short portfolio becomes risk-free. This is statistical arbitrage: if two stocks are highly correlated ( $\rho\approx1$ ), a long-short position in them has nearly zero variance while capturing the mean-reversion of their spread.

As $\rho\to-1$ : $\sigma_{LS}^2\to4\sigma^2$ — maximum variance, the two assets always move opposite each other, amplifying the long-short swing.

Two assets have equal variance $\sigma^2$ and correlation $\rho$ , giving covariance matrix $\Sigma=\sigma^2\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}$ . Compute the variance of the long-short portfolio $\mathbf{w}=(1,-1)^T$ as a function of $\rho$ . Explain what happens as $\rho\to1$ and connect this to statistical arbitrage.

08 · Chapter Summary

Concept	Key Formula
Quadratic form	$Q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}=\sum_{ij}a_{ij}x_ix_j$
Diagonal of $A$	Coefficient of $x_i^2$ in $\mathbf{x}^TA\mathbf{x}$
Off-diagonal of $A$	Half the coefficient of $x_ix_j$ (for $i\neq j$ )
Classify form	PD/PSD/ND/NSD/Indefinite by sign of $Q(\mathbf{x})$ for all $\mathbf{x}$
Sample covariance	$S=\frac{1}{n-1}X_c^TX_c$ ; always PSD; PD when $n>p$
Portfolio variance	$\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ ; quadratic form in weights
Spectral decomposition	$\Sigma=Q\Lambda Q^T$ ; $\sigma_p^2=\sum_k\lambda_kf_k^2$ ; $\mathbf{f}=Q^T\mathbf{w}$
Correlation matrix	$\mathbf{C}=D^{-1}\Sigma D^{-1}$ ; entries in $[-1,1]$ ; ones on diagonal
$\rho=1$ assets	Covariance matrix singular; long-short has zero variance
Nearest PSD fix	Replace negative eigenvalues with $\epsilon>0$ ; $\hat{\Sigma}=Q\Lambda_+Q^T$

Next: Chapter 16 — Matrix Decompositions (LU) introduces the LU factorisation as the matrix encoding of Gaussian elimination, providing an efficient method for solving multiple linear systems with the same coefficient matrix.