Chapter 02

Medium

Diversification and Covariance

00 · Symbol Glossary

$R_p$R sub p — portfolio return

The return on a portfolio. For two assets, $R_p=wR_1+(1-w)R_2$ . For $N$ assets, $R_p=\sum_{i=1}^N w_i R_i$ with $\sum_i w_i=1$ . A weight outside $[0,1]$ is a short sale; the algebra does not change.

$\mu_p$mu sub p — portfolio mean

$\mu_p=\mathbb{E}[R_p]$ . Means add with the weights: $\mu_p=\sum_i w_i\mu_i$ . Correlation does not appear.

$\sigma_p^2$sigma sub p squared — portfolio variance

$\sigma_p^2=\mathrm{Var}(R_p)$ . For two assets, $w^2\sigma_1^2+(1-w)^2\sigma_2^2+2w(1-w)\mathrm{Cov}(R_1,R_2)$ . In general, the quadratic form $\mathbf{w}^T\Sigma\mathbf{w}$ .

$\mathrm{Cov}(R_i,R_j)$Covariance of two returns

$\mathrm{Cov}(R_i,R_j)=\mathbb{E}[(R_i-\mu_i)(R_j-\mu_j)]$ . Diagonal case $i=j$ is the variance. Units are return squared. The definition and the sample matrix $S$ are Linear Algebra, Chapter 15.

$\rho_{ij}$rho — correlation

$\rho_{ij}=\mathrm{Cov}(R_i,R_j)/(\sigma_i\sigma_j)$ , a pure number in $[-1,1]$ . The matrix of all $\rho_{ij}$ is the correlation matrix; ones sit on its diagonal.

$\Sigma$Sigma — covariance matrix

Symmetric matrix with $\Sigma_{ij}=\mathrm{Cov}(R_i,R_j)$ . Positive semidefinite, so $\mathbf{w}^T\Sigma\mathbf{w}\ge 0$ for every weight vector. Positive definite matrices, used to get a unique variance minimum, are Linear Algebra Chapter 14 and the optimizer in Chapter 03.

$\mathbf{w}^T\Sigma\mathbf{w}$Portfolio variance as a quadratic form

The $N$ -asset variance. Same object as $\mathbf{x}^T A\mathbf{x}$ in Linear Algebra Chapter 15, with the weight vector in the slot of $\mathbf{x}$ and $\Sigma$ in the slot of $A$ .

$\bar{c}$Average pairwise covariance

The mean of the off-diagonal entries that an equal-weight book actually holds. Equal-weight variance tends to $\bar{c}$ as $N\to\infty$ , which is the risk diversification does not remove.

01 · Two Positions, One Book

Chapter 01 priced a single return by its mean and its variance. A book is a sum. Fix a weight $w$ in asset 1 and put the residual weight $1-w$ in asset 2:

R_p = w R_1 + (1-w) R_2

Expectation is linear, and linearity does not consult the joint distribution. The portfolio mean is the same weighted sum of the two means:

\mu_p = w\mu_1 + (1-w)\mu_2

Variance is not linear. Expanding $\mathrm{Var}(wR_1+(1-w)R_2)$ by bilinearity of covariance — the same identity behind $\mathbf{x}^TA\mathbf{x}$ in Linear Algebra Chapter 15 — gives three terms, two from the diagonals and one from the cross:

\sigma_p^2 = w^2\sigma_1^2 + (1-w)^2\sigma_2^2 + 2w(1-w)\,\mathrm{Cov}(R_1,R_2)

The cross term is the whole subject of this chapter. It is large when the two residuals tend to share a sign, small when they do not, and negative when they tend to offset. Individual volatilities never see that sign.

The mean is blind to co-movement

Changing $\rho$ moves $\sigma_p$ and leaves $\mu_p$ where it is. On a plot with volatility on the horizontal axis and mean on the vertical, correlation slides a portfolio sideways. It does not lift it. Any pitch that "diversification raises expected return" is smuggling in a different weight, a different asset, or a different mean.

02 · A Desk Calculation

Running example for the rest of the chapter. Asset 1: $\mu_1=0.05$ , $\sigma_1=0.10$ . Asset 2: $\mu_2=0.11$ , $\sigma_2=0.20$ . Hold them half and half, $w=0.5$ .

Step-by-step — Half-and-half variance at three correlations

Mean, once. $\mu_p=0.5(0.05)+0.5(0.11)=0.08$ . Store it. Every correlation below prints this same $8\%$ mean.

Variances. $\sigma_1^2=0.01$ , $\sigma_2^2=0.04$ . At $w=0.5$ the diagonal contribution is always

(0.5)^2(0.01)+(0.5)^2(0.04)=0.0025+0.0100=0.0125

The correlation enters only through the covariance.

$\rho=0.5$ . $\mathrm{Cov}=0.5\cdot 0.10\cdot 0.20=0.010$ . The cross term is $2(0.5)(0.5)(0.010)=0.005$ .

\sigma_p^2=0.0125+0.005=0.0175, \qquad \sigma_p=\sqrt{0.0175}\approx 0.1323

About $13.2\%$ volatility against an $8\%$ mean.

$\rho=1$ . $\mathrm{Cov}=0.020$ . Cross term $2(0.5)(0.5)(0.020)=0.010$ .

\sigma_p^2=0.0125+0.010=0.0225, \qquad \sigma_p=\sqrt{0.0225}=0.15

This is exactly $0.5\cdot 0.10+0.5\cdot 0.20$ . Perfect positive correlation makes volatility itself a weighted average.

$\rho=-1$ . $\mathrm{Cov}=-0.020$ . Cross term $-0.010$ .

\sigma_p^2=0.0125-0.010=0.0025, \qquad \sigma_p=\sqrt{0.0025}=0.05

Same weights, same means, volatility cut from $15\%$ to $5\%$ . The gap is the cross term, nothing else.

Example — A mix quieter than the quieter asset

Keep the same two vols and move to $w=0.8$ with $\rho=0$ , so the covariance is zero. Then

\sigma_p^2=(0.8)^2(0.01)+(0.2)^2(0.04)=0.0064+0.0016=0.0080

and $\sigma_p=\sqrt{0.0080}\approx 0.0894$ , under the $10\%$ vol of asset 1. The mean is $0.8(0.05)+0.2(0.11)=0.062$ .

Repeat the identical weights at $\rho=0.5$ , where $\mathrm{Cov}=0.010$ :

\sigma_p^2=0.0080+2(0.8)(0.2)(0.010)=0.0080+0.0032=0.0112

and $\sigma_p=\sqrt{0.0112}\approx 0.1058$ , which is above $10\%$ . Holding both names paid only while the covariance was small enough to offset the vol imported from asset 2.

03 · Correlation Is Scaled Covariance

Covariance still carries the units of the two returns. A covariance of $0.010$ is "large" next to vols of $10\%$ and $20\%$ , and negligible next to vols of $80\%$ . Correlation strips the scale:

Definition — Correlation

For $\sigma_1>0$ and $\sigma_2>0$ ,

\rho=\rho_{12}=\frac{\mathrm{Cov}(R_1,R_2)}{\sigma_1\sigma_2}, \qquad \mathrm{Cov}(R_1,R_2)=\rho\,\sigma_1\sigma_2

Substituting the second display into the two-asset variance gives the form used on a desk:

\sigma_p^2 = w^2\sigma_1^2+(1-w)^2\sigma_2^2+2w(1-w)\rho\,\sigma_1\sigma_2

The bound $-1\le\rho\le 1$ is Cauchy–Schwarz applied to the centred returns, the same inner-product fact as in the covariance chapter. Equality is not a mysterious edge case. It is an algebraic relation between the two random variables:

$\rho=1$ if and only if $R_2=a+bR_1$ almost surely for some $b>0$ . One return is a positively levered copy of the other, plus a constant.
$\rho=-1$ if and only if the same holds with $b<0$ . The copy is levered in the opposite direction.

Anything strictly inside $(-1,1)$ means neither return is an affine function of the other. There is residual variation to diversify.

❌ Treating a small correlation as a small covariance

A sheet shows $\rho=0.20$ between two equity books and the risk manager marks the pair "barely related." Each book has volatility $40\%$ , so

\mathrm{Cov}=0.20\cdot 0.40\cdot 0.40=0.032

On a half-and-half mix the cross term alone is $2(0.5)(0.5)(0.032)=0.016$ , which is $4$ volatility points once you take the square root of a variance that was otherwise near $0.08$ . The correlation looked modest because each volatility in the denominator was large. The covariance, which is what $\sigma_p^2$ adds, did not shrink with them.

Consequence: hedge ratios and risk contributions are computed from covariances. Quote $\rho$ when you want a scale-free comparison. Feed $\mathrm{Cov}$ , or $\rho\sigma_i\sigma_j$ , into the variance formula.

04 · The Curve in the Volatility–Mean Plane

Put volatility on the horizontal axis and mean on the vertical. The two assets of the running example are the points $(0.10,\ 0.05)$ and $(0.20,\ 0.11)$ . Every weight $w$ produces one further point $(\sigma_p(w),\ \mu_p(w))$ . The mean is always the straight interpolation

\mu_p=0.11-0.06\,w

so the vertical coordinate is locked once $w$ is chosen. The horizontal coordinate depends on $\rho$ . Read the picture as $w$ runs from $0$ (pure asset 2) to $1$ (pure asset 1).

Perfect positive correlation. With $\rho=1$ and $w\in[0,1]$ ,

\sigma_p=w\sigma_1+(1-w)\sigma_2=0.20-0.10\,w

Both $\sigma_p$ and $\mu_p$ are affine in $w$ , so the plot is the straight segment joining $(0.20,\ 0.11)$ to $(0.10,\ 0.05)$ . The half-and-half book sits on that chord at $(0.15,\ 0.08)$ . There is no bulge. A weight outside $[0,1]$ continues the same line: $w=1.2$ is $120\%$ long asset 1 and $20\%$ short asset 2, and both formulas still apply. The point lies past asset 1, at higher volatility. Constraints that forbid the short are Chapter 10.

Perfect negative correlation. With $\rho=-1$ the volatility collapses to an absolute value,

\sigma_p=\lvert w\sigma_1-(1-w)\sigma_2\rvert=\lvert 0.30\,w-0.20\rvert

This is zero at $w=2/3$ : the long mix $2/3$ in the $10\%$ asset and $1/3$ in the $20\%$ asset. The mean there is $(2/3)(0.05)+(1/3)(0.11)=0.07$ . The picture is two straight rays meeting at the kink $(0,\ 0.07)$ . One ray connects the kink to $(0.10,\ 0.05)$ , the other connects the kink to $(0.20,\ 0.11)$ . The half-and-half book is at $(0.05,\ 0.08)$ , on the ray toward asset 2. Past either asset, a short sale walks back out along the same ray and volatility rises again. You can lever a perfect hedge; you cannot make variance negative.

Correlation strictly inside $(-1,1)$ . The map $w\mapsto\sigma_p(w)$ is now the square root of a quadratic that stays positive for every $w$ . As $w$ runs over the reals, $(\sigma_p,\mu_p)$ traces one branch of a hyperbola, opening to the right, with a leftmost point at the global minimum-variance mix. Restricting to $w\in[0,1]$ shows only the arc between the two asset points.

That arc bows to the left of the $\rho=1$ chord. The half-and-half numbers from Section 02 are four points at the same height $\mu=0.08$ :

$\rho$	$\sigma_p$ at $w=0.5$	Where it sits
$1$	$0.150$	On the chord
$0.5$	$\sqrt{0.0175}\approx 0.132$	Left of the chord
$0$	$\sqrt{0.0125}\approx 0.112$	Further left
$-1$	$0.050$	On the lower ray

The horizontal gap from the chord to the arc, at fixed mean, is the diversification benefit in volatility units. Lower $\rho$ deepens the bow. The arc never crosses to the left of the $\rho=-1$ rays: those rays are the Cauchy–Schwarz boundary. For $\rho=0.5$ in this example the bow does not get left of asset 1's $10\%$ vol anywhere on the long-only arc. For $\rho=0$ it does — the $w=0.8$ mix of Section 02 prints $8.9\%$ . Beating the safer asset's volatility is a statement about how small $\rho$ is relative to $\sigma_1/\sigma_2$ , checked by computing $\sigma_p$ , not by counting positions.

Locating the leftmost point subject to a required mean, and deciding which arc is worth holding, is the optimization in Chapters 03 and 04. This chapter only needs the geometry: $\rho=1$ is a line, $\rho=-1$ is a broken line through a zero-vol kink, and every interior correlation is a curve bowed toward lower volatility at the same mean.

05 · N Assets, One Quadratic Form

Let $R_i$ have mean $\mu_i$ , and let $\Sigma$ be the covariance matrix, $\Sigma_{ij}=\mathrm{Cov}(R_i,R_j)$ . A portfolio is a weight vector $\mathbf{w}$ with $\sum_{i=1}^N w_i=1$ . The return $R_p=\sum_i w_i R_i$ has

\mu_p=\sum_{i=1}^N w_i\mu_i, \qquad \sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}=\sum_{i=1}^N\sum_{j=1}^N w_i\,\Sigma_{ij}\,w_j

The double sum is the two-asset formula written without privilege for a pair. Diagonal entries contribute $w_i^2\sigma_i^2$ . Each unordered pair $\{i,j\}$ contributes twice, once as $(i,j)$ and once as $(j,i)$ , which is why the two-asset cross term carries a $2$ .

Definition — Portfolio variance

If $\Sigma$ is the covariance matrix of the asset returns and $\mathbf{w}$ is the weight vector, portfolio variance is the quadratic form $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ . Because every covariance matrix is symmetric and positive semidefinite, the form is nonnegative for every $\mathbf{w}$ . A negative number out of a risk system means the matrix that was typed is not a covariance matrix.

Count the terms before interpreting them. The double sum has $N^2$ summands: $N$ variances and $N(N-1)$ covariances. At $N=30$ that is $30$ diagonal entries and $870$ off-diagonal entries. A risk report that lists thirty volatilities has inspected one thirtieth of the matrix that actually produces $\sigma_p^2$ .

Step-by-step — Three-asset variance, both as a matrix product and as six terms

Inputs. Weights $\mathbf{w}=(0.4,\ 0.4,\ 0.2)^T$ and

\Sigma=\begin{pmatrix}0.04 & 0.01 & 0.02\\ 0.01 & 0.01 & 0\\ 0.02 & 0 & 0.09\end{pmatrix}

Rows and columns are assets 1, 2, 3. The matrix is symmetric. Asset 2 is uncorrelated with asset 3.

Apply $\Sigma$ to $\mathbf{w}$ .

\Sigma\mathbf{w}=\begin{pmatrix}0.4\cdot 0.04+0.4\cdot 0.01+0.2\cdot 0.02\\ 0.4\cdot 0.01+0.4\cdot 0.01+0.2\cdot 0\\ 0.4\cdot 0.02+0.4\cdot 0+0.2\cdot 0.09\end{pmatrix}=\begin{pmatrix}0.024\\ 0.008\\ 0.026\end{pmatrix}

Dot with $\mathbf{w}$ .

\sigma_p^2=0.4(0.024)+0.4(0.008)+0.2(0.026)=0.0096+0.0032+0.0052=0.0180

So $\sigma_p=\sqrt{0.0180}\approx 0.1342$ .

Open the double sum to see who paid. Diagonal: $0.4^2(0.04)+0.4^2(0.01)+0.2^2(0.09)=0.0064+0.0016+0.0036=0.0116$ . Cross terms: $2(0.4)(0.4)(0.01)+2(0.4)(0.2)(0.02)+2(0.4)(0.2)(0)=0.0032+0.0032+0=0.0064$ . Total $0.0116+0.0064=0.0180$ . About a third of this book's variance is covariance, and the zero block between assets 2 and 3 contributed nothing.

Short sales, one line

A coordinate of $\mathbf{w}$ may be negative. Then $R_p=\sum_i w_i R_i$ and $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ are unchanged as formulas: the short is just a negative weight in the same quadratic form. Which shorts a mandate allows is a constraint problem, taken up in Chapter 10, not a change to covariance.

06 · What an Equal-Weight Book Keeps

Split each return, for intuition only, into a piece shared with the others and a piece private to the name:

R_i=\mu_i+\beta_i F+\varepsilon_i

Take $F$ as a common factor, $\mathrm{Cov}(\varepsilon_i,F)=0$ , and $\mathrm{Cov}(\varepsilon_i,\varepsilon_j)=0$ for $i\neq j$ . Then the off-diagonal of $\Sigma$ is entirely factor covariance, $\Sigma_{ij}=\beta_i\beta_j\mathrm{Var}(F)$ , while the diagonal adds the private variance: $\Sigma_{ii}=\beta_i^2\mathrm{Var}(F)+\mathrm{Var}(\varepsilon_i)$ . The private piece is what people call idiosyncratic. The factor piece is systematic: it is still there in every portfolio that has not set its factor exposure to zero. A later chapter builds the single-index model properly. The only claim used here is the accounting inside $\Sigma$ .

The cleanest place to watch the private piece die is an equal-weight book of $N$ names that share one volatility $\sigma$ and one pairwise correlation $\rho$ . Then $w_i=1/N$ , every diagonal term is $\sigma^2/N^2$ , and there are $N$ of them. Every off-diagonal term is $\rho\sigma^2/N^2$ , and there are $N(N-1)$ of them. Adding,

\sigma_p^2=\frac{1}{N}\sigma^2+\left(1-\frac{1}{N}\right)\rho\sigma^2=\rho\sigma^2+\frac{(1-\rho)\sigma^2}{N}

Step-by-step — Equal-weight volatility as the book grows

Fix the names. $\sigma=0.30$ and $\rho=0.40$ , so single-name variance is $0.09$ and pairwise covariance is $0.40\cdot 0.09=0.036$ . The formula specialises to

\sigma_p^2=0.036+\frac{0.054}{N}

$N=1$ . $\sigma_p^2=0.090$ , $\sigma_p=0.30$ . The "portfolio" is the asset.

$N=2$ . $\sigma_p^2=0.036+0.027=0.063$ , $\sigma_p=\sqrt{0.063}\approx 0.251$ . One extra name, same correlation, removes about five volatility points.

$N=10$ and $N=100$ . At $N=10$ , $\sigma_p^2=0.036+0.0054=0.0414$ , $\sigma_p\approx 0.203$ . At $N=100$ , $\sigma_p^2=0.036+0.00054=0.03654$ , $\sigma_p\approx 0.191$ .

The limit. As $N\to\infty$ the second term vanishes and $\sigma_p^2\to 0.036$ , so $\sigma_p\to 0.30\sqrt{0.40}\approx 0.190$ . Everything past the first handful of names is a slow crawl toward that floor. The floor equals the common covariance, not zero.

Read the two terms in the display. $(1-\rho)\sigma^2/N$ is the equal-weight average of the idiosyncratic variances; it is $O(1/N)$ . $\rho\sigma^2$ is the covariance every pair still shares; it does not depend on $N$ . Adding the 50th name at the same $\rho$ barely moves a book that already holds 49. It does nothing to the floor.

If the pairwise covariances are not identical, the same limit holds with $\rho\sigma^2$ replaced by the average off-diagonal covariance $\bar{c}$ : equal-weight variance tends to $\bar{c}$ . Diversification drives the contribution of the diagonal to zero. It converges to the covariances you refused to make small.

07 · Diversify Is a Covariance Statement

The operable sentence is: portfolio variance is $\mathbf{w}^T\Sigma\mathbf{w}$ , so a change of book changes risk only by changing the weights on the entries of $\Sigma$ . "Own more names" is a special case that works when the new names arrive with small covariances against the book you already have, and when their weights are large enough for those entries to matter.

❌ Averaging the volatilities

A two-asset sleeve, $w=0.5$ , vols $20\%$ and $40\%$ , correlation $0.25$ , is reported at $30\%$ risk because $(20+40)/2=30$ .

Why it breaks: $30\%$ is the $\rho=1$ answer from Section 03. At $\rho=0.25$ the covariance is $0.25\cdot 0.20\cdot 0.40=0.020$ , and

\sigma_p^2=(0.5)^2(0.04)+(0.5)^2(0.16)+2(0.5)(0.5)(0.020)=0.01+0.04+0.01=0.06

so $\sigma_p=\sqrt{0.06}\approx 0.245$ . The report overstated volatility by about five and a half points, which will mis-size every limit that reads it.

Consequence: average the vols only on a day when you are prepared to defend $\rho=1$ . Otherwise compute the quadratic form.

❌ Counting names

A 20-name equal-weight book with $\sigma=0.30$ and $\rho=0.40$ already sits at

\sigma_p^2=0.036+\frac{0.054}{20}=0.0387, \qquad \sigma_p\approx 0.197

against a floor of $0.190$ . Replacing one name by a near-clone of another ( $\rho$ near $1$ with a name you already hold, ordinary $\rho$ with the rest) changes $N$ by zero and changes the relevant row of $\Sigma$ a lot. Adding a 21st clone does increase $N$ and still imports almost a full covariance row. Either trade leaves $\sigma_p$ essentially on the floor already computed.

Consequence: the diversification ratio you can still earn is the gap from $0.197$ down toward $0.190$ , unless a new name comes in with a genuinely smaller covariance against this book. The headcount is not the state variable. The matrix is.

Example — Where a new name actually moves the number

Start from asset 2 alone in the running example: $\sigma=0.20$ , $\mu=0.11$ . Add asset 1 at weight $0.5$ with $\rho=0.5$ . Section 02 gives $\sigma_p\approx 0.132$ and $\mu_p=0.08$ . Volatility fell, and so did the mean, because asset 1's mean is lower. The covariance statement is only the volatility half. Whether the trade is attractive at the new mean is a preference over $(\sigma_p,\mu_p)$ , and that comparison is Chapter 03. The input it will use is the variance just computed, not the fact that the book now contains two tickers.

08 · Exercises

EXERCISE 2.1

The mean does not use $\rho$ . For the variance, compute $\mathrm{Cov}=\rho\sigma_1\sigma_2$ separately at each correlation, then add $w^2\sigma_1^2+(1-w)^2\sigma_2^2+2w(1-w)\mathrm{Cov}$ . The $\rho=1$ volatility should match $w\sigma_1+(1-w)\sigma_2$ .

$\mu_p=0.5(0.06)+0.5(0.14)=0.10$ at either correlation.

Variances: $\sigma_1^2=0.04$ , $\sigma_2^2=0.16$ . Diagonal contribution at $w=0.5$ : $0.25(0.04)+0.25(0.16)=0.01+0.04=0.05$ .

At $\rho=0.25$ , $\mathrm{Cov}=0.25\cdot 0.20\cdot 0.40=0.020$ , and the cross term is $2(0.5)(0.5)(0.020)=0.010$ . So $\sigma_p^2=0.05+0.010=0.060$ and $\sigma_p=\sqrt{0.060}\approx 0.2449$ .

At $\rho=1$ , $\mathrm{Cov}=0.080$ , cross term $=0.040$ , $\sigma_p^2=0.090$ , $\sigma_p=0.30$ . The weighted vol is $0.5(0.20)+0.5(0.40)=0.30$ , matching. The diversification gap at this weight is $0.30-\sqrt{0.060}\approx 0.055$ of volatility, entirely from replacing covariance $0.080$ with $0.020$ .

Asset 1 has mean $0.06$ and volatility $0.20$ . Asset 2 has mean $0.14$ and volatility $0.40$ . For the half-and-half book $w=0.5$ , compute $\mu_p$ and $\sigma_p$ at $\rho=0.25$ and again at $\rho=1$ . Check the $\rho=1$ result against the weighted average of the two volatilities, and state the volatility gap between the two correlations.

EXERCISE 2.2

At $\rho=-1$ , $\sigma_p=\lvert w\sigma_1-(1-w)\sigma_2\rvert$ . Set the inside to zero to place the kink. The mean is still $w\mu_1+(1-w)\mu_2$ . The long-only picture is two segments meeting at that kink; $w$ outside $[0,1]$ continues them.

(a) $w(0.10)=(1-w)(0.30)$ gives $0.10w=0.30-0.30w$ , so $0.40w=0.30$ and $w=0.75$ . The mean is $0.75(0.04)+0.25(0.10)=0.030+0.025=0.055$ . The kink is the point $(0,\ 0.055)$ .

(b) Asset 1 is $(0.10,\ 0.04)$ , asset 2 is $(0.30,\ 0.10)$ . One straight segment joins $(0.10,\ 0.04)$ to $(0,\ 0.055)$ as $w$ runs from $1$ to $0.75$ . The other joins $(0,\ 0.055)$ to $(0.30,\ 0.10)$ as $w$ runs from $0.75$ to $0$ .

(c) At $w=0.5$ , $\mathrm{Cov}=-0.10\cdot 0.30=-0.030$ . Then $\sigma_p^2=(0.5)^2(0.01)+(0.5)^2(0.09)+2(0.5)(0.5)(-0.030)=0.0025+0.0225-0.015=0.010$ , so $\sigma_p=0.10$ . The mean is $0.5(0.04)+0.5(0.10)=0.07$ . The point $(0.10,\ 0.07)$ lies on the segment from the kink to asset 2.

(d) $w=1.25$ means $125\%$ long asset 1 and $25\%$ short asset 2. Then $\sigma_p=\lvert 1.25(0.10)-(-0.25)(0.30)\rvert=\lvert 0.125+0.075\rvert=0.20$ , which is past asset 1 and larger than $0.10$ . The mean is $1.25(0.04)+(-0.25)(0.10)=0.050-0.025=0.025$ .

Two assets have $\sigma_1=0.10$ , $\mu_1=0.04$ , $\sigma_2=0.30$ , $\mu_2=0.10$ , and $\rho=-1$ . (a) Find the weight $w$ in asset 1 that makes $\sigma_p=0$ , and the mean at that weight. (b) Describe the long-only curve in the volatility–mean plane: the two endpoints and the kink. (c) Compute $(\sigma_p,\mu_p)$ at $w=0.5$ and say which segment it lies on. (d) Compute $(\sigma_p,\mu_p)$ at $w=1.25$ and place that point relative to asset 1.

EXERCISE 2.3

Compute $\Sigma\mathbf{w}$ one row at a time, then dot with $\mathbf{w}$ . As a check, expand $w_i^2\Sigma_{ii}$ and $2w_iw_j\Sigma_{ij}$ for $i<j$ . The pair with a zero covariance must contribute nothing.

$\Sigma\mathbf{w}=(0.4\cdot 0.04+0.4\cdot 0.01+0.2\cdot 0.02,\ 0.4\cdot 0.01+0.4\cdot 0.01+0,\ 0.4\cdot 0.02+0+0.2\cdot 0.09)^T=(0.024,\ 0.008,\ 0.026)^T$ .

$\sigma_p^2=0.4(0.024)+0.4(0.008)+0.2(0.026)=0.0096+0.0032+0.0052=0.0180$ , so $\sigma_p=\sqrt{0.0180}\approx 0.1342$ .

Diagonal contribution: $0.16(0.04)+0.16(0.01)+0.04(0.09)=0.0064+0.0016+0.0036=0.0116$ . Covariance contribution: $2(0.4)(0.4)(0.01)+2(0.4)(0.2)(0.02)+2(0.4)(0.2)(0)=0.0032+0.0032+0=0.0064$ . The assets 2 and 3 pair adds zero. Covariances are $0.0064/0.0180\approx 36\%$ of portfolio variance. Of the nine entries in the double sum, six are off-diagonal; after symmetry they are three unordered pairs, one of which is idle.

Let $\mathbf{w}=(0.4,\ 0.4,\ 0.2)^T$ and

\Sigma=\begin{pmatrix}0.04 & 0.01 & 0.02\\ 0.01 & 0.01 & 0\\ 0.02 & 0 & 0.09\end{pmatrix}

Compute $\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$ and $\sigma_p$ . Split the variance into the diagonal part and the covariance part, and say what the zero entry $\Sigma_{23}$ contributed.

EXERCISE 2.4

With equal vols and a common correlation, $\sigma_p^2=\rho\sigma^2+(1-\rho)\sigma^2/N$ . The limit as $N\to\infty$ is the covariance $\rho\sigma^2$ , so the limiting volatility is $\sigma\sqrt{\rho}$ . Compute the $\rho=0.90$ floor the same way and compare floors, not headcounts.

Single-name variance $\sigma^2=0.0625$ . At $\rho=0.36$ the pairwise covariance is $0.36\cdot 0.0625=0.0225$ , and

\sigma_p^2=0.0225+\frac{0.0400}{N}

(a) $N=4$ : $\sigma_p^2=0.0225+0.0100=0.0325$ , $\sigma_p=\sqrt{0.0325}\approx 0.1803$ . $N=16$ : $\sigma_p^2=0.0225+0.0025=0.0250$ , $\sigma_p=\sqrt{0.0250}\approx 0.1581$ .

(b) The limit is $0.0225$ , hence $\sigma_p=0.25\sqrt{0.36}=0.15$ . Of the original variance $0.0625$ , the fraction that remains is $0.0225/0.0625=0.36=\rho$ . The fraction $1-\rho=0.64$ is the idiosyncratic share, and it is the only share that $N$ can kill.

(c) At $\rho=0.90$ the floor is $0.90\cdot 0.0625=0.05625$ , so $\sigma_p\to 0.25\sqrt{0.90}\approx 0.2372$ . A 16-name book at $\rho=0.36$ is already at about $15.8\%$ vol, under that floor. Raising $N$ further at $\rho=0.90$ cannot get below $23.7\%$ . The binding input is the correlation that sets the floor, not the number of lines in the position blotter.

Every asset in an equal-weight book has volatility $0.25$ . (a) If every pairwise correlation equals $0.36$ , compute $\sigma_p^2$ and $\sigma_p$ at $N=4$ and at $N=16$ . (b) Give the limiting variance and the limiting volatility as $N\to\infty$ , and the fraction of single-name variance that remains. (c) Repeat the limit at pairwise correlation $0.90$ , and say which book is riskier: 16 names at correlation $0.36$ , or infinitely many names at correlation $0.90$ .

09 · Chapter Summary

Object	What it does
$\mu_p=\sum_i w_i\mu_i$	Means add. Correlation never enters.
Two-asset $\sigma_p^2$	$w^2\sigma_1^2+(1-w)^2\sigma_2^2+2w(1-w)\mathrm{Cov}(R_1,R_2)$
$\rho$	Scale-free covariance, $\mathrm{Cov}=\rho\sigma_i\sigma_j$ , and $-1\le\rho\le 1$
$\rho=\pm 1$	Straight lines in the volatility–mean plane; $\rho=-1$ kinks at a zero-vol mix
$-1<\rho<1$	The long-only mixes bow left of the $\rho=1$ chord: same mean, less volatility
$\sigma_p^2=\mathbf{w}^T\Sigma\mathbf{w}$	$N$ variance terms and $N(N-1)$ covariance terms
Equal-weight limit	$\sigma_p^2\to\rho\sigma^2$ when vols and correlations are common; the floor is the covariance
A new name	Changes risk through its row of $\Sigma$ , not through the headcount

Next: Chapter 03 — Mean-Variance Optimization, which takes this variance as the objective and chooses $\mathbf{w}$ subject to a target mean.