Finance-Theory · Unit 16 · Video 6 · Interactive Practice

Average Covariance, and Why Correlations Jump

IKey Formulas

Formula	Name	What it says
$E[R_p] = \omega_1\mu_1 + \cdots + \omega_n\mu_n$	Expected portfolio return	Weighted average of the means
$\mathrm{Var}[R_p] = \sum_{i=1}^{n}\omega_i^2\sigma_i^2 + \sum_{i \neq j}\omega_i\omega_j\,\mathrm{Cov}[R_i, R_j]$	Portfolio variance, $n$ assets	$n$ variances on the diagonal, $n^2 - n$ covariances off it
$\mathrm{Var}[R_p] = \tfrac{1}{n}\,\overline{\sigma^2} + \tfrac{n-1}{n}\,\overline{\mathrm{Cov}}$	Equal weights, $\omega_i = \tfrac{1}{n}$	$\overline{\sigma^2}$ = average variance, $\overline{\mathrm{Cov}}$ = average of the $n^2 - n$ covariances
$\mathrm{Var}[R_p] \approx \overline{\mathrm{Cov}}$	Large equally weighted portfolio	$\tfrac{1}{n} \to 0$ and $\tfrac{n-1}{n} \to 1$

Key Insight: There are only $n$ variances but $n^2 - n$ covariances, so in a large, evenly spread portfolio the covariances decide the risk, and covariances are set by how people behave.

IICounting the Cells

With weights $\tfrac{1}{n}$ , the $n$ diagonal variances fade while the $n^2 - n$ covariances take over the total.

Number of stocks n 10

IIIThe Pool That Looked Safe

A 1,000-loan pool's risk is its average covariance, so when correlation jumps, the pool's risk jumps with it.

Time

💡 Nothing about any single loan changed: each loan's $\sigma$ stays $15\%$ throughout, and only $\rho$ moved, because a national house-price fall pushed borrowers everywhere to default together.

IVA 1% Event, or a Pattern?

One crowded Wednesday looks like random error; the same week crowded every year is behavior everyone coordinates on.

Years observed 1

💡 Someone who knows the calendar needs no data at all: a distribution measures what we do not know, and economic understanding of behavior is what replaces it.

VQuiz Questions

Problem 1 · Counting Covariance Terms

Given: a portfolio of $n = 100$ stocks. Find the number of terms in the sum $\sum_{i \neq j}\omega_i\omega_j\,\mathrm{Cov}[R_i, R_j]$ , and the number of different covariance values that must be estimated.

How many terms are in the sum over i ≠ j?

How many different covariance values must be estimated?

✅ Correct! The table has

n^2 - n = 9{,}900

off-diagonal cells, but

\mathrm{Cov}[R_i, R_j] = \mathrm{Cov}[R_j, R_i]

, so they hold only

4{,}950

different values.

❌ Close, but…

4{,}950

counts each pair of stocks once. The sum over

i \neq j

visits both

(i, j)

and

(j, i)

, so every pair appears twice.

❌ Not quite.

n^2 = 10{,}000

counts every cell, including the

100

variances on the diagonal, which the condition

i \neq j

excludes.

❌ Not quite. Count the off-diagonal cells of the

100 \times 100

table: all cells minus the diagonal.

❌ Close, but…

9{,}900

is the number of cells, and the cells

(i, j)

and

(j, i)

hold the same covariance, so each value is counted twice.

❌ Not quite.

5{,}050 = \tfrac{100 \times 101}{2}

also counts the

100

variances; the question asks only for covariances.

❌ Not quite. Each different covariance sits in two mirror-image cells, so halve the number of off-diagonal cells.

Show solution

The $n \times n$ table has $n^2 = 10{,}000$ cells. The diagonal holds the $n = 100$ variances, so the sum over $i \neq j$ has

n^2 - n = 10{,}000 - 100 = 9{,}900 \text{ terms.}

Because $\mathrm{Cov}[R_i, R_j] = \mathrm{Cov}[R_j, R_i]$ , the cells above the diagonal mirror those below it, so the number of different covariances is

\frac{9{,}900}{2} = 4{,}950.

The covariance cells make up $\tfrac{9{,}900}{10{,}000} = 99\%$ of the table, which is why they, and not the variances, drive a large portfolio's risk.

Problem 2 · Five Stocks Is Not "Large"

Given: an equally weighted portfolio of $n = 5$ stocks (illustrative), each with a standard deviation of $12\%$ per month, and every pair with correlation $0.35$ . Find the portfolio variance, in decimals.

✅ Correct!

\tfrac{1}{5}(0.0144) + \tfrac{4}{5}(0.00504) = 0.006912

, an SD of about

8.31\%

per month.

❌ Close, but…

0.00504

is the average covariance alone. The approximation

\mathrm{Var}[R_p] \approx \overline{\mathrm{Cov}}

needs large

n

; at

n = 5

the term

\tfrac{1}{5}\,\overline{\sigma^2}

is far from negligible.

❌ Close, but… the average covariance is multiplied by

\tfrac{n-1}{n} = \tfrac{4}{5}

, not by

1

❌ Not quite.

0.00288

is only the variance part,

\tfrac{1}{5}(0.0144)

; the covariance part is missing.

Show solution

Every stock has variance $0.12^2 = 0.0144$ , so $\overline{\sigma^2} = 0.0144$ . Every pair has covariance $\rho\,\sigma_i\sigma_j = 0.35 \times 0.12 \times 0.12 = 0.00504$ , so $\overline{\mathrm{Cov}} = 0.00504$ .

\mathrm{Var}[R_p] = \tfrac{1}{5}(0.0144) + \tfrac{4}{5}(0.00504) = 0.00288 + 0.004032 = 0.006912

The standard deviation is $\sqrt{0.006912} \approx 0.0831$ , about $8.31\%$ per month. The large- $n$ shortcut would give $0.00504$ , understating the variance by about $27\%$ : with five stocks the variances still matter.

Problem 3 · When the Loans Default Together

Given: an equally weighted pool of $1{,}000$ loans (illustrative). Each loan's return has a standard deviation of $12\%$ , and history puts the average correlation between loans at $0.05$ . In a national downturn the average correlation rises to $0.6$ while each loan's standard deviation stays at $12\%$ . Find the pool's standard deviation before and after.

Pool SD at ρ = 0.05

Pool SD at ρ = 0.6

✅ Correct! The pool's SD rises from about

2.71\%

to about

9.30\%

, roughly

3.4

times, with no change in any single loan.

❌ Not quite.

0.38\%

is the pool's SD if the loans were truly uncorrelated (

\rho = 0

): only the tiny

\tfrac{1}{1000}\,\overline{\sigma^2}

term would remain. At

\rho = 0.05

the covariance term dominates.

❌ Close, but… that is the variance written as a percentage. Take the square root to get the standard deviation.

❌ Not quite.

\rho \times \sigma

is not the pool's SD. The average covariance is

\rho\,\sigma^2

, and the SD is the square root of the variance.

❌ Not quite.

12\%

is a single loan's SD; the pool would be that risky only at

\rho = 1

❌ Not quite. Use

\mathrm{Var} = \tfrac{1}{1000}(0.0144) + \tfrac{999}{1000}(0.05)(0.0144)

, then take the square root.

❌ Not quite. Use

\mathrm{Var} = \tfrac{1}{1000}(0.0144) + \tfrac{999}{1000}(0.6)(0.0144)

, then take the square root.

Show solution

Each loan has variance $0.12^2 = 0.0144$ ; the average covariance is $\rho \times 0.0144$ .

Before ( $\rho = 0.05$ , average covariance $0.00072$ ):

\mathrm{Var} = \tfrac{1}{1000}(0.0144) + \tfrac{999}{1000}(0.00072) = 0.0000144 + 0.00071928 = 0.00073368

\mathrm{SD} = \sqrt{0.00073368} \approx 0.0271 = 2.71\%

After ( $\rho = 0.6$ , average covariance $0.00864$ ):

\mathrm{Var} = 0.0000144 + \tfrac{999}{1000}(0.00864) = 0.0000144 + 0.00863136 = 0.00864576

\mathrm{SD} = \sqrt{0.00864576} \approx 0.0930 = 9.30\%

The ratio is $9.30 / 2.71 \approx 3.4$ . In both cases the variance is almost exactly the average covariance, so the pool is only as safe as the correlation assumption: when borrowers start to default together, the "diversified" pool's risk more than triples.

Problem 4 · Which Lever Moves the Risk?

Given: an equally weighted portfolio of $n = 25$ stocks (illustrative) with average variance $\overline{\sigma^2} = 0.0144$ and average covariance $\overline{\mathrm{Cov}} = 0.00504$ , so $\mathrm{Var}[R_p] = 0.0054144$ . A manager can either halve every stock's variance (covariances unchanged) or halve every covariance (variances unchanged). Find the new portfolio variance in each case.

Every variance halved

Every covariance halved

✅ Correct! Halving the variances cuts the portfolio variance by about

5\%

; halving the covariances cuts it by about

45\%

❌ Not quite.

0.00271

halves the whole variance, which would need both the variances and the covariances halved.

❌ Close, but… that value belongs to the other lever. Halving the variances changes only the

\tfrac{1}{25}\,\overline{\sigma^2}

term; halving the covariances changes only the

\tfrac{24}{25}\,\overline{\mathrm{Cov}}

term.

❌ Not quite. That is only the term that changed; add back the term that stayed the same.

❌ Not quite. Recompute

\tfrac{1}{25}(0.0072) + \tfrac{24}{25}(0.00504)

❌ Not quite. Recompute

\tfrac{1}{25}(0.0144) + \tfrac{24}{25}(0.00252)

Show solution

Start from $\mathrm{Var}[R_p] = \tfrac{1}{25}(0.0144) + \tfrac{24}{25}(0.00504) = 0.000576 + 0.0048384 = 0.0054144$ .

Variances halved ( $\overline{\sigma^2} = 0.0072$ ):

\tfrac{1}{25}(0.0072) + 0.0048384 = 0.000288 + 0.0048384 = 0.0051264

Covariances halved ( $\overline{\mathrm{Cov}} = 0.00252$ ):

0.000576 + \tfrac{24}{25}(0.00252) = 0.000576 + 0.0024192 = 0.0029952

The first lever removes $0.000288$ , about $5.3\%$ of the variance; the second removes $0.0024192$ , about $44.7\%$ . With $600$ covariance cells against $25$ variance cells, how the holdings move together, not how volatile each one is, controls the portfolio's risk.

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