Finance-Theory · Unit 16 · Video 1 · Interactive Practice

Portfolio Variance: Why Risk Lives in the Covariances

IKey Formulas

Formula	Name	What it says
$\mu_p = \text{E}[R_p] = \omega_1\mu_1 + \omega_2\mu_2 + \cdots + \omega_n\mu_n$	Portfolio mean	Follows from the accounting identity $R_p = \omega_1 R_1 + \cdots + \omega_n R_n$ : the mean is the weighted average of the means
$\sigma_i^2 = \text{Var}[R_i] = \text{E}[(R_i - \mu_i)^2]$	Variance of one security	The standard deviation $\sigma_i$ is its square root: the risk coordinate on the mean–SD graph
$\text{Var}[R_p] = \displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n} \omega_i\omega_j\sigma_{ij}$	Portfolio variance	Add up all $n^2$ cells of the table; a cell with $i = j$ is $\omega_i^2\sigma_i^2$
$\sigma_{ij} = \text{Cov}[R_i, R_j] = \sigma_i\sigma_j\rho_{ij}$	Covariance	The two standard deviations times the correlation; it is negative when $\rho_{ij} < 0$

Key Insight: The table has $n$ variances on its diagonal and $n^2 - n = n(n-1)$ covariances off it, so as $n$ grows a portfolio's risk depends mainly on how its stocks move together.

IIEvery Cell of the Table

Portfolio variance is the sum of all nine cells, and a negative covariance subtracts from it (illustrative stocks; the slider's range is limited so the inputs stay consistent).

Weights A / B / C ρ Birch–Cedar −0.3

IIICounting the Cells

An $n$ -stock table has $n$ variance cells and $n^2 - n$ covariance cells, so covariances soon dominate.

Stocks (n) 6

IVMean Averages, Risk Does Not

The portfolio's mean is the weighted average of the means; where does its standard deviation land? (Illustrative means: Alder $0.6\%$ , Birch $0.9\%$ , Cedar $1.4\%$ per month.)

Portfolio Your guess for σ_p 11.0%

VQuiz Questions

Problem 1 · A Month's Realized Return

Given: An illustrative portfolio holds 20% Alder, 30% Birch and 50% Cedar. Last month Alder returned $3\%$ , Birch $-2\%$ and Cedar $4\%$ . Find the portfolio's realized return for the month.

✅ Correct!

R_p = \sum_i \omega_i R_i

is an accounting identity: the return you earned is the weighted average of what each stock earned.

❌ Not quite. That is the plain average

(3 - 2 + 4)/3

. Cedar carries half the money, so its

4\%

counts for half.

❌ Not quite. Birch lost

2\%

: its term is

0.3 \times (-2\%) = -0.6\%

, which subtracts.

❌ Not quite. That adds the three returns without weights. Each return is multiplied by its weight, and the weights sum to one.

❌ Not quite. Weight each stock's return by its share of the portfolio.

Show solution

The weights sum to one: $0.2 + 0.3 + 0.5 = 1$ . The realized return is the weighted sum of the realized returns:

R_p = 0.2 \times 3\% + 0.3 \times (-2\%) + 0.5 \times 4\% = 0.6\% - 0.6\% + 2.0\% = 2.0\%

Taking expected values of the same line gives $\mu_p = \sum_i \omega_i \mu_i$ : the identity holds month by month, and the mean formula follows from it.

Problem 2 · One Off-Diagonal Cell

Given: In an illustrative portfolio, Alder has weight $0.20$ and monthly SD $5\%$ , Birch has weight $0.30$ and monthly SD $8\%$ , and their correlation is $\rho_{AB} = -0.3$ . Find the value of the single cell $\omega_A\omega_B\sigma_{AB}$ in the portfolio's variance table, in $\%^2$ .

✅ Correct!

0.20 \times 0.30 \times 5 \times 8 \times (-0.3) = -0.72

: a negative covariance, so this cell lowers the portfolio's variance.

❌ Not quite.

-12

is the covariance

\sigma_{AB} = 5 \times 8 \times (-0.3)

itself. The cell also carries the two weights,

\omega_A\omega_B = 0.06

❌ Close, but…

-1.44

is the two mirror cells

\omega_A\omega_B\sigma_{AB}

and

\omega_B\omega_A\sigma_{BA}

together. The question asks for one cell.

❌ Not quite. The correlation is negative, so the covariance and the cell are negative too.

❌ Not quite. Multiply both weights, both standard deviations and the correlation.

Show solution

First the covariance: $\sigma_{AB} = \sigma_A\sigma_B\rho_{AB} = 5 \times 8 \times (-0.3) = -12\ \%^2$ .

The cell carries both weights:

\omega_A\omega_B\sigma_{AB} = 0.20 \times 0.30 \times (-12) = 0.06 \times (-12) = -0.72\ \%^2

The mirror cell $\omega_B\omega_A\sigma_{BA}$ holds the same value, so the Alder–Birch pair contributes $2 \times (-0.72) = -1.44\ \%^2$ to the sum of all cells.

Problem 3 · The Whole Table

Given: An illustrative portfolio holds 25% Alder, 25% Birch and 50% Cedar, with monthly SDs $5\%$ , $8\%$ and $12\%$ and correlations $\rho_{AB} = 0.2$ , $\rho_{AC} = 0.5$ , $\rho_{BC} = -0.3$ . Find the portfolio's monthly variance and standard deviation.

What is Var[R_p]?

What is SD[R_p]?

✅ Correct! All nine cells give

42.8625\ \%^2

, so

\sigma_p \approx 6.55\%

per month, well below the weighted average of the SDs,

9.25\%

❌ Close, but… That counts each covariance once. Each pair fills two cells,

(i, j)

and

(j, i)

, so each off-diagonal value appears twice.

❌ Not quite. That is the weighted average of the variances,

\sum_i \omega_i\sigma_i^2

. Portfolio variance is not that: each variance enters with

\omega_i^2

, and the covariance cells must be added.

❌ Not quite. The Birch–Cedar cells are negative,

\rho_{BC} = -0.3

, so they subtract from the total.

❌ Not quite. Add all nine cells

\omega_i\omega_j\sigma_i\sigma_j\rho_{ij}

❌ Not quite.

9.25\%

is the weighted average of the SDs. Standard deviations do not average; take the square root of the portfolio variance.

❌ Close, but…

\sqrt{41.5625} \approx 6.45

uses only the diagonal. The covariance cells add a net

+1.3\ \%^2

❌ Not quite.

\sqrt{94.25} \approx 9.71

is the root of the weighted average of the variances, which is not the portfolio variance.

❌ Not quite. Take the square root of the portfolio variance.

Show solution

Diagonal (weighted variances, $\omega_i^2\sigma_i^2$ ):

0.25^2 \times 25 + 0.25^2 \times 64 + 0.5^2 \times 144 = 1.5625 + 4 + 36 = 41.5625

Off-diagonal (each pair appears twice, $2\,\omega_i\omega_j\sigma_i\sigma_j\rho_{ij}$ ):

Alder–Birch: $2 \times 0.25 \times 0.25 \times 5 \times 8 \times 0.2 = 1$
Alder–Cedar: $2 \times 0.25 \times 0.5 \times 5 \times 12 \times 0.5 = 7.5$
Birch–Cedar: $2 \times 0.25 \times 0.5 \times 8 \times 12 \times (-0.3) = -7.2$

\text{Var}[R_p] = 41.5625 + 1 + 7.5 - 7.2 = 42.8625\ \%^2 \approx 42.86\ \%^2

\text{SD}[R_p] = \sqrt{42.8625} \approx 6.55\%\ \text{per month}

Problem 4 · Adding a 31st Stock

Given: A portfolio already holds 30 stocks, and Dogwood is added as a 31st. Find how many covariance cells the variance table gains.

✅ Correct! Dogwood gets a new row and a new column: 61 new cells, of which only one is its own variance.

❌ Close, but… Dogwood forms 30 new pairs, but each pair fills two cells of the table, one in Dogwood's row and one in its column.

❌ Close, but…

31^2 - 30^2 = 61

counts every new cell, including Dogwood's own variance on the diagonal.

❌ Not quite.

31 \times 30 = 930

is the total number of covariance cells in the new table, not the number added.

❌ Not quite. Compare the covariance count

n(n-1)

before and after.

Show solution

Covariance cells in an $n$ -stock table: $n^2 - n = n(n-1)$ .

31 \times 30 - 30 \times 29 = 930 - 870 = 60

Equivalently: the table grows from $30^2 = 900$ to $31^2 = 961$ cells, 61 new ones. One is Dogwood's variance; the other 60 are its covariances with the 30 stocks already held, 30 in its row and 30 in its column. However volatile Dogwood is on its own, its variance is 1 cell of 961.

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