Finance-Theory · Unit 16 · Video 8 · Interactive Practice

Three Stocks, One Bullet: Long Horizons and the Road to Equilibrium

IKey Formulas

Formula	Name	What it says
$\omega_1 + \omega_2 + \omega_3 = 1$	Portfolio weights	Fractions of the portfolio; a negative weight is a short sale
$E[R_P] = \omega_1\mu_1 + \omega_2\mu_2 + \omega_3\mu_3$	Portfolio mean	The weighted average of the three means
$\mathrm{var}(R_P) = \sum_{i=1}^{3}\sum_{j=1}^{3}\omega_i\omega_j\sigma_{ij}$	Portfolio variance	Every cell of the $3 \times 3$ matrix, times the weights for its row and column
$= \omega_1^2\sigma_1^2 + \omega_2^2\sigma_2^2 + \omega_3^2\sigma_3^2$ $+ 2\omega_1\omega_2\sigma_{12} + 2\omega_1\omega_3\sigma_{13} + 2\omega_2\omega_3\sigma_{23}$	Cells grouped	3 diagonal variances, plus the 6 off-diagonal covariances paired into 3 terms

Key Insight: With three or more stocks, none of them need lie on the minimum-variance boundary. For GM, IBM and Motorola the bullet passes strictly northwest of all three, and its inputs are long-horizon estimates (15–20 years), not forecasts of tomorrow.

IIMonthly Noise over a 12-Year Holding

Over $h$ months the mean grows like $h$ but the standard deviation only like $\sqrt{h}$ .

Months held (

h

) 12

💡 Illustrative holding (the fictional stock Nimbus: $0.8\%$ mean and $5\%$ SD per month), months independent with the same variance every month, compounding ignored: the annualizing rules (mean $\times 12$ , SD $\times \sqrt{12}$ ) applied to $h$ months. Buffett's $5 billion Goldman Sachs investment of September 2008 was under water within weeks, on a 15–20-year horizon.

IIIEvery Cell of the Matrix

The portfolio variance adds all nine cells: each weight pair times its variance or covariance.

ω_GM 0.60 ω_IBM 0.20 ω_MOT = 1 − ω_GM − ω_IBM 0.20

IVThe Bullet Lies Northwest of All Three

Each stock sits inside the three-stock bullet: some portfolio beats it on return, and some on risk (short sales allowed: weights may be negative).

Stock

💡 If every investor holds a portfolio on this upper branch, the result is not self-defeating: everybody following the theory produces an equilibrium, the Capital Asset Pricing Model, which the coming units build.

VQuiz Questions

Problem 1 · Expected Return of a Three-Stock Mix

Given (illustrative): three fictional stocks, monthly returns in percent.

Stock	Mean	SD	Nimbus	Orrery	Pylon
Nimbus	$0.8$	$5$	$25$	$10.5$	$15.0$
Orrery	$1.1$	$7$	$10.5$	$49$	$28.0$
Pylon	$1.6$	$10$	$15.0$	$28.0$	$100$

With weights $\omega_N = 0.5$ , $\omega_O = 0.3$ , $\omega_P = 0.2$ , find the portfolio's expected monthly return.

✅ Correct!

0.5 \times 0.8 + 0.3 \times 1.1 + 0.2 \times 1.6 = 0.40 + 0.33 + 0.32 = 1.05

, in percent per month.

❌ That is the equal-weighted average.

(0.8 + 1.1 + 1.6)/3 \approx 1.17

ignores the weights; Nimbus carries half the portfolio.

❌ The weights are on the wrong stocks.

0.2 \times 0.8 + 0.3 \times 1.1 + 0.5 \times 1.6 = 1.29

puts half the money in Pylon, not Nimbus.

❌ Not quite.

0.8 + 1.1 + 1.6 = 3.5

adds the means without weighting them; the weights sum to 1, so the answer must lie between

0.8

and

1.6

Show solution

The portfolio mean is the weighted average of the three means:

E[R_P] = \omega_N\mu_N + \omega_O\mu_O + \omega_P\mu_P = 0.5(0.8) + 0.3(1.1) + 0.2(1.6)

= 0.40 + 0.33 + 0.32 = 1.05

The expected return is 1.05% per month. Only the means and the weights enter; the covariances matter for the variance, not for the mean.

Problem 2 · Adding Up the Nine Cells

Given: the same three stocks and weights $\omega_N = 0.5$ , $\omega_O = 0.3$ , $\omega_P = 0.2$ (variances and covariances in $\%^2$ from the table in Problem 1). Find the portfolio variance, then compare the portfolio with Nimbus alone.

What is $\mathrm{var}(R_P)$ ?

Compared with holding only Nimbus ( $0.8\%$ mean, $5\%$ SD), this portfolio…

✅ Correct!

\mathrm{var}(R_P) = 24.17

, so

\sigma_P = \sqrt{24.17} \approx 4.92\%

, below Nimbus's

5\%

, with a mean of

1.05\%

above its

0.8\%

: Nimbus is not on the minimum-variance boundary.

❌ The factor 2 is missing. Each covariance sits in two cells of the matrix (

\sigma_{NO}

and

\sigma_{ON}

), so each pair term is

2\omega_i\omega_j\sigma_{ij}

❌ The diagonal weights must be squared. A diagonal cell is

\omega_i \cdot \omega_i \cdot \sigma_i^2

: its row weight times its column weight, so

0.5^2 \times 25 = 6.25

, not

0.5 \times 25

❌ That is only the diagonal.

6.25 + 4.41 + 4.00 = 14.66

leaves out the six covariance cells, which add

9.51

❌ Check the variance. Add all nine cells: three squared-weight variances and three doubled covariance terms.

❌ Mixing in riskier stocks can still cut risk. With correlations of

0.3

and

0.4

, the covariance cells are small enough that

\sigma_P \approx 4.92\%

, below Nimbus's

5\%

❌ SDs do not average out. The portfolio's SD comes from all nine cells:

\sigma_P = \sqrt{24.17} \approx 4.92\%

, not

5\%

❌ Everything needed is in the table. The covariances (

10.5

15.0

28.0

) are given, so

\sigma_P \approx 4.92\%

can be set directly against Nimbus's

5\%

❌ Compare the numbers. Take

\sigma_P = \sqrt{\mathrm{var}(R_P)}

and set mean against mean, SD against SD.

Show solution

Diagonal cells (squared weight × variance):

0.5^2(25) + 0.3^2(49) + 0.2^2(100) = 6.25 + 4.41 + 4.00 = 14.66

Off-diagonal cells, paired (2 × weight × weight × covariance):

2(0.5)(0.3)(10.5) + 2(0.5)(0.2)(15.0) + 2(0.3)(0.2)(28.0) = 3.15 + 3.00 + 3.36 = 9.51

\mathrm{var}(R_P) = 14.66 + 9.51 = 24.17\ (\%^2), \qquad \sigma_P = \sqrt{24.17} \approx 4.92\%

From Problem 1 the mean is $1.05\%$ . Against Nimbus ( $0.8\%$ , $5\%$ ) the portfolio has a higher mean and a lower SD, so it dominates Nimbus.

Problem 3 · Less Risk at Orrery's Mean

Given: the same three stocks. Build a portfolio of Nimbus and Pylon only ( $\omega_O = 0$ ) whose expected return equals Orrery's $1.1\%$ per month. Find the weight on Pylon and the portfolio's SD.

What weight on Pylon gives a mean of $1.1\%$ ?

What is the portfolio's SD?

✅ Correct! At Orrery's own mean of

1.1\%

this mix carries about

5.56\%

SD against Orrery's

7\%

, so Orrery is not on the minimum-variance boundary either.

❌ Not a ratio of means.

1.1/1.6 \approx 0.6875

ignores Nimbus's

0.8\%

; solve

0.8(1 - \omega_P) + 1.6\,\omega_P = 1.1

❌ Divide by the spread in means.

1.1 - 0.8 = 0.3

is the extra return needed; each unit of weight moved to Pylon adds

1.6 - 0.8 = 0.8

, so

\omega_P = 0.3/0.8

❌ An even split overshoots.

0.5(0.8) + 0.5(1.6) = 1.2

, above Orrery's

1.1

; solve

0.8(1 - \omega_P) + 1.6\,\omega_P = 1.1

❌ Check the weight. Set

0.8(1 - \omega_P) + 1.6\,\omega_P = 1.1

and solve for

\omega_P

❌ Risk is not a weighted average of SDs.

0.625(5) + 0.375(10) = 6.875

would hold only with correlation

+1

; with

\sigma_{NP} = 15.0

the variance is smaller.

❌ The covariance term is missing. Add

2(0.625)(0.375)(15.0) \approx 7.03

to the two diagonal terms before taking the square root.

❌ That is the variance.

30.86

is in

\%^2

; the SD is its square root.

❌ Check the SD. Compute the two-stock variance with the covariance term, then take the square root.

Show solution

Step 1 — Match the mean. With $\omega_N = 1 - \omega_P$ :

0.8(1 - \omega_P) + 1.6\,\omega_P = 0.8 + 0.8\,\omega_P = 1.1 \quad\Rightarrow\quad \omega_P = \frac{0.3}{0.8} = 0.375,\ \ \omega_N = 0.625

Step 2 — Variance. Only the Nimbus and Pylon cells are non-zero:

\mathrm{var}(R_P) = 0.625^2(25) + 0.375^2(100) + 2(0.625)(0.375)(15.0)

= 9.765625 + 14.0625 + 7.03125 = 30.859375\ (\%^2)

\sigma_P = \sqrt{30.859375} \approx 5.56\%

Same mean as Orrery, about $1.44$ percentage points less SD: Orrery lies to the right of the boundary, inside the bullet.

Problem 4 · A Fourth Stock

Given: the minimum-variance boundary of Nimbus, Orrery and Pylon (short sales allowed). A fourth fictional stock is added and the boundary is recomputed from all four. Which statement must be true?

✅ Correct! Every three-stock portfolio is still available with weight

0

on the new stock, so the least variance at each mean can only fall or stay the same.

❌ Adding an option never removes one. The old portfolios remain available (weight

0

on the new stock), so the least variance at any mean cannot rise, however risky the new stock is.

❌ Not with three or more stocks. As with GM, IBM and Motorola, an individual stock can sit strictly inside the bullet; nothing forces the new one onto the boundary.

❌ Too narrow. The new stock can be held long or short in any portfolio, so it can cut variance at means below its own as well; the only guarantee is that nothing gets worse.

Show solution

The boundary is defined as the least variance at each level of expected return. With four stocks the candidate portfolios include every three-stock portfolio (put $\omega_4 = 0$ ), so at each mean the minimum is taken over a larger set:

\sigma_{\text{4 stocks}}(\mu) \le \sigma_{\text{3 stocks}}(\mu) \quad \text{for every mean } \mu

The boundary moves left or stays put; it never moves right. This is why the case for portfolio theory grows stronger as stocks are added, and why an individual stock, old or new, need not lie on the boundary at all.

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