Finance-Theory · Unit 16 · Video 2 · Interactive Practice

Two Stocks, One Covariance: Why the Risk-Return Trade-Off Is a Bullet

IKey Formulas

Formula	Name	What it says
$E[R_p] = \omega_a \mu_a + \omega_b \mu_b$ , with $\omega_a + \omega_b = 1$	Two-asset expected return	The mean is a plain weighted average of the two means
$\text{Var}[R_p] = \omega_a^2 \sigma_a^2 + \omega_b^2 \sigma_b^2 + 2\,\omega_a \omega_b \,\text{Cov}[R_a, R_b]$	Two-asset variance	The sum of a two-by-two table: two diagonal cells plus two identical off-diagonal cells, hence the factor $2$
$\text{Var}[R_p] = \omega_a^2 \sigma_a^2 + \omega_b^2 \sigma_b^2 + 2\,\omega_a \omega_b \,\sigma_a \sigma_b \,\rho_{ab}$ , with $-1 \le \rho_{ab} \le 1$	Correlation form	With both weights positive, a higher $\rho_{ab}$ means a higher variance
$\mu_{\text{annual}} \approx 12\,\mu_{\text{monthly}}, \quad \sigma_{\text{annual}} = \sqrt{12}\,\sigma_{\text{monthly}}$	Annualizing	Compounding ignored; months uncorrelated with a constant variance, so the variance of a year is $12\,\sigma^2_{\text{monthly}}$

Key Insight: Means average; risks do not. Half General Motors and half Motorola earns $0.5(1.08) + 0.5(1.75) = 1.415\%$ a month (the video's table rounds it to $1.42\%$ ), exactly halfway, but its standard deviation is about $6.68\%$ , short of the halfway $7.98\%$ . The cross term, with $\rho = 0.37 < 1$ , is what bends the trade-off into a bullet.

IIThe Two-by-Two Variance Table

Portfolio variance is the sum of four cells, and the two off-diagonal cells are always the same number.

Weight on Motorola, ω_MOT 0.50 Correlation, ρ 0.37

IIIThe Weights Hidden Along the Bullet

Every point on the General Motors–Motorola curve is one choice of weights, yet neither axis shows them.

Weight on Motorola, ω_MOT 0.50

IVAnnualizing: Twelve Times the Mean, Root Twelve Times the Risk

With uncorrelated months of equal variance, variance grows with the number of months, so standard deviation grows only with its square root.

Months, n 12 Monthly figures

💡 For scale: the volatility index (VIX), which measures expected stock-market volatility, closed at $80.86$ on 20 November 2008 and stood at $17.71$ on 16 September 2026, so the video's roughly $40\%$ a year is a great deal of risk for a single holding in calm markets.

VQuiz Questions

Problem 1 · A 60/40 Mix

Given: two fictional stocks (illustrative numbers, per month): Ember has $\mu = 0.9\%$ and $\sigma = 5\%$ ; Flint has $\mu = 1.6\%$ and $\sigma = 11\%$ ; $\rho = 0.25$ . A portfolio holds $60\%$ Ember and $40\%$ Flint. Find its expected return and its standard deviation.

Expected return per month?

Standard deviation per month?

✅ Correct! The mean is the weighted average,

1.18\%

, but the risk, about

5.91\%

, is well below the weighted average of the two standard deviations,

7.40\%

❌ Not quite.

1.25\%

is the simple average of

0.9\%

and

1.6\%

; the weights are

0.6

and

0.4

, not one half each.

❌ Weights swapped.

1.32\%

puts

0.6

on Flint; the

60\%

weight belongs to Ember.

❌ Not quite. The weights are squared only in the variance; the mean uses

\omega

itself, and

0.6 + 0.4 = 1

❌ Not quite. Use

E[R_p] = \omega_E \mu_E + \omega_F \mu_F

❌ That is the straight-line answer.

0.6(5) + 0.4(11) = 7.40

averages the standard deviations, which is right only when

\rho = 1

; here the cross term is smaller.

❌ Take the square root.

34.96

is the variance in

\%^2

; the standard deviation is its square root.

❌ Missing the factor of 2. The cross term appears in both off-diagonal cells of the two-by-two table, so it is

2 \times 0.6 \times 0.4 \times 5 \times 11 \times 0.25 = 6.6

, not

3.3

❌ Not quite. Build the variance from the four cells, then take the square root.

Show solution

Mean:

E[R_p] = 0.6(0.9) + 0.4(1.6) = 0.54 + 0.64 = 1.18\%

Variance (in $\%^2$ , returns in percent per month):

\text{Var}[R_p] = 0.6^2(5^2) + 0.4^2(11^2) + 2(0.6)(0.4)(5)(11)(0.25) = 9 + 19.36 + 6.6 = 34.96

\sigma_p = \sqrt{34.96} \approx 5.91\%

The risk is about $1.49$ points below the weighted average $0.6(5) + 0.4(11) = 7.40\%$ , because $\rho = 0.25 < 1$ .

Problem 2 · Short One, Lever the Other

Given: the same Ember and Flint (illustrative: $0.9\%$ / $5\%$ and $1.6\%$ / $11\%$ per month, $\rho = 0.25$ ). You short Ember and invest the proceeds in Flint: $\omega_{\text{Flint}} = 1.2$ , $\omega_{\text{Ember}} = -0.2$ . Find the portfolio's monthly standard deviation.

✅ Correct! With one weight negative, the cross term is subtracted:

174.24 + 1 - 6.6 = 168.64

, so

\sigma_p \approx 12.99\%

❌ Check the sign of the cross term.

2\,\omega_E \omega_F

2(-0.2)(1.2) = -0.48

, so the cross term enters with a minus sign; adding

6.6

gives

181.84

and

13.48\%

❌ That averages the standard deviations.

1.2(11) - 0.2(5) = 12.20

would be the risk only if

\rho = 1

; with

\rho = 0.25

, build the variance from the four cells.

❌ The cross term is missing.

\sqrt{174.24 + 1} \approx 13.24

leaves out

2(-0.2)(1.2)(5)(11)(0.25) = -6.6

❌ Not quite. Use the correlation form with

\omega_E = -0.2

Show solution

The weights still sum to one: $1.2 + (-0.2) = 1$ .

\text{Var}[R_p] = 1.2^2(11^2) + (-0.2)^2(5^2) + 2(1.2)(-0.2)(11)(5)(0.25)

= 174.24 + 1 - 6.6 = 168.64, \qquad \sigma_p = \sqrt{168.64} \approx 12.99\%

The squared weights are positive whatever the sign, but the product $\omega_E \omega_F$ is negative, so the cross term subtracts. The expected return is $1.2(1.6) + (-0.2)(0.9) = 1.92 - 0.18 = 1.74\%$ , above Flint's own $1.6\%$ , and the risk is above Flint's $11\%$ : the portfolio sits past Flint on the curve.

Problem 3 · From Monthly to Annual

Given: a fictional fund (illustrative) has a monthly mean of $1.1\%$ and a monthly standard deviation of $7\%$ . Assume the months are uncorrelated with one another and the monthly variance is constant. Find the annual standard deviation and the annual mean with compounding.

Annual standard deviation?

Annual mean, with compounding?

✅ Correct! Variance scales by

12

, so risk scales by

\sqrt{12}

\approx 24.25\%

a year; the compounded mean is

(1.011)^{12} - 1 \approx 14.03\%

❌ Square root of the wrong thing.

\sqrt{12 \times 7} \approx 9.17

; the variance is

12 \times 7^2 = 588\ \%^2

, and

\sqrt{588} = \sqrt{12} \times 7

❌ That treats the months as moving in lockstep. Multiplying the standard deviation by

12

assumes every pair of months is perfectly correlated; with uncorrelated months it is the variance that is multiplied by

12

❌ Wrong direction. A year is riskier than a month: multiply by

\sqrt{12}

, do not divide.

❌ Not quite. Start from

\text{Var}[R_1 + \dots + R_{12}] = 12\,\sigma^2_{\text{monthly}}

❌ That ignores compounding.

12 \times 1.1\% = 13.2\%

is the simple version; with compounding, add one, raise to the twelfth power and subtract one.

❌ Subtract the one.

(1.011)^{12} \approx 1.1403

is the growth factor; the return is

1.1403 - 1

❌ Check the units.

(1.011)^{12} - 1 \approx 0.1403

is a decimal:

14.03\%

, not

1.14\%

❌ Not quite. Use

(1 + \mu_{\text{monthly}})^{12} - 1

with

\mu

as a decimal.

Show solution

Risk. With uncorrelated months every covariance term in the variance of the sum drops out, and each month contributes the same $7^2 = 49\ \%^2$ :

\text{Var}[R_1 + \dots + R_{12}] = 12 \times 49 = 588\ \%^2, \qquad \sigma_{\text{annual}} = \sqrt{588} = \sqrt{12} \times 7 \approx 24.25\%

Mean. Ignoring compounding, $12 \times 1.1\% = 13.2\%$ . With compounding:

(1 + 0.011)^{12} - 1 \approx 1.1403 - 1 = 0.1403 \approx 14.03\%

Problem 4 · When Correlation Rises

Given: Ember and Flint as above (illustrative), but suppose their correlation rises from $0.25$ to $0.65$ . Find what happens to the standard deviation of each portfolio: (i) $60\%$ Ember, $40\%$ Flint (was about $5.91\%$ ); (ii) $-20\%$ Ember, $120\%$ Flint (was about $12.99\%$ ).

(i) The 60/40 portfolio?

(ii) The short-Ember portfolio?

✅ Correct! Higher correlation raises variance only when both weights are positive; with a short position the cross term carries a minus sign, so higher correlation lowers the variance.

❌ Check the sign. Both weights are positive, so

2\,\omega_E \omega_F \sigma_E \sigma_F = 26.4 > 0

multiplies

\rho

: a higher

\rho

adds variance, it does not subtract it.

❌ Correlation does enter. Unless one weight is zero,

\rho

multiplies the cross term

2\,\omega_E \omega_F \sigma_E \sigma_F

, so changing it changes the variance.

❌ Not with a short position. Here

2\,\omega_E \omega_F \sigma_E \sigma_F = 2(-0.2)(1.2)(5)(11) = -26.4

, so raising

\rho

makes the cross term more negative and lowers the variance.

❌ Not quite. Recompute the variance of the 60/40 portfolio with

\rho = 0.65

❌ Not quite. Recompute the variance of the short-Ember portfolio with

\rho = 0.65

, keeping

\omega_E = -0.2

Show solution

Only the cross term changes. Its coefficient $2\,\omega_E \omega_F \sigma_E \sigma_F$ decides the direction.

(i) $2(0.6)(0.4)(5)(11) = 26.4 > 0$ :

\text{Var}[R_p] = 9 + 19.36 + 26.4(0.65) = 9 + 19.36 + 17.16 = 45.52, \qquad \sigma_p \approx 6.75\%

(ii) $2(-0.2)(1.2)(5)(11) = -26.4 < 0$ :

\text{Var}[R_p] = 1 + 174.24 - 26.4(0.65) = 175.24 - 17.16 = 158.08, \qquad \sigma_p \approx 12.57\%

Both weights positive: higher $\rho$ , more risk. One weight negative: higher $\rho$ , less risk, because the short position now offsets more of the long one.

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