Finance-Theory · Unit 14 · Video 2 · Interactive Practice

Correlation, the Normal Curve, and Fat Tails

IKey Formulas

Formula	Name	What it tells you
$\mathrm{Cov}[R_i, R_j] = E\big[(R_i - \mu_i)(R_j - \mu_j)\big]$	Covariance	Positive when the returns tend to move together, negative when they tend to move in opposite directions
$\mathrm{Corr}[R_i, R_j] = \dfrac{\mathrm{Cov}[R_i, R_j]}{\sigma_i\,\sigma_j}$	Correlation $\rho$	Scale-free: always $-1 \le \rho \le +1$
$\mu \pm z\,\sigma$	Confidence interval for a normal $R$	$z = 1.00,\ 1.65,\ 1.96,\ 2.58$ gives $68\%,\ 90\%,\ 95\%,\ 99\%$

Key Insight: For a normal return, $\mu$ and $\sigma$ fix every probability; real monthly returns, like General Motors', have fatter tails than the fitted normal curve predicts.

IICovariance as Signed Rectangles

Each scenario contributes the signed area $(R_A - \mu_A)(R_B - \mu_B)$ ; covariance is their probability-weighted average.

IIITightness, Not Slope

Correlation measures how tightly the months hug a line, not how steep that line is.

ρ 0.50 Birch returns

IVConfidence Bands Against a Fat-Tailed Series

A normal with the same $\mu$ and $\sigma$ predicts how many months fall outside each band; fat tails break the widest promise.

Band

💡 General Motors' monthly returns show the same pattern: a reasonable fit through the middle, and more months below $-15\%$ and $-20\%$ than the fitted normal allows.

VQuiz Questions

Problem 1 · Covariance and Correlation from Scenarios

Given: next year's returns on two illustrative stocks depend on the economy: recession (probability $0.3$ ) Aster $-8\%$ , Birch $0\%$ ; normal (probability $0.5$ ) Aster $4\%$ , Birch $2\%$ ; boom (probability $0.2$ ) Aster $12\%$ , Birch $10\%$ . Find the covariance and the correlation.

What is the covariance?

What is the correlation?

✅ Correct! Aster and Birch sit below their means together in the recession and above them together in the boom, so

\mathrm{Cov} = 22

and

\rho = 22/26 \approx 0.85

: a strongly positively correlated pair.

❌ Not quite. The scenarios are not equally likely: weight each product by

0.3

0.5

0.2

, not by

\tfrac{1}{3}

❌ Not quite.

98

is the plain sum of the products; covariance is an expectation, so each product is multiplied by its probability.

❌ Check the sign. In the recession Aster is

10

points below its mean and Birch

3

points below its; in the boom both are above. Two deviations of the same sign give a positive product, so the returns move together.

❌ Not quite. Divide by the standard deviations

\sigma_A\sigma_B

, not by the variances

\sigma_A^2\sigma_B^2

❌ Impossible value. A correlation can never exceed

+1

; this one comes from the equal-weight covariance

32

❌ Not quite. Find each mean first, then average the products of the deviations using the scenario probabilities.

❌ Not quite. Divide the covariance by

\sigma_A \sigma_B

Show solution

Means:

\mu_A = 0.3(-8) + 0.5(4) + 0.2(12) = 2\%, \qquad \mu_B = 0.3(0) + 0.5(2) + 0.2(10) = 3\%

Deviations: Aster $-10,\ 2,\ 10$ ; Birch $-3,\ -1,\ 7$ .

\mathrm{Cov} = 0.3(-10)(-3) + 0.5(2)(-1) + 0.2(10)(7) = 9 - 1 + 14 = 22\ (\%^2)

Standard deviations (probability-weighted, like the covariance):

\sigma_A^2 = 0.3(100) + 0.5(4) + 0.2(100) = 52, \quad \sigma_A \approx 7.21\%

\sigma_B^2 = 0.3(9) + 0.5(1) + 0.2(49) = 13, \quad \sigma_B \approx 3.61\%

\rho = \frac{22}{\sqrt{52}\,\sqrt{13}} = \frac{22}{\sqrt{676}} = \frac{22}{26} \approx 0.85

Problem 2 · Rescaling a Return

Given: for two illustrative assets, $\sigma_{\text{Aster}} = 6\%$ , $\sigma_{\text{Birch}} = 5\%$ and $\mathrm{Cov} = 18\ (\%^2)$ . A new asset, Birch 2x, pays exactly twice Birch's return in every period. Find the Aster–Birch correlation, then the covariance and correlation of Aster with Birch 2x.

Correlation of Aster and Birch

Aster with Birch 2x

✅ Correct! Doubling a return doubles every deviation, so the covariance doubles, but

\sigma

doubles too and the correlation stays at

0.60

: correlation ignores scale.

❌ Not quite.

18/5 = 3.6

divides by only one standard deviation, and no correlation can exceed

1

❌ Not quite.

18/(36 \times 25) = 0.02

divides by the variances; the denominator is

\sigma_A\sigma_B = 6 \times 5

❌ Not quite.

18

is the covariance, in

\%^2

; the correlation is scale-free and lies between

-1

and

+1

❌ Close, but… the covariance does double, but so does Birch 2x's standard deviation (

10\%

), and a correlation above

1

is impossible.

❌ Not quite. Each deviation of Birch 2x from its mean is twice Birch's, so every product in the covariance doubles.

❌ Not quite. Only one of the two returns is doubled, so the covariance doubles; it is the variance of Birch 2x that quadruples.

❌ Not quite.

\rho = \mathrm{Cov}/(\sigma_A\sigma_B)

❌ Not quite. Track what doubling does to the deviations and to

\sigma

Show solution

\rho_{\text{Aster, Birch}} = \frac{18}{6 \times 5} = \frac{18}{30} = 0.60

Birch 2x has mean $2\mu_B$ , so its deviation in every period is $2(R_B - \mu_B)$ :

\mathrm{Cov}[R_A, 2R_B] = E\big[(R_A - \mu_A)\cdot 2(R_B - \mu_B)\big] = 2 \times 18 = 36\ (\%^2)

Its standard deviation is $2 \times 5\% = 10\%$ , so

\rho = \frac{36}{6 \times 10} = 0.60

The fitted line through the scatter becomes twice as steep, but the points hug it exactly as tightly as before.

Problem 3 · An Annual Confidence Interval

Given: an illustrative stock's annual return is normal with $\mu = 9\%$ and $\sigma = 21\%$ . Find the $95\%$ confidence interval, and the probability that a given year's return falls below its lower end.

95% interval

P(return below the lower end)

✅ Correct! The interval is

9\% \pm 41.16\%

, and by symmetry half of the excluded

5\%

lies below it: about one year in

40

on average.

❌ Not quite.

1.65\sigma

gives the

90\%

band; the

95\%

band uses

1.96\sigma

❌ Close, but…

\pm 2\sigma

is the rounded rule of thumb and covers about

95.4\%

; the

95\%

band is exactly

\pm 1.96\sigma

❌ Not quite.

2.58\sigma

gives the

99\%

band.

❌ Not quite.

5\%

lies outside the band in total, split equally between the two tails.

❌ Not quite.

0.5\%

is one tail of the

99\%

band (

2.58\sigma

); the lower end here is

1.96\sigma

below the mean.

❌ Not quite.

95\%

is the probability of landing inside the band, not below it.

❌ Not quite. Use

\mu \pm 1.96\sigma

❌ Not quite.

95\%

lies inside; the rest is split between two symmetric tails.

Show solution

1.96 \times 21\% = 41.16\%

9\% - 41.16\% = -32.16\%, \qquad 9\% + 41.16\% = 50.16\%

The normal is symmetric, so the $5\%$ outside the band splits into $2.5\%$ below $-32.16\%$ and $2.5\%$ above $50.16\%$ . A return below $-32.16\%$ happens in about one year in $1/0.025 = 40$ .

Problem 4 · Testing the Normal Model

Given: $480$ months ( $40$ years) of an illustrative stock's returns have sample mean $0.7\%$ and sample standard deviation $4.8\%$ . The normal curve with this same $\mu$ and $\sigma$ is laid over the histogram. Find how many months the normal predicts below $\mu - 2.58\sigma$ , then interpret the data, which contain $11$ such months.

Months predicted below μ − 2.58σ

What do 11 such months show?

✅ Correct! The normal expects

2.4

such months;

11

is more than four times that. This is the fat-tail pattern of General Motors' monthly returns: the normal model understates the risk of very large losses.

❌ Not quite.

4.8

1\%

480

, the count outside the

99\%

band on both sides together; only the lower tail is asked.

❌ Not quite.

24

5\%

480

, the count outside the

95\%

band;

2.58\sigma

belongs to the

99\%

band.

❌ Not quite.

12

2.5\%

480

, one tail of the

95\%

band (

1.96\sigma

); here the cutoff is

2.58\sigma

❌ Not quite. For a normal return,

\mu

and

\sigma

fix every probability, including how often a return falls outside each band.

❌ Not quite.

\sigma

was measured from these same

480

months, so the fitted curve already has the data's variance; the misfit is in the shape, not the spread.

❌ Not quite. The mean was measured from these same months; the question is how often extreme months occur, which is about the shape of the tails.

❌ Not quite. The

99\%

band leaves

1\%

outside, split between the two tails.

❌ Not quite. Compare

11

with the normal's prediction.

Show solution

$\mu \pm 2.58\sigma$ is the $99\%$ band, so $1\%$ lies outside it, $0.5\%$ in each tail. The cutoff is $0.7\% - 2.58 \times 4.8\% \approx -11.68\%$ .

\text{predicted months below} = 0.005 \times 480 = 2.4

The data contain $11$ , more than four times as many. Because the fitted normal already matches the sample mean and variance, the gap says the distribution itself has a fat left tail: very large monthly losses are materially more likely than the normal model admits.

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