Finance-Theory · Unit 16 · Video 7 · Interactive Practice

Why Diversification Hits a Floor: Average Covariance and the Efficient Frontier

IKey Formulas

Formula	Name	What it says
$\text{Var}[R_p] = \dfrac{1}{n}\,\overline{\sigma^2} + \dfrac{n-1}{n}\,\overline{\text{Cov}}$	Equal weights $\omega_i = \tfrac{1}{n}$	$\overline{\sigma^2}$ is the average of the $n$ variances, $\overline{\text{Cov}}$ the average of the $n(n-1)$ covariances
$\text{Cov}[R_i, R_j] = \rho_{ij}\,\sigma_i\,\sigma_j$	Covariance from correlation	Video example: $0.40 \times 0.10 \times 0.10 = 0.004$ (decimal returns, monthly)
$\text{Var}[R_p] \to \overline{\text{Cov}}$ as $n \to \infty$	The floor: systematic (market) risk	$\tfrac{1}{n} \to 0$ and $\tfrac{n-1}{n} \to 1$ ; in the video $\sigma_p \to \sqrt{0.004} \approx 6.3\%$ per month
$\dfrac{\overline{\sigma^2} - \overline{\text{Cov}}}{\overline{\sigma^2}}$	Share of a stock's variance diversification removes	Video example: $(0.01 - 0.004)/0.01 = 0.6$ removed, $0.4$ systematic

Key Insight: However many stocks you add, the variance of an equally weighted portfolio cannot fall below the average covariance. That floor is systematic risk, and the only way to carry less of it is to leave stocks for Treasury bills or cash.

IIAdding Stocks: Risk Falls to a Floor

With equal weights, portfolio risk falls quickly, then flattens at $\sqrt{\overline{\text{Cov}}}$ and never goes below it.

Average SD Correlation ρ 0.30 Stocks n 1

💡 The correlation is not a constant of nature: estimated over a long history it already contains past crises (October 1987, LTCM in August 1998, March 2000), but a crisis under way now can push correlations up and lift the floor.

IIIHold Only the Upper Branch

Each boundary point carries the least risk for its expected return; only the branch above the minimum-variance point is efficient.

Expected return 1.30%

💡 With $n$ assets every choice of weights is one point and the feasible portfolios fill a bullet-shaped region; the curve is its left edge, and the conclusion is the same. Where on the upper branch to sit depends on how much risk you will accept for more expected return.

IVQuiz Questions

Problem 1 · Twenty-Five Stocks

Given: an illustrative set of stocks with average standard deviation $20\%$ per year and average correlation $0.15$ between any two. You hold $25$ of them in equal weights. Find the portfolio's standard deviation per year.

✅ Correct!

\text{Var}[R_p] = 0.0016 + 0.00576 = 0.00736

, so

\sigma_p = \sqrt{0.00736} \approx 8.58\%

: already close to the floor of about

7.75\%

❌ Not quite. You added the full average covariance

0.006

. With equal weights it is multiplied by

\tfrac{n-1}{n} = \tfrac{24}{25}

, giving

0.00576

❌ Close, but that is the floor.

\sqrt{0.006} \approx 7.75\%

is the limit as

n \to \infty

; with

25

stocks the term

\tfrac{1}{25} \times 0.04 = 0.0016

has not vanished yet.

❌ Not quite.

20\%/\sqrt{25} = 4\%

is the answer only if the correlation is zero. Here every pair has covariance

0.15 \times 0.20 \times 0.20 = 0.006

, and those

n(n-1)

terms dominate.

❌ Not quite. Work in decimals:

\overline{\sigma^2} = 0.04

and

\overline{\text{Cov}} = 0.006

, then apply the equal-weight formula with

n = 25

Show solution

Average variance and average covariance, in decimals per year:

\overline{\sigma^2} = 0.20^2 = 0.04, \qquad \overline{\text{Cov}} = 0.15 \times 0.20 \times 0.20 = 0.006

Equal weights with $n = 25$ :

\text{Var}[R_p] = \frac{1}{25}(0.04) + \frac{24}{25}(0.006) = 0.0016 + 0.00576 = 0.00736

\sigma_p = \sqrt{0.00736} \approx 0.0858 = 8.58\%

A single stock carries $20\%$ ; twenty-five of them in equal weights carry about $8.58\%$ , only about $0.83$ percentage points above the floor $\sqrt{0.006} \approx 7.75\%$ .

Problem 2 · Shares Live in Variance

Given: for the same illustrative stocks (average SD $20\%$ per year), a very large equally weighted portfolio settles at a standard deviation of $\sqrt{0.006} \approx 7.75\%$ per year. Find the share of a typical single stock's risk, measured as variance, that diversification eliminates.

✅ Correct!

(0.04 - 0.006)/0.04 = 0.85

: diversification removes

85\%

of the variance, and the systematic

15\%

stays.

❌ Close, but reversed.

0.006/0.04 = 15\%

is the systematic share, the part that stays. The question asks for the part removed.

❌ Not quite: that compares standard deviations.

1 - 7.75/20 \approx 61.3\%

uses SDs, but the decomposition is additive only in variance. Square first:

0.0775^2 \approx 0.006

against

0.20^2 = 0.04

❌ Not quite.

7.75/20 \approx 38.7\%

is a ratio of standard deviations, and it describes what stays, not what is removed. Compare variances:

0.006

against

0.04

❌ Not quite. Convert both risks to variances before comparing them.

Show solution

The large portfolio's variance is the floor, the average covariance: $\left(\sqrt{0.006}\right)^2 = 0.006$ . A typical single stock's variance is $0.20^2 = 0.04$ .

\text{systematic share} = \frac{0.006}{0.04} = 0.15, \qquad \text{eliminated share} = \frac{0.04 - 0.006}{0.04} = 0.85

So $85\%$ of a single stock's variance is diversifiable. Comparing standard deviations instead ( $7.75\%$ against $20\%$ ) gives $61.3\%$ , which understates it: the total splits into "systematic + diversifiable" only in variance.

Problem 3 · How Many Stocks for a Target?

Given: the same illustrative stocks: average SD $20\%$ per year, average correlation $0.15$ . Find the smallest number $n$ of stocks, in equal weights, that brings the portfolio's standard deviation down to $8\%$ per year.

✅ Correct!

\text{Var}[R_p] = 0.006 + 0.034/n

, and

0.034/n \le 0.0004

needs

n \ge 85

; at

n = 85

the variance is exactly

0.0064

, an SD of

8\%

❌ Not quite.

20\%/\sqrt{n} \le 8\%

treats the stocks as uncorrelated. With

n = 7

the covariance term still contributes

\tfrac{6}{7} \times 0.006

and

\sigma_p \approx 10.4\%

❌ Not quite. You split the standard deviation as floor plus

(20\% - 7.75\%)/n

. That split holds for variances, not standard deviations: at

n = 49

\sigma_p \approx 8.18\%

❌ Close, but not the smallest. Writing

0.04/n + 0.006

drops the factor

\tfrac{n-1}{n}

on the covariance. With it,

85

stocks already reach

8\%

;

100

give about

7.96\%

❌ Not quite. Write the variance as floor plus a

\tfrac{1}{n}

term and solve for

n

Show solution

With $\overline{\sigma^2} = 0.04$ and $\overline{\text{Cov}} = 0.15 \times 0.20^2 = 0.006$ , regroup the equal-weight formula:

\text{Var}[R_p] = \frac{0.04}{n} + \frac{n-1}{n}(0.006) = 0.006 + \frac{0.04 - 0.006}{n} = 0.006 + \frac{0.034}{n}

The target SD $8\%$ means a variance of $0.08^2 = 0.0064$ :

0.006 + \frac{0.034}{n} \le 0.0064 \;\Longrightarrow\; \frac{0.034}{n} \le 0.0004 \;\Longrightarrow\; n \ge 85

Check: $n = 85$ gives $0.006 + 0.0004 = 0.0064$ , so $\sigma_p = 8\%$ ; $n = 84$ gives about $8.003\%$ . The last stretch is slow because the floor, $\approx 7.75\%$ , is only a quarter of a point below the target.

Problem 4 · Is the Safer Stock Efficient?

Given: illustrative stocks Lumen ( $1.0\%$ mean, $6\%$ SD per month) and Marlow ( $1.4\%$ , $9\%$ ), correlation $0.1$ . An investor holds Lumen alone because it is the less risky of the two. Decide whether that portfolio is on the efficient frontier.

✅ Correct! Mixing in a little Marlow raises the mean and lowers the risk at the same time, because the correlation is only

0.1

. Lumen alone sits below the minimum-variance point, on the branch nobody should hold.

❌ Not quite. The minimum-variance portfolio is a mix, not the least risky single stock:

50/50

already has

\sigma_p \approx 5.65\% \lt 6\%

❌ Not quite. Lumen is on the boundary (with two stocks, every mix is), but the boundary has two branches. Only the upper one, from the minimum-variance point up, is efficient.

❌ Not quite. Marlow is on the upper branch, so it is efficient, but so is every mix between the minimum-variance point and Marlow. Which one to hold depends on how much risk you accept.

Show solution

Try $\omega_{\text{Lumen}} = \omega_{\text{Marlow}} = 0.5$ (returns in percent, variances in $\%^2$ ):

E[R_p] = 0.5(1.0) + 0.5(1.4) = 1.20\%

\text{Var}[R_p] = 0.5^2(36) + 0.5^2(81) + 2(0.5)(0.5)(0.1)(6)(9) = 9 + 20.25 + 2.7 = 31.95

\sigma_p = \sqrt{31.95} \approx 5.65\%

This mix has a higher mean ( $1.20\% \gt 1.0\%$ ) and a lower SD ( $5.65\% \lt 6\%$ ) than Lumen alone, so it dominates Lumen. A dominated portfolio cannot be on the efficient frontier: Lumen sits on the lower branch of the minimum-variance boundary, below the minimum-variance point.

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