Finance-Theory · Unit 14 · Video 1 · Interactive Practice

Why a Ten Percent Loss Needs a Yardstick: Returns, the Risk Premium, and Standard Deviation

IKey Formulas

Formula	Name	What it says
$R_{it} \equiv \dfrac{D_{it} + P_{it} - P_{it-1}}{P_{it-1}} = \dfrac{D_{it} + P_{it}}{P_{it-1}} - 1$	Return, dividends included	Dividend plus price change, over the previous period's price. The price change alone understates the return by the dividend yield $D_{it}/P_{it-1}$ .
$\text{E}[R_{it}]$ $R_{it} - r_f$ $\text{E}[R_{it}] - r_f$	Expected return Excess return Risk premium	The risk premium is an average; each period's excess return is one realization of it and can land far above or below it. $r_f$ is the Treasury-bill rate.
$\mu_i \equiv \text{E}[R_{it}]$ $\sigma_i^2 \equiv \text{E}[(R_{it} - \mu_i)^2]$ , $\ \sigma_i = \sqrt{\sigma_i^2}$	Population mean, variance, standard deviation	True values nobody observes. $\sigma_i$ is in the return's units (% per year); $\sigma_i^2$ is in $(\%)^2$ per year.
$\hat{\mu}_i = \dfrac{1}{T}\displaystyle\sum_{t=1}^{T} R_{it}$ $\hat{\sigma}_i^2 = \dfrac{1}{T-1}\displaystyle\sum_{t=1}^{T} (R_{it} - \hat{\mu}_i)^2$	Sample estimators	Computed from $T$ historical returns. The mean divides by $T$ , the variance by $T - 1$ , and $\hat{\sigma}_i = \sqrt{\hat{\sigma}_i^2}$ .

Key Insight: Standard deviation measures fluctuation in either direction: a return $10$ points above the mean counts exactly as much as one $10$ points below it. From 1900 to 2025, US equities earned about $9.8\%$ a year compounded against about $3.4\%$ for T-bills, a risk premium on the order of $6$ points a year.

IIReturn with the Dividend Included

Each year's return stacks a price change and a dividend yield, both measured against the previous year's price.

IIISample Mean and Sample Standard Deviation

Six annual returns give an average, and the typical distance of a return from that average is $\hat{\sigma}$ .

IVWhy the Sample Variance Divides by $T - 1$

Among all possible centres $c$ , the sample mean makes the sum of squared deviations smallest.

💡 Because the sample mean minimises the sum, $\sum_t (R_t - \hat{\mu})^2$ is never larger than the sum around the unknown true mean, $\sum_t (R_t - \mu)^2$ , and it is strictly smaller whenever $\hat{\mu} \ne \mu$ , which is almost always. Squared deviations measured from $\hat{\mu}$ therefore understate the spread. Dividing by $T - 1$ instead of $T$ makes up for that shortfall on average, which is what makes $\hat{\sigma}^2$ unbiased.

VQuiz Questions

Problem 1 · A Year's Return with a Dividend

Given: You buy a share of Kestrel Mining (illustrative) at $50. During the year it pays a dividend of $1.50, and the year ends with the share at $46. Find the year's return $R_t$ .

✅ Correct! The $1.50 dividend recovers 3 of the 8 points the price lost, so the year's return is

-5\%

❌ That is the price change alone. The return includes the dividend paid during the year: add

D_t

to the numerator.

❌ Close, but check the denominator. The return is measured against the price you paid,

P_{t-1}

, not the year-end price

P_t

❌ That is the dividend yield alone. The price change of

-8\%

belongs in the return too.

❌ Not quite. Use

R_t = (D_t + P_t - P_{t-1})/P_{t-1}

Show solution

With $P_{t-1} = 50$ , $D_t = 1.50$ and $P_t = 46$ (in dollars):

R_t = \frac{D_t + P_t - P_{t-1}}{P_{t-1}} = \frac{1.50 + 46 - 50}{50} = \frac{-2.50}{50} = -0.05 = -5.0\%

Split into its two parts: the price change is $-4/50 = -8.0\%$ and the dividend yield is $1.50/50 = 3.0\%$ , and $-8.0\% + 3.0\% = -5.0\%$ .

Problem 2 · Premium or Realization?

Given: Kestrel Mining's annual returns over the last four years were $18\%$ , $-6\%$ , $23\%$ and $1\%$ (illustrative), and the Treasury-bill rate was $r_f = 2.5\%$ throughout. Find the estimate of Kestrel's risk premium $\text{E}[R] - r_f$ from this sample.

✅ Correct! The sample mean

\hat{\mu} = 9\%

estimates

\text{E}[R]

; subtracting

r_f = 2.5\%

leaves an estimated premium of

6.5

points.

❌ That is the estimated expected return. The risk premium is what the stock earns on average above the risk-free rate: subtract

r_f

❌ That is last year's excess return, $1\% - 2.5\%$ . A single year's excess return is one realization; the premium is its average.

❌ That is the first year's excess return, $18\% - 2.5\%$ . One year's excess return can land far from the premium; average all four.

❌ Not quite. Estimate

\text{E}[R]

by the sample mean, then subtract

r_f

Show solution

Estimate the expected return by the sample mean:

\hat{\mu} = \frac{18 + (-6) + 23 + 1}{4} = \frac{36}{4} = 9\%

\widehat{\text{E}[R] - r_f} = \hat{\mu} - r_f = 9\% - 2.5\% = 6.5 \text{ percentage points}

The four yearly excess returns are $15.5$ , $-8.5$ , $20.5$ and $-1.5$ points: individual realizations scattered above and below the premium, and their average is again $6.5$ .

Problem 3 · Six Years to a Standard Deviation

Given: Kestrel Mining's annual returns over six years (illustrative) were $21\%$ , $-13\%$ , $26\%$ , $7\%$ , $-4\%$ and $23\%$ . Find the sample standard deviation $\hat{\sigma}$ .

✅ Correct!

\hat{\sigma}^2 = 1280/5 = 256\ (\%)^2

, and its square root is

16\%

a year.

❌ Close, but check the divisor. Dividing the sum of squared deviations by

T = 6

gives about

213.3

; the sample variance divides by

T - 1 = 5

❌ That averages the distances without squaring them. The variance averages squared deviations (over

T - 1

), and

\hat{\sigma}

is its square root.

❌ That is the variance, in $(\%)^2$ . Take the square root to get back to percent per year.

❌ Not quite. Find

\hat{\mu}

, square each deviation, divide the sum by

T - 1

, then take the root.

Show solution

\hat{\mu} = \frac{21 - 13 + 26 + 7 - 4 + 23}{6} = \frac{60}{6} = 10\%

Deviations from $\hat{\mu}$ : $11, -23, 16, -3, -14, 13$ (they sum to $0$ ). Their squares: $121, 529, 256, 9, 196, 169$ .

\sum_t (R_t - \hat{\mu})^2 = 1280, \qquad \hat{\sigma}^2 = \frac{1280}{6 - 1} = 256\ (\%)^2, \qquad \hat{\sigma} = \sqrt{256} = 16\%

Dividing by $T = 6$ would give about $213.3\ (\%)^2$ and about $14.6\%$ , understating the risk.

Problem 4 · A Big Gain or a Big Loss

Given: Two illustrative four-year stretches of Kestrel Mining returns. Stretch A: $2\%$ , $2\%$ , $2\%$ , $26\%$ . Stretch B: $14\%$ , $14\%$ , $14\%$ , $-10\%$ . Which stretch has the larger sample standard deviation?

✅ Correct! Both have mean

8\%

, and B's deviations are A's with the signs flipped. Squaring erases the sign, so the two are equally risky by this measure.

❌ Not quite. Standard deviation counts a return far below the mean exactly as much as one equally far above it; it does not single out losses.

❌ Deviations are measured from the mean, not from zero. Both stretches have mean

8\%

: A's big year is

18

points above it, B's is

18

points below it.

❌ Not quite. The standard deviation uses only the returns and their own mean;

r_f

matters for excess returns and the premium, not for

\hat{\sigma}

❌ Not quite. Compute each stretch's mean and deviations.

Show solution

Stretch A: $\hat{\mu} = 32/4 = 8\%$ ; deviations $-6, -6, -6, +18$ .

Stretch B: $\hat{\mu} = 32/4 = 8\%$ ; deviations $+6, +6, +6, -18$ .

\sum_t (R_t - \hat{\mu})^2 = 36 + 36 + 36 + 324 = 432 \text{ for both}, \qquad \hat{\sigma}^2 = \frac{432}{4 - 1} = 144\ (\%)^2, \qquad \hat{\sigma} = 12\%

A's surprise is a gain and B's a loss, but squaring treats them identically. Telling them apart takes a different statistic, such as skewness (A is positively skewed, B negatively).

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Why a Ten Percent Loss Needs a Yardstick: Returns, the Risk Premium, and Standard Deviation

IKey Formulas

IIReturn with the Dividend Included

IIISample Mean and Sample Standard Deviation

IVWhy the Sample Variance Divides by T−1T - 1T−1

VQuiz Questions

IVWhy the Sample Variance Divides by $T - 1$