Part I · Returns and risk · Chapter 1A

What a return is

Simple returns, log returns and compounding

Part I builds the four ideas that every later chapter stands on: what a return is, what shape returns take, what an edge is, and how to weigh return against risk. They are not optional preliminaries but the load-bearing assumptions under every claim that follows. Skip them, and every later chapter misleads you in subtle and expensive ways.

This chapter takes the first idea, defining the simple return and the log return and showing why professionals calculate with the second. Along the way it teaches the mathematics the rest of the book needs: logarithms, the number e, compounding, and the cost that volatility imposes on growth.

In plain terms

A simple return, such as "the price rose 20%", is the natural way to talk about a move, but it behaves awkwardly. A rise of 20% followed by a fall of 16.67% brings you back exactly where you started, so the same round trip gets two different sizes. A log return measures moves so that a round trip is exactly $+x$ followed by $-x$ and the returns of successive days add up to the return of the whole period. Most of the mathematics in this book is done in log returns for those two reasons. The convenience costs one new tool, the logarithm, and brings one surprise: volatility lowers the rate at which money compounds even when the average return stays the same.

1 · The simple return

Write $S_t$ for the price of an asset at the end of period $t$ and $S_{t-1}$ for its price one period earlier. The period can be a day, a week or a year: the subscript simply counts periods. The simple return over the period is the change in price divided by the starting price:

\begin{aligned} R_t &= \frac{S_t - S_{t-1}}{S_{t-1}} \\ &= \frac{S_t}{S_{t-1}} - 1 \end{aligned}

(1A.1)

A stock that goes from $100 to $120 has $R = 120/100 - 1 = 0.20$ , a simple return of 20%.

The ratio $S_t/S_{t-1}$ , which equals $1 + R_t$ , is the growth factor: what each dollar at the start of the period became by its end. A 20% gain is a growth factor of 1.20, and a 10% loss is one of 0.90.

Compounding: returns multiply

Hold a position for two periods and each dollar grows twice, first by $1 + R_1$ and then by $1 + R_2$ . This multiplying of growth factors is compounding. Over $n$ periods, the total simple return comes from the product of all $n$ growth factors:

\begin{aligned} 1 + R_{\text{total}} &= \prod_{t=1}^{n} (1 + R_t) \\ &= (1+R_1)\cdots(1+R_n) \end{aligned}

(1A.2)

The symbol $\prod$ means "multiply together", in the way that $\sum$ means "add together".

Worked example: Up 10%, then down 10%

Step 1: write each growth factor. A 10% gain is a growth factor of 1.10 and a 10% loss a growth factor of 0.90.

Step 2: multiply them. Together the two periods give a growth factor of $1.10 \times 0.90 = 0.99$ .

Step 3: subtract 1. The total simple return is $0.99 - 1 = -0.01$ , a loss of 1%.

The two simple returns, $+10\%$ and $-10\%$ , add to zero, yet the account ended the two periods 1% down. Simple returns do not add over time: they compound.

A round trip that is not symmetric

Suppose a stock at $100 rises to $120 and then falls back to $100. Measured as simple returns, the two legs of this round trip have different sizes:

\begin{aligned} \text{up:} \quad & \tfrac{120 - 100}{100} = +20.0\% \\ \text{down:} \quad & \tfrac{100 - 120}{120} = -16.67\% \end{aligned}

(1A.3)

The asymmetry is real, not a mistake: a 20% gain followed by a 16.67% loss does return to the starting price, because the loss is measured from the higher base of $120. Real or not, the asymmetry makes simple returns awkward to reason with.

2 · Logarithms and the number e

The fix for both problems, the multiplying and the asymmetry, is the logarithm, of which this section teaches the little you need.

Where e comes from

Suppose a bank pays 100% interest a year on $1. Paid once, at the end of the year, the interest turns the dollar into $2. Paid twice, at 50% each half-year, it makes $1.5 \times 1.5 = \$2.25$ , because the second payment earns interest on the first. The more often the interest is paid, the more the dollar grows:

interest paid	growth of $1 in a year
once, $n = 1$	$(1 + 1)^1 = 2$
twice, $n = 2$	$(1 + 1/2)^2 = 2.25$
monthly, $n = 12$	$(1 + 1/12)^{12} = 2.6130$
daily, $n = 365$	$(1 + 1/365)^{365} = 2.7146$

The growth does not run away, though: as the payments become more frequent, the result settles on a fixed number, 2.71828 to five decimal places, which is called e. Interest paid in infinitely small, infinitely frequent amounts is continuous compounding. At a rate $r$ a year for $t$ years, one dollar compounded continuously grows to

e^{rt}

(1A.4)

At 5% for 10 years, one dollar grows to $e^{0.05 \times 10} = e^{0.5} = 1.6487$ , or about $1.65.

The natural logarithm

The natural logarithm undoes $e$ : the natural logarithm of $x$ , written $\ln x$ , is the power to which you must raise $e$ to get $x$ .

y = e^{x} \quad\Longleftrightarrow\quad x = \ln y

(1A.5)

So $\ln 1 = 0$ , because $e^0 = 1$ , and $\ln e = 1$ . The logarithm of a number below 1 is negative, as in $\ln 0.5 = -0.6931$ . No power of $e$ is zero or negative, so the logarithm of zero or of a negative number does not exist.

Three rules follow from the rule for powers, $e^a \times e^b = e^{a+b}$ , and this book uses all three:

\begin{aligned} \ln(ab) &= \ln a + \ln b \\ \ln(a/b) &= \ln a - \ln b \\ \ln(a^k) &= k \ln a \end{aligned}

(1A.6)

The first rule is the one that matters most: a logarithm turns multiplication into addition. Compounding multiplies growth factors, so the logarithm of a compounded total is a sum of logarithms.

Near 1, the logarithm is almost a straight line

For a small number $x$ , the value of $\ln(1 + x)$ is very close to $x$ itself:

$x$	$\ln(1 + x)$	gap
0.01	0.00995	0.00005
0.05	0.04879	0.00121
0.20	0.18232	0.01768
0.50	0.40547	0.09453

Because the gap grows roughly with the square of $x$ , a sharper approximation subtracts a squared term. It is accurate enough for everything in this chapter:

\ln(1 + x) \approx x - \frac{x^2}{2}

(1A.7)

For $x = 0.05$ it gives $0.05 - 0.00125 = 0.04875$ , against the true value of 0.04879. It is the start of what mathematicians call the Taylor series of $\ln(1 + x)$ , a sum of powers of $x$ that approximates the logarithm ever more closely as terms are added.

Why ln(1 + x) is close to x − x²/2

Step 1: expand $e^x$ for a small $x$ . Replace the 100% rate of the compounding table by a small rate $x$ , so that $e^x$ is the limit of $(1 + x/n)^n$ as $n$ grows. Multiplying out the first few terms of that power, as in $(1 + a)^2 = 1 + 2a + a^2$ , and letting $n$ grow gives the series

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

(1A.8)

For a small $x$ , each term is much smaller than the one before it.

Step 2: guess the logarithm. Try $y = x - x^2/2$ and compute $e^y$ from Step 1, keeping the terms up to $x^2$ :

\begin{aligned} e^y &\approx 1 + y + \tfrac{y^2}{2} \\ &= 1 + x - \tfrac{x^2}{2} + \tfrac{x^2}{2} + \cdots \\ &= 1 + x + \cdots \end{aligned}

(1A.9)

The terms left out all contain $x^3$ or a higher power of $x$ .

Step 3: read off the answer. Since $e^y \approx 1 + x$ , the definition of the logarithm gives $y \approx \ln(1 + x)$ . The error is of the order of $x^3$ , which for $x = 0.05$ is about 0.0001.

3 · The log return

The log return over one period is the natural logarithm of the growth factor:

\begin{aligned} r_t &= \ln\!\left(\frac{S_t}{S_{t-1}}\right) \\ &= \ln S_t - \ln S_{t-1} \end{aligned}

(1A.10)

The second line uses the rule for the logarithm of a ratio. Since the growth factor $S_t/S_{t-1}$ is also $1 + R_t$ , either kind of return converts into the other:

\begin{aligned} r_t &= \ln(1 + R_t) \\ R_t &= e^{r_t} - 1 \end{aligned}

(1A.11)

For small moves the two are nearly equal, because $\ln(1 + x)$ is close to $x$ near zero. A 1% daily rise is a simple return of 0.0100 and a log return of 0.00995. For large moves they part company: a 20% rise is a log return of only 0.1823.

Property 1: log returns are symmetric

Measure the round trip from $100 to $120 and back again, this time in log returns:

\begin{aligned} \text{up:} \quad & \ln(120/100) = +0.1823 \\ \text{down:} \quad & \ln(100/120) = -0.1823 \end{aligned}

(1A.12)

The two legs have the same size and opposite signs, because going back exactly undoes going forward. This symmetry matters when you build intuition about volatility, which is, after all, a measure of the size of moves irrespective of their direction.

The figure below measures a round trip both ways, with a slider that sets the size of the up move from 1% to 50%, starting at 20%. The top pair of bars shows the two legs as simple returns and the bottom pair shows them as log returns, both written as percentages. Drag the slider down to 1%, then up to 50%.

up move 20.0%

path = $100 → $120.00 → $100

simple returns = +20.00% · −16.67%

log returns = +18.23% · −18.23%

The same round trip, measured two ways. A price rises from $100 by the up move on the slider and falls back to $100. The bars are the size of each leg: as simple returns the fall is smaller than the rise, as log returns the two are equal and opposite. The larger the move, the wider the gap between the simple legs.

At 1% the four bars are almost the same length: simple returns of +1.00% and −0.99%, and log returns of ±1.00%. At 50% the simple legs are +50% and −33.33%, while the log legs are still equal and opposite at ±40.55%. The larger the move, the wider the gap between the two simple legs.

Property 2: log returns add over time

Where the total simple return over $n$ periods is the product in (1A.2), the total log return is a sum. To see it, write out each period's log return with (1A.10) and add them:

\begin{aligned} \sum_{t=1}^{n} r_t &= (\ln S_1 - \ln S_0) \\ &\quad + (\ln S_2 - \ln S_1) \\ &\quad + \cdots \\ &\quad + (\ln S_n - \ln S_{n-1}) \\ &= \ln S_n - \ln S_0 \end{aligned}

(1A.13)

Every intermediate price appears once with a plus sign and once with a minus sign and cancels, leaving the log return of the whole period:

r_{\text{total}} = \ln\!\left(\frac{S_n}{S_0}\right) = \sum_{t=1}^{n} r_t

(1A.14)

This is the additivity property: the total log return is the sum of the period-by-period log returns. The proof takes one line because a telescoping chain of price ratios collapses to a single ratio, whose logarithm is one logarithm. It is the single most useful property in quantitative finance.

Worked example: Up 10%, then down 10%, in log returns

Step 1: convert each simple return. The two periods have log returns of $\ln 1.10 = 0.0953$ and $\ln 0.90 = -0.1054$ .

Step 2: add them. The total log return is $0.0953 - 0.1054 = -0.0101$ .

Step 3: convert back. The total simple return is $e^{-0.0101} - 1 = -0.0100$ , a loss of 1%.

This matches the compounded answer from the earlier worked example, found here by adding instead of multiplying. The log returns also show at a glance why the account lost: in log terms, the fall of 10% is a bigger move than the rise of 10%.

4 · Why additivity matters

Additivity is more than a convenience of notation, because it makes four practical calculations clean.

Volatility grows with the square root of time

From introductory statistics you know that the variance of a sum of independent random quantities is the sum of their variances. A multi-day log return is a sum of daily log returns, so if the days are independent of each other, the variance of the total is

\sigma^2_{\text{total}} = \sum_{t=1}^{n} \sigma^2_t

(1A.15)

where $\sigma_t$ is the standard deviation of the log return on day $t$ . If every day has the same variance $\sigma^2_{\text{daily}}$ , the $n$ -day variance is $n\,\sigma^2_{\text{daily}}$ , whose square root is the $n$ -day standard deviation:

\sigma_{n\text{-day}} = \sigma_{\text{daily}} \sqrt{n}

(1A.16)

This is the famous square-root-of-time rule: risk grows with the square root of the horizon, not in proportion to it. Four days carry twice the standard deviation of one day, not four times. The rule works cleanly only for log returns, and only under independence.

The mean behaves differently, because the mean of a sum is the sum of the means whether or not the days are independent. The $n$ -day mean is simply $n$ times the daily mean. Means grow with $n$ while standard deviations grow with $\sqrt{n}$ , a difference that Chapter 1D uses to explain why a strategy's Sharpe ratio rises with how often it trades.

Annualizing is one multiplication

A year has about 252 trading days, so (1A.16) converts the standard deviation of daily log returns into a yearly one:

\sigma_{\text{annual}} = \sigma_{\text{daily}} \times \sqrt{252}

(1A.17)

Since $\sqrt{252} = 15.87$ , you multiply daily figures by about 15.9, weekly figures by $\sqrt{52} = 7.21$ and monthly ones by $\sqrt{12} = 3.46$ . Under the independence assumption these are exact, not approximations. The yearly standard deviation of log returns is what markets call volatility, written $\sigma$ without a subscript.

Worked example: From a daily move to annual volatility

Step 1: take the daily standard deviation. Suppose an index's daily log returns have a standard deviation of 1.26%.

Step 2: multiply by the square root of 252. The annual figure is $0.0126 \times 15.87 = 0.200$ .

Step 3: read the result. The annual volatility is 20%. Turned around, a volatility of 20% means a typical daily move of about $20\% / 15.87 = 1.26\%$ .

Stress tests become a sum

Suppose you want the cumulative loss over a five-day stress event. In log returns you add the five daily numbers, while with simple returns you must multiply five growth factors. The tempting shortcut of adding the five simple returns goes wrong, most of all when moves are large, which is exactly when you most want the answer to be accurate.

Take five days of −4% each: adding the simple returns says −20%, but the truth is $0.96^5 - 1 = -18.46\%$ . In log returns the five days sum to $5 \times \ln 0.96 = -0.2041$ , and converting back gives $e^{-0.2041} - 1 = -18.46\%$ directly.

The normal curve fits log returns

The classical Black-Scholes model, which Part VII derives, assumes that prices follow geometric Brownian motion, which in continuous time means that log returns are normally distributed. A sum of independent normal quantities is itself normal, so the log return over $n$ periods is also normal, with no correction needed. Its mean is $n$ times one period's mean and its variance $n$ times one period's variance.

Simple returns cannot do this, because a normal distribution gives some probability to every value, including a simple return below −100%, which would mean a negative price. A log return can be any number at all, and because $e^{r}$ is positive for every $r$ , a normal log return always gives a positive price. A product of normal growth factors is not normal either, so simple returns lose the clean rule for adding periods.

5 · Variance drag

Part VII writes geometric Brownian motion in the notation of calculus as

\frac{dS}{S} = \mu\, dt + \sigma\, dW

(1A.18)

Read it as a statement about a very short time $dt$ : the simple return over that instant, $dS/S$ , has a drift $\mu\, dt$ and a random part $\sigma\, dW$ whose variance is $\sigma^2 dt$ . Here $\mu$ is the arithmetic mean return per year and $\sigma$ is the volatility.

The log return over the same instant is also normal, but its mean is $(\mu - \tfrac{1}{2}\sigma^2)\,dt$ , not $\mu\, dt$ . The term $-\tfrac{1}{2}\sigma^2$ is called the Itô drift correction, or variance drag. Over a year it sets the growth rate of the log price at

G = \mu - \frac{\sigma^2}{2}

(1A.19)

where $G$ is the continuously compounded growth rate, the mean log return per year. Part VII returns to this correction when it derives the Black-Scholes equation.

Why the logarithm takes something away

The correction arises because the logarithm curve bends downward: it is a concave function. When you take the logarithm of an uncertain quantity, you do not simply get the logarithm of its mean, because you also pick up a negative term that grows with the variance.

To see it in a two-outcome example, suppose each year's growth factor is 1.20 or 0.80 with equal probability, so that the mean growth factor is 1.00 and the mean simple return is zero. The logarithms tell a different story: $\ln 1.20 = 0.1823$ and $\ln 0.80 = -0.2231$ average to $-0.0204$ , so a typical year loses about 2% in log terms. Compounding one good year and one bad year confirms it, since $1.20 \times 0.80 = 0.96$ is a loss of 4% over two years, or about 2% a year.

Worked example: A 10% mean at 20% volatility

Step 1: write the inputs as decimals. Here $\mu = 0.10$ and $\sigma = 0.20$ , so $\sigma^2 = 0.04$ .

Step 2: apply (1A.19). The growth rate of the log price is

\begin{aligned} G &= \mu - \tfrac{1}{2}\sigma^2 \\ &= 0.10 - 0.5 \times 0.04 \\ &= 0.08 \end{aligned}

(1A.20)

Step 3: read the result. A stock with a 10% expected simple return per year and 20% volatility has an expected log return of only 8% a year. The 2% gap between the expected simple return and the expected log return is variance drag.

Where −σ²/2 comes from

Step 1: cut the year into short steps. Take steps of length $\Delta t$ , a small fraction of a year, so that over one step the simple return $R$ has mean $\mu \Delta t$ and variance $\sigma^2 \Delta t$ .

Step 2: approximate the step's log return. By (1A.7), $\ln(1 + R) \approx R - R^2/2$ for a small step.

Step 3: take the mean. For the mean of $R^2$ , use the fact that the mean of the square of any random quantity is its variance plus the square of its mean:

\begin{aligned} \mathbb{E}[R^2] &= \operatorname{Var}(R) + (\mathbb{E}[R])^2 \\ &= \sigma^2 \Delta t + \mu^2 \Delta t^2 \end{aligned}

(1A.21)

Here $\mathbb{E}[\,\cdot\,]$ is the mean over all outcomes, weighted by their probabilities, and $\operatorname{Var}(\cdot)$ is the variance, both developed in Chapter 1B. Putting the pieces together gives the step's mean log return:

\begin{aligned} \mathbb{E}[\ln(1+R)] &\approx \mu\Delta t - \tfrac{1}{2}\sigma^2\Delta t \\ &\quad - \tfrac{1}{2}\mu^2\Delta t^2 \end{aligned}

(1A.22)

Step 4: drop the tiny term. The last term carries $\Delta t^2$ , which for a daily step is $1/252^2 = 1/63{,}504$ , so the term is negligible. The terms that the approximation in Step 2 left out also shrink faster than $\Delta t$ , and they vanish over many small steps.

Step 5: add up the year. Log returns add, so the year's mean log return is the sum over its $1/\Delta t$ steps:

\begin{aligned} G &= \frac{1}{\Delta t}\left(\mu - \frac{\sigma^2}{2}\right)\Delta t \\ &= \mu - \frac{\sigma^2}{2} \end{aligned}

(1A.23)

In the continuous-time model of Part VII the result is exact.

What drag does to an account

The practical consequence is large: with everything else equal, a lower-volatility strategy compounds to more terminal wealth than a higher-volatility strategy with the same arithmetic mean return. The arithmetic mean is the average of the yearly returns, while the compound growth rate, which is what your account actually earns over many years, is that mean less half the variance. This is also the mathematical engine behind the Kelly criterion, which Chapter 1C introduces and Chapter 3D develops.

The drag also explains a sequence that looks harmless. Suppose returns alternate between +50% and −50%, so that the arithmetic mean is zero. Each pair of periods still multiplies wealth by $1.5 \times 0.5 = 0.75$ . The geometric mean return, which is what compounding actually delivers, is therefore

\sqrt{1.5 \times 0.5} - 1 = -13.4\%

(1A.24)

That is a loss of 13.4% per period, or 25% per pair of periods. After 10 pairs, $100 has shrunk to $100 \times 0.75^{10} = \$5.63$ : the sequence grinds toward zero although its average return is zero. Formula (1A.19) predicts a log growth of $0 - 0.5^2/2 = -0.125$ per period, against an exact $\ln\sqrt{0.75} = -0.144$ . For moves this large the approximation is loose, erring on the kind side.

The figure below simulates wealth paths, with four sliders: the arithmetic mean $\mu$ (starting at 10%), the volatility $\sigma$ (30%), the horizon (10 years) and a seed that re-draws the random shocks (42). Six paths are drawn, each twice: once compounding simple returns and once compounding log returns from the same shocks, which is why every pair of lines nearly coincides. The dashed curve, $e^{\mu t}$ , is what the arithmetic mean promises after $t$ years. The solid curve is $e^{Gt}$ , growth at the rate $G$ of (1A.19).

arithmetic mean μ 10%

volatility σ 30%

horizon (years) 10y

seed 42

variance drag σ²/2 = 4.50%

geometric G = 5.50%

gap A − G = 4.50pp

10-yr drag (compounded) = 56.8%

A strategy with arithmetic mean 10% and volatility 30% compounds at only 5.50% a year.

What compounding takes from the arithmetic mean. Six simulated wealth paths with arithmetic mean μ and volatility σ, each compounded twice from the same daily shocks, once as simple returns and once as log returns, which is why every pair of lines nearly coincides. The paths gather around e^(Gt), where G = μ − σ²/2, rather than the e^(μt) the arithmetic mean promises, and the gap widens with volatility. Holding μ fixed, doubling volatility from 20% to 40% raises the drag from 2% to 8% a year.

At the starting settings the readouts show a variance drag of 4.50%, so a 10% arithmetic mean compounds at only 5.50% a year. The paths gather around the solid curve rather than the dashed one. The last readout compounds the gap between them: after 10 years the arithmetic promise stands 56.8% above the typical path. Now hold $\mu$ at 10% and move $\sigma$ from 20% to 40%: the drag rises from 2% to 8% a year, because doubling volatility quadruples the drag. Push $\sigma$ to 45% and $G$ turns negative, so that a strategy with a 10% average return loses money over time on a typical path.

One subtlety explains the dashed curve. Averaged across very many paths, final wealth still comes to $e^{\mu t}$ , because a few very lucky paths pull the average up. The typical path, the median, grows only at $G$ . You hold one account and live one path, so $G$ is the number that counts for you.

6 · The convention in this book

For the rest of this book, "return" means log return unless the text says otherwise. Volatility, the Sharpe ratio, variance and drift all assume log returns under the hood. When you see $\sigma_{\text{daily}} \times \sqrt{252}$ used to annualize, that is a log-return convention. Almost every formula in derivatives pricing is more natural in log space.

The one place where simple returns persist is in reporting to people who do not work in log returns. A client wants to hear "the strategy returned 12% this year", not "the log return was 0.1133". That is fine, provided you translate at the boundary and not in the middle of your math.

In practice

Your statements speak in simple returns. A brokerage statement or a fund's report quotes simple returns, because dollars are simple. Do your own arithmetic in log returns, whether adding days, scaling volatility or summing a stress path, then convert at the end with $R = e^{r} - 1$ .
Discount any average return by half its variance. A reported arithmetic average of yearly returns overstates what an account compounded, so a record that averages 10% a year with 20% volatility compounds at about 8%.
Doubling time comes from the growth rate $G$ . Money growing at a continuous rate $G$ doubles when $e^{Gt} = 2$ , that is, after $t = \ln 2 / G = 0.693/G$ years. At $G = 8\%$ that is about 8.7 years, and at the 5.50% of the figure's starting settings it is about 12.6 years.
Volatility costs money even with no fee attached. Adding volatility to a portfolio without adding mean return lowers what the portfolio compounds to. The cost never appears on a statement as a line item, only as a lower balance.

When it fails

The square-root-of-time rule assumes that one period's return is independent of the last, with the same variance in every period. When returns carry over from one day to the next, a pattern that Chapter 3B models, $\sigma\sqrt{n}$ misstates multi-day risk. It also misstates it when volatility itself changes, as it does under the dealer mechanics of Part VIII.

The drag formula (1A.19) rests on the same model and on small moves. For large moves the further terms of the logarithm's expansion come into play, and they make the drag larger when returns have a long left tail or unusually many extreme values. Those are the shapes that Chapter 1B describes. The +50%/−50% sequence shows the direction of the error: the formula predicted a log growth of −12.5% a period, and the truth was −14.4%.

Decision rule

Compare strategies by $G$ , not by $\mu$ . Compute $\mu - \sigma^2/2$ for each one and rank them on that.
Accept extra volatility only for enough extra mean. Moving from a volatility $\sigma_A$ to a higher $\sigma_B$ costs $(\sigma_B^2 - \sigma_A^2)/2$ a year of compound growth. Going from 15% to 30% volatility costs $(0.09 - 0.0225)/2 = 0.034$ , about 3.4 percentage points a year, so take the more volatile strategy only if its arithmetic mean is higher by more than that.
Compute in log returns and report in simple returns. Convert once, at the end, and never in the middle of a calculation.

Common mistake: judging growth by the average return

An average of simple returns says nothing reliable about what an account compounds to. The +50%/−50% sequence averages zero and loses 25% of the account every two periods. Always ask for the compound (geometric) return, or compute it with (1A.19).

Common mistake: adding log returns across positions

Log returns add over time, not across positions. Dollars add across the positions in a portfolio, so a portfolio's simple return is the weighted average of its positions' simple returns, $R_p = \sum_i w_i R_i$ , where $w_i$ is the share of capital in position $i$ . Put $50 in a stock that rises 50% and $50 in one that falls 50%, and you end with $\$75 + \$25 = \$100$ , a portfolio return of zero. Averaging the log returns instead, $(0.405 - 0.693)/2 = -0.144$ , wrongly reports a loss of 13.4%.

Familiar ground

You already read effects on a log scale. A hazard ratio of 2.0 and one of 0.5 are equal and opposite effects, which is why a forest plot draws its axis on a log scale. On that scale the two sit the same distance either side of 1, because $\ln 2 = +0.693$ and $\ln 0.5 = -0.693$ . A meta-analysis pools log hazard ratios rather than the ratios themselves, for the same reason this book uses log returns. On the ratio scale, effects combine by multiplying. On the log scale they add, which is what the machinery of variances and normal curves is built for.

What to carry forward

A simple return, $R = S_t/S_{t-1} - 1$ , is how returns are reported, and over time simple returns compound by multiplying growth factors.
A log return, $r = \ln(S_t/S_{t-1})$ , is symmetric (a round trip is $+x$ then $-x$ ) and additive over time (the total is the sum of the parts).
Additivity gives the square-root-of-time rule for independent returns, so annual volatility is daily volatility times $\sqrt{252}$ .
Volatility drags on compounding: the growth rate of wealth is $G = \mu - \sigma^2/2$ , not the arithmetic mean $\mu$ .
In this book "return" means log return unless the text says otherwise, and simple returns appear only at the boundary.

7 · Check yourself

Question 1

A share goes from $50 to $60 and back to $50. What are the simple return and the log return of each leg?

Answer

The simple returns are $60/50 - 1 = +20\%$ up and $50/60 - 1 = -16.67\%$ down. The log returns are $\ln(60/50) = +0.1823$ up and $\ln(50/60) = -0.1823$ down, so only the log returns are equal and opposite. The numbers match the $100 example because only the ratio of the two prices matters.

Question 2

Where does an index that started at $200 end, after five days with log returns of +0.010, −0.020, +0.015, −0.005 and +0.010?

Answer

Log returns add, so the total is $0.010 - 0.020 + 0.015 - 0.005 + 0.010 = 0.010$ . The final price is $200 \times e^{0.010} = 200 \times 1.01005 = \$202.01$ , a simple return over the five days of $e^{0.010} - 1 = 1.005\%$ .

Question 3

A stock's daily log returns have a standard deviation of 0.8%. What are its weekly (five-day) and annual volatility, assuming independent days?

Answer

The weekly figure is $0.008 \times \sqrt{5} = 0.008 \times 2.236 = 1.79\%$ , and the annual figure is $0.008 \times \sqrt{252} = 0.008 \times 15.87 = 12.7\%$ .

Question 4

Which compounds faster: strategy A, with an arithmetic mean return of 12% and volatility of 25%, or strategy B, with a mean of 10% and volatility of 15%?

Answer

Neither does: for A, $G = 0.12 - 0.25^2/2 = 0.12 - 0.03125 = 0.08875$ , and for B, $G = 0.10 - 0.15^2/2 = 0.10 - 0.01125 = 0.08875$ . Both compound at about 8.9% a year: A's extra 2 points of mean are used up exactly by its extra drag, so B earns the same growth with less volatility.

Question 5

You put half your money in a fund that returns +20% and half in one that returns −10%. What does the portfolio return, and what goes wrong if you average the log returns?

Answer

Dollars add across positions, so use simple returns: $0.5 \times 0.20 + 0.5 \times (-0.10) = +5\%$ . Averaging the logs gives $(\ln 1.20 + \ln 0.90)/2 = (0.1823 - 0.1054)/2 = 0.0385$ , which converts to $e^{0.0385} - 1 = 3.9\%$ . That answer is wrong, because log returns add over time, not across positions.