Part I · Returns and risk · Chapter 1B

The shape of returns

Distributions, skew and fat tails

Where we are

Chapter 1A defined a return and showed how volatility, the standard deviation of returns, grows with time and drags on compounding. This chapter asks what else a set of returns has besides a mean and a standard deviation. The answer, skew and fat tails, explains how a strategy that sells options can look safe for years and then give back a year of gains in a single bad month.

What you need first

Log returns, volatility and the square-root-of-time rule (Chapter 1A).
From introductory statistics: the mean, the standard deviation and the normal curve.

In plain terms

Two strategies can have the same average return and the same standard deviation and still be completely different bets. One wins and loses in moderate amounts, spread evenly around its average. The other wins a little almost every month and, now and then, loses a lot. The mean and the standard deviation cannot tell these two apart. Two further numbers can: the skew, which measures lopsidedness, and the kurtosis, which measures how much weight sits in the extreme tails. Selling options produces the second, lopsided shape by construction, and that shape is where option sellers lose serious money.

1 · What a return distribution is

A return distribution is the set of possible outcomes of a strategy, each weighted by its probability. Every strategy you trade has one, whether or not you describe it that way.

Start with a toy: suppose you and a friend bet on a coin, where heads means you pay your friend $1 and tails means your friend pays you $1. Your return distribution has two outcomes, +$1 with probability 50% and −$1 with probability 50%. It is boring but instructive, because every distribution in this book is built the same way: a list of outcomes and a probability for each, with the probabilities adding to 1.

Now consider selling a 30-delta put on SPX for $3 of premium. A put is a contract that pays its holder if the index ends below a set price, the strike, at expiry. The seller collects the premium up front and owes the payout. Chapter 5C teaches delta, but for now read a 30-delta put as one whose strike sits far enough below the index that it expires worthless most of the time. The distribution of a short 30-delta put looks roughly like this:

outcome	probability	result
"win": the option expires worthless	roughly 70%	+$3
small loss: a modest pullback	roughly 20%	−$2 to −$10
medium loss: a correction	roughly 8%	−$10 to −$30
tail loss: a crash	roughly 2%	−$30 to −$100 or more

The probabilities add to 100%, as they must, but the key point is the shape: most of the time you win a little, and occasionally you lose a lot. This is what a negatively skewed distribution looks like, and it characterizes almost every short-volatility strategy: any strategy that earns money when markets stay calm, usually by selling options.

When outcomes can take any value in a range, as returns can, the list becomes a smooth curve called a density, like the normal curve you know. The area under the curve between two values is the probability of landing between them, and the whole area is 1.

2 · Four numbers describe a distribution

For any return distribution, four numbers tell you most of what you need to know. They are its first four moments, and each answers an intuitive question.

All four are averages, so they need one piece of notation first. The expected value of a quantity $X$ , written $\mathbb{E}[X]$ , is its probability-weighted average over all outcomes:

\mathbb{E}[X] = \sum_i p_i\, x_i

(1B.1)

Here $x_i$ is the $i$ -th possible outcome and $p_i$ its probability. For the coin, the expected value is $\mathbb{E}[X] = 0.5 \times 1 + 0.5 \times (-1) = 0$ . Chapter 1C builds on this definition, because the expected value of a trade is what this book means by edge.

The mean

The first moment is the mean, $\mu = \mathbb{E}[R]$ , where $R$ is the return. It answers the question: what is the average return? This is the number every backtest tab brags about. By itself it tells you almost nothing, because wildly different distributions can produce the same mean.

The variance

The second moment about the mean is the variance, $\sigma^2 = \mathbb{E}[(R - \mu)^2]$ , the average squared distance from the mean. Its square root, $\sigma$ , is the standard deviation. It answers the question: how spread out are the returns around the mean? Variance is the engine of the Sharpe ratio of Chapter 1D, which divides the mean by the standard deviation. A high mean with a low variance produces a high Sharpe ratio, and a high mean with a high variance a mediocre one.

The skewness

The third moment, standardized by $\sigma^3$ , is the skewness:

\gamma_3 = \mathbb{E}\!\left[\left(\frac{R - \mu}{\sigma}\right)^{3}\right]

(1B.2)

It answers the question: is the distribution symmetric, or is one tail longer than the other? Dividing by $\sigma$ first turns each outcome into a number of standard deviations from the mean, so the skewness has no units. Cubing keeps the sign of each deviation and exaggerates the large ones. A long tail of large losses therefore makes the skewness negative, and a long tail of large gains makes it positive.

For a true normal (Gaussian) distribution, the skewness is zero. For SPX daily returns over multi-decade samples, it is approximately −0.7: left-skewed, with the left tail, the loss tail, longer than the right. The CBOE PUT Index is a benchmark for systematic monthly at-the-money put-writing on SPX, which Cboe has tracked since June 1986. Its skewness on monthly returns is roughly −2.1, and Oleg Bondarenko's analysis of the index puts it at approximately −2.09. The PUT Index has a skew approximately three times worse than the underlying it sells protection on. Selling premium does not eliminate skew; it concentrates it.

Other measurements run in the same direction: one puts the skewness of SPX itself at about −0.81, and a short far out-of-the-money put can reach −5 or worse. Treat the comparison between the PUT Index and SPX as rough, because the −2.09 is measured on monthly returns and the −0.7 on daily ones. Skewness changes with the length of the period it is measured over.

The kurtosis

The fourth moment, standardized by $\sigma^4$ , is the kurtosis:

\gamma_4 = \mathbb{E}\!\left[\left(\frac{R - \mu}{\sigma}\right)^{4}\right]

(1B.3)

It answers the question: how fat are the tails relative to a normal distribution? The fourth power makes every deviation positive and exaggerates the extreme ones even more than the cube does, so the kurtosis is driven almost entirely by the rare outcomes far from the mean. For a true normal distribution the kurtosis is exactly 3. Some texts use excess kurtosis, the kurtosis minus 3, so that zero means normal. This book uses the kurtosis itself, as the formulas in Chapter 1D do.

For SPX daily returns since 1928, the kurtosis is roughly 15 to 20, depending on the sample: five to seven times the normal value. Another estimate puts it around 18, six times normal, and the PUT Index has even higher kurtosis. Returns are massively fatter-tailed than a normal distribution.

moment	symbol	what it measures	why it matters
mean	$\mu$	the average outcome	the thing you are trying to maximize
variance	$\sigma^2$	spread around the mean	how uncertain the outcome is
skewness	$\gamma_3$	asymmetry	whether bad outcomes are bigger than good ones
kurtosis	$\gamma_4$	tail fatness	how likely extreme events are

A normal distribution is fully described by its mean and variance: its skewness is always 0 and its kurtosis always 3. For symmetric, thin-tailed returns like these, the first two moments are enough. Real financial returns almost never look normal, and for them, which means essentially every options strategy, all four matter.

Worked example: The four moments of a put spread

Suppose a put spread collects a $1 credit and can lose at most $9. A put spread sells one put and buys a cheaper one further from the index, which caps the loss, and Chapter 12C covers it. For this example, suppose it keeps the $1 in 90% of months and loses the full $9 in the other 10%.

Step 1: the mean. By (1B.1), the mean is $\mu = 0.9 \times 1 + 0.1 \times (-9) = 0.9 - 0.9 = 0$ , so the trade is fair: it neither makes nor loses money on average.

Step 2: the variance. The squared distances from the mean are $1^2 = 1$ and $9^2 = 81$ , so $\sigma^2 = 0.9 \times 1 + 0.1 \times 81 = 9$ and $\sigma = 3$ .

Step 3: standardize each outcome. Measured in standard deviations from the mean, the win is $+1/3$ and the loss is $-9/3 = -3$ .

Step 4: the skewness. Formula (1B.2) weights the cube of each standardized outcome by its probability:

\begin{aligned} \gamma_3 &= 0.9 \times (1/3)^3 \\ &\quad + 0.1 \times (-3)^3 \\ &= 0.033 - 2.7 = -2.67 \end{aligned}

(1B.4)

Step 5: the kurtosis. Formula (1B.3) does the same with the fourth power:

\begin{aligned} \gamma_4 &= 0.9 \times (1/3)^4 \\ &\quad + 0.1 \times (-3)^4 \\ &= 0.011 + 8.1 = 8.11 \end{aligned}

(1B.5)

The skewness of −2.67 says the loss sits three standard deviations below the mean while the win sits only a third of one above it. The kurtosis of 8.11, an excess of 5.11, says the variance is dominated by one rare outcome: 8.1 of the 9 comes from the loss.

For any bet with just two outcomes, the skewness depends only on the probability $p$ of the winning outcome, not on the sizes of the win and the loss:

\gamma_3 = \frac{1 - 2p}{\sqrt{p(1-p)}}

(1B.6)

At $p = 0.9$ this gives $-0.8/0.3 = -2.67$ , as in the worked example. At $p = 0.95$ it gives $-0.9/0.218 = -4.13$ , and at $p = 0.5$ it gives zero. The higher the win rate of a two-outcome bet, the more negative its skew. The sizes of the win and the loss decide something else, the mean, which is the subject of Chapter 1C.

Why a two-outcome bet's skew depends only on its win rate

Step 1: write the bet as a coin. Let $B$ be 1 with probability $p$ and 0 otherwise. A bet that wins $a$ or loses $b$ is $X = -b + (a + b)B$ : a shift and a positive stretch of $B$ .

Step 2: shifts and stretches do not change skewness. Standardizing subtracts the mean, which removes the shift, and divides by the standard deviation, which removes the stretch. So $X$ and $B$ have the same skewness.

Step 3: compute it for $B$ . The mean of $B$ is $p$ and its variance is $p(1-p)$ . Its third central moment, the average cubed distance from its mean, is

\begin{aligned} & p(1-p)^3 + (1-p)(-p)^3 \\ &= p(1-p)\left[(1-p)^2 - p^2\right] \\ &= p(1-p)(1 - 2p) \end{aligned}

(1B.7)

Step 4: standardize. Dividing by $\sigma^3 = [p(1-p)]^{3/2}$ gives the skewness, $(1 - 2p)/\sqrt{p(1-p)}$ .

3 · Why selling options produces negative skew

The put spread is not a special case, since small frequent gains and rare large losses are the structural shape of all premium-selling strategies. A typical short volatility strategy splits its trades into three groups:

Most trades, 70% to 90% of them, make a small positive return: the premium, less minor mark-to-market swings.
Some trades, 5% to 20% of them, make a moderate negative return: the position moved against you and you closed it for a loss.
A few trades, 1% to 5% of them, make a large negative return: a market shock blew through your strikes.

The mean of this distribution is the premium minus the expected loss in the tail. If the premium is high enough, the mean is positive. But the median, the middle outcome, and the mode, the most common one, are both positive, while the mean can be barely positive or negative, because the tail dominates it. Picture the bulk of the probability sitting just right of zero, in a tall peak of frequent small wins, with a long thin tail stretching far to the left. The mean is dragged left by that tail; the median stays right of zero.

The payoff makes the shape

Sell an option and you collect the premium up front. That caps the right tail: your maximum gain is the premium. Your maximum loss is much larger than the premium. That is especially true for puts on equity indices, where the underlying can drop 30% or more and you owe the difference between the strike and the index level.

Asymmetric payoffs produce asymmetric distributions. Capped gains and uncapped (or merely large) losses inevitably create negative skew, and there is no way to construct a short-option position without it. The pattern is general:

Lottery tickets have positive skew: small probable losses and rare large gains.
Selling insurance has negative skew: small probable gains and rare large losses.
Most short volatility strategies have negative skew, often severe.

The seller of an option is an insurer: the buyer pays a premium to be protected against a large move, and the seller is paid to carry that move's risk. Whether the premium is enough to pay for it is an empirical question, which Chapter 10C takes up as the variance risk premium.

What the premium has to pay for

If you sell premium, you collect a small expected gain in exchange for accepting negative skew and high kurtosis. The premium has to compensate you for three things:

The expected loss in the tail.
The illiquidity of unwinding the position in stress, when everyone wants out at once.
The path-dependent risk of a margin breach before prices revert, which Chapter 1E develops.

If the premium is too small to cover these, which is often true for retail-sized accounts, you have a strategy with a positive median outcome but a negative mean outcome. You win most of the time and lose money over the long run.

4 · Same mean, same spread, different shape

The mean and standard deviation of two strategies can match exactly while their shapes differ completely. Chapter 1D defines the Sharpe ratio properly; for this example, read it as the mean return divided by its standard deviation.

Worked example: Two strategies with a Sharpe ratio of 0.5

Step 1: set the targets. Both strategies have a mean return of 5% a year and a standard deviation of 10% a year, so both have a Sharpe ratio of $5/10 = 0.5$ . By the rules of Chapter 1A, the monthly mean is $5\%/12 = 0.417\%$ and the monthly standard deviation is $10\%/\sqrt{12} = 2.89\%$ .

Step 2: strategy A is roughly normal. Its returns spread evenly around the mean, with a worst month in a long record of about −8%, which is 2.9 standard deviations below the monthly mean. You can plan around a month like that.

Step 3: strategy B wins 95% of months. It gains 1.08% in 95% of months and loses 12.17% in the other 5%. Its monthly mean is $0.95 \times 1.08 - 0.05 \times 12.17 = 0.417\%$ , the same as A's. By the variance formula, its monthly standard deviation is $\sqrt{0.95 \times 0.05} \times (1.08 + 12.17) = 2.89\%$ , also the same.

Step 4: compare the shapes. A has a skewness of 0 and a kurtosis of 3. By (1B.6), B has a skewness of −4.13, and its kurtosis works out at 18.05. B's worst month is −12.17%, far worse than A's −8%. In real short volatility strategies the loss months also cluster: when you lose, you often lose several months in a row.

The Sharpe ratio does not see the difference. This is not a small problem: it is the central reason that naive risk measures underestimate the danger of short volatility strategies, and Chapter 1D develops it fully.

The kurtosis of strategy B shows what fat tails mean in practice. Its loss month sits $(-12.17 - 0.417)/2.89 = -4.36$ standard deviations from its mean. A normal model gives a month that bad a probability of about 1 in 154,000, once in roughly 12,800 years, yet strategy B has one every 20 months.

5 · Fat tails: rare events are not as rare as the normal curve says

Kurtosis measures how often "rare" events happen, and the lesson it carries is structural. On October 19, 1987, the S&P 500 closed down 22.6% in a single day. Measured against a normal model with volatility calibrated to the preceding decade, the fall was about 22 standard deviations. Another reckoning, using a pre-1987 normal model, puts it at about 25. A move of 25 standard deviations has a normal probability of roughly $10^{-137}$ , and even a move of 22 has a probability of only about $10^{-107}$ .

That number is so absurd it should have ended faith in normal models for good. Give the model every chance: count every second in the lifetime of the universe, about $4 \times 10^{17}$ , across perhaps $10^5$ markets. That is $4 \times 10^{22}$ trading opportunities, and the chance of seeing the crash even once is about $4 \times 10^{22} \times 10^{-137} = 4 \times 10^{-115}$ , still essentially zero. The event happened anyway. Anyone who told you in 1986 that the normal model was an acceptable approximation owed you an apology by the next day.

Sigma counts, the number of standard deviations a move represents, get unintuitive past about four. Here are the normal model's odds for a single day's fall of each size, with the waiting time at 252 trading days a year:

a fall of	normal probability on a given day	about once every
3 standard deviations	1 in 740	2.9 years
4 standard deviations	1 in 31,500	125 years
5 standard deviations	1 in 3.5 million	14,000 years
6 standard deviations	1 in 1.0 billion	4 million years
22 standard deviations	1 in $10^{107}$	never, in practice
25 standard deviations	1 in $10^{137}$	never, in practice

Yet the 22-sigma event happened. So did an 18-sigma move during the 2008 financial crisis, a 14-sigma move on COVID Monday in March 2020, and a 19-sigma single-day move in the VIX during Volmageddon in 2018. The tail events that matter most are Black Monday in 1987, October 2008, Volmageddon in 2018 and March 2020. None of them is a fluke against the right baseline: each is routine against an empirically calibrated baseline that admits fat tails. So-called five-sigma events, which the normal model puts at once in 14,000 years, actually happen every few years in financial markets.

The lesson is not that markets are mysterious or that quantitative analysts are foolish. The lesson is that the normal (Gaussian) assumption is wrong. Returns have fat tails and high kurtosis, and "once-per-century" events happen several times in a career.

For options strategies this matters a great deal, because the tail of the loss distribution is where you actually lose serious money. Underestimating kurtosis means underestimating tail risk. Any sizing model that assumes normal tails under-sizes its stress reserves by orders of magnitude.

6 · Measuring the tail

If the standard deviation cannot describe the tail, you need numbers that describe it directly, and two are standard.

Value at risk at the 95% level, written $\text{VaR}_{95}$ , is the 5th percentile of the return distribution: the return that is worse than 95% of outcomes. It answers the question: how bad is the loss at the 5th percentile?

Expected shortfall at the 95% level, also called conditional value at risk (CVaR), is the average of the returns in that worst 5%:

\text{ES}_{95} = \mathbb{E}\left[R \mid R \le \text{VaR}_{95}\right]

(1B.8)

The vertical bar means "given that", so the expected shortfall is the mean return of the months that are at or beyond the 5th percentile. Where VaR asks how bad the loss at the 5th percentile is, ES asks: given that you hit the 5th percentile or worse, how bad is it on average? ES is the measure Basel III adopted for bank capital after the 2008 crisis, precisely because VaR is silent about the size of losses beyond its threshold. For short volatility sizing, ES is the single most useful number you can compute.

Worked example: The tails of strategies A and B

Step 1: strategy A's VaR. For a normal distribution, the 5th percentile sits 1.645 standard deviations below the mean, so $\text{VaR}_{95} = 0.417\% - 1.645 \times 2.89\% = -4.33\%$ .

Step 2: strategy A's ES. For a normal distribution, the worst 5% average 2.063 standard deviations below the mean, a factor equal to $\varphi(1.645)/0.05$ , where $\varphi$ is the height of the standard normal curve, so $\text{ES}_{95} = 0.417\% - 2.063 \times 2.89\% = -5.54\%$ .

Step 3: strategy B's VaR and ES. Exactly 5% of B's months are losses of 12.17%, so both its 5th percentile and the average beyond it are −12.17%.

Step 4: compare the two. With the same mean and the same standard deviation, B's expected shortfall is more than twice A's: −12.17% against −5.54% a month.

The figure below draws one distribution whose shape a slider sets. The slider is labeled shape and runs from −4 to +4, starting at −1.5: it sets the shape of the curve rather than the skewness itself. The "skew" readout reports the skewness that the shape produces, which stays between about −0.8 and +0.8. The horizontal axis is in units of σ. The shaded bars left of −1.645σ mark the loss tail, the dashed line is the mean, and the solid line is the expected shortfall of the worst 5% of outcomes. The readouts give the mean, the standard deviation, the skewness, a Sharpe ratio computed as the mean over the standard deviation, and $\text{ES}_{95}$ .

Start at −0.5 and drag the slider slowly to −4, watching the standard deviation and the ES together.

shape −1.5

mean μ = −0.264

std σ = 0.748

skew = −0.30

Sharpe (blind to skew) = −0.35

ES₉₅ = −1.94

A thinner spread over the same tail. A skew-normal return distribution whose shape the slider sets, in units of σ; the skew readout gives the skewness that shape produces. The bars left of −1.645σ mark the loss tail, the dashed line is the mean, and the solid line is the expected shortfall, ES₉₅, the average of the worst 5%. Drag the shape from −0.5 toward −4: the standard deviation falls by about a third while ES₉₅ hardly moves, so a summary by mean and spread reads less risk where the tail is as bad as it was.

As you drag, the standard deviation falls from about 0.93 to about 0.63, which a mean-and-spread summary reads as a third less risk. The expected shortfall hardly moves, staying near −1.9: the worst 5% of outcomes are as bad as they were. Divided by the distribution's own standard deviation, the expected shortfall has grown from about 2 to about 3. The Sharpe readout moves too, because this figure's mean shifts as the shape changes, but it is built from the mean and the standard deviation alone. Nothing in it registers that the tail stayed put while the spread shrank. Now set the slider to +3: the same standard deviation as at −3, 0.65, comes with an ES of only −0.73, because the long tail is now on the side of the gains.

Stress scenarios, not percentiles

This is also why quantitative risk managers talk about stress scenarios rather than VaR percentiles. A stress scenario asks a concrete question: what happens if SPX falls 20% in three days and the VIX goes from 15 to 60? You can answer it with explicit calculations from an option's Greeks, which Part V teaches. A VaR percentile asks what the 99th-percentile loss is, and its answer depends entirely on the tail thickness you assumed. The first question survives bad assumptions; the second is hostage to them.

In practice

Your statement shows the median, not the mean. Month after month, the profit and loss (P&L) on a short put's statement shows the typical outcome: the premium kept. The mean, which decides whether the strategy makes money, lives in the rare months your statement has not shown you yet.
Short records hide the tail. Take the short put's roughly 2% monthly chance of a crash. A backtest of 36 monthly returns contains no crash at all with probability $0.98^{36} = 48\%$ , so about half of all three-year backtests of that strategy show no left tail. Their skewness and kurtosis look harmless because the sample never reached the tail.
Size to the tail, not to the spread. The standard deviation of a short volatility position understates what it can lose. Size it from its expected shortfall and its losses in explicit stress scenarios.
Skew has no line on the statement. It costs nothing until the month it costs everything, which is why it must be priced before the trade and not discovered after it.

When it fails

Moments estimated from data are unstable. The kurtosis of SPX runs from 15 to 20 depending on the sample, and one extreme day can move it a long way, because the fourth power gives that day enormous weight. A sample with no crash in it reports a mild skew and a kurtosis near normal, whatever the true distribution is.

Even four moments describe a distribution only to a first approximation. Two strategies can share all four and still differ in the tail, and a historical expected shortfall is only as bad as the worst history in the sample. That is why the decision rule below leans on stress scenarios as well as on measured moments.

Decision rule

Assume fat tails. Put a kurtosis between 10 and 20 into any tail model, and treat any "30-sigma" event as roughly a 1-in-30-years event.
Size to survive a large move. Size your largest single position so that a 4-standard-deviation move, measured against the empirical distribution rather than the normal one, does not threaten the survival of the account.
Ask stress questions. Before sizing a short volatility position, compute its $\text{ES}_{95}$ and its loss in explicit stress scenarios. Chapter 1D turns this into a full stress test.
Judge a premium-selling strategy by its mean. A high win rate and a positive median say nothing about whether it makes money.

Common mistake: "defined-risk spreads eliminate the skew problem"

A spread caps the loss, but the cap is still much larger than the gain. A put spread with a $1 credit and a $9 maximum loss risks nine times what it can make on a single trade. At a 90% win rate its skewness is −2.67, as the worked example showed.

Aggregating many trades narrows the skew but does not remove it. For $n$ independent trades, the skewness of the total is the single trade's skewness divided by $\sqrt{n}$ : over 100 independent spreads, $-2.67/10 = -0.27$ . But spreads on the same index open at the same time are not independent, because one crash hits them all together, so the portfolio keeps much of a single trade's skew. The structural issue is the asymmetric payoff. Spreads do not fix an asymmetric payoff; they just bound the worst case. Chapter 1D returns to this with the iron condor.

Common mistake: "I win 80% of my trades, so I have edge"

Win rate alone is meaningless, and you can engineer a high win rate with a bad expected value trivially. Take a bet that wins $0.10 and loses $1. It only breaks even if it wins 10 times for every loss, a win rate of $10/11 = 90.9\%$ . At an 80% win rate it loses $0.8 \times 0.10 - 0.2 \times 1 = -\$0.12$ a trade. What matters is the full distribution: each outcome's size times its probability, including the tail. Chapter 1C makes this precise.

Familiar ground

You know a skewed distribution from survival analysis. When a trial has a tail of long-term survivors, a few patients live far longer than the rest, and they pull mean survival above median survival. Trial reports lead with the median partly because it is robust to that tail. A premium seller's monthly results are the mirror image. Most months are small wins, so the median is positive, while a long left tail of rare large losses drags the mean below it, sometimes below zero. Judging a short volatility record by its typical month is like judging a survival curve by its median alone: the tail that decides the outcome is exactly what that number leaves out.

What to carry forward

A return distribution lists each outcome with its probability, and four moments summarize it: the mean, the variance, the skewness and the kurtosis.
Selling options caps the gain at the premium and leaves a large loss, so every short-option position is negatively skewed: a positive median can sit beside a negative mean.
Real returns have fat tails: SPX daily returns have a kurtosis of roughly 15 to 20 against the normal's 3, and "impossible" moves recur several times in a career.
The standard deviation cannot see the tail. Expected shortfall and explicit stress scenarios can, and they are the numbers to size by.

7 · Check yourself

Question 1

A bet wins $2 with probability 0.8 and loses $8 with probability 0.2. Compute its mean, standard deviation, skewness and kurtosis.

Answer

The mean is $0.8 \times 2 - 0.2 \times 8 = 0$ and the variance is $0.8 \times 4 + 0.2 \times 64 = 16$ , so the standard deviation is $\sigma = 4$ . Standardized, the win is $+0.5$ and the loss $-2$ . The skewness is $0.8 \times 0.125 + 0.2 \times (-8) = -1.5$ , which matches $(1 - 1.6)/\sqrt{0.16} = -1.5$ from (1B.6). The kurtosis is $0.8 \times 0.0625 + 0.2 \times 16 = 3.25$ , so this bet is strongly skewed yet has a kurtosis close to the normal's 3: the two moments measure different things.

Question 2

Why is every short-option position negatively skewed, whatever its strike?

Answer

The seller's gain is capped at the premium collected, while the loss can be many times larger. Capped gains beside large losses always put the long tail on the loss side. A spread bounds the loss, but the bound is still far larger than the premium.

Question 3

Two strategies have the same mean and standard deviation. Strategy B loses 12.17% in 5% of months; strategy A is normal with a monthly standard deviation of 2.89%. Which has the larger expected shortfall at 95%, and by how much?

Answer

Strategy B does. Its worst 5% of months all lose 12.17%, so its $\text{ES}_{95}$ is −12.17%, against A's $0.417\% - 2.063 \times 2.89\% = -5.54\%$ . B's expected shortfall is more than twice A's, although a summary by mean and standard deviation calls the two identical.

Question 4

A normal model gives a 5-standard-deviation daily fall a probability of 1 in 3.5 million. What should you conclude if you see several in one career?

Answer

That the model's tails are too thin, not that you have been unlucky. At 252 trading days a year, 1 in 3.5 million days is once in about 14,000 years. Seeing several in a career means the true distribution puts far more weight in the tail, which is what a kurtosis of 15 to 20 describes.

Question 5

You backtest a strategy that has a 2% chance each month of a large loss, using 36 months of data. How likely is it that your backtest contains no such loss? And if you hold 25 of these positions, independent of each other, what happens to the skewness of the total?

Answer

The chance of 36 clean months is $0.98^{36} = 0.48$ , about a coin flip. For 25 independent positions, the skewness of the total is the single position's divided by $\sqrt{25} = 5$ . Positions on one index during one crash are not independent, though, and then the total keeps most of the single position's skew.