Case study: lognormal prices and fat-tailed returns

Normal log returns make lognormal prices

Case study 1.4.7 showed that log returns add across time. Over $T$ years, $\ln(S_T / S_0)$ is the sum of the daily log returns, with $S_t$ the price at time $t$. If the daily log returns are independent and normal, their sum is normal (lesson 5.4.3), so $\ln S_T$ is normal and $S_T$ is lognormal (lesson 5.4.4).

Case study 2.6.5 took this as its model: $S_T = S_0\,e^{(r - \sigma^2/2)T + \sigma\sqrt{T}\,Z}$, with $Z$ standard normal. Here $r$ is the continuous interest rate and $\sigma$ the annual volatility, both decimals, and $T$ is in years.

The variance $\sigma^2 T$ grows in proportion to time. With 252 trading days a year, a day's log return has standard deviation $\frac{0.2}{\sqrt{252}} \approx 1.26\%$. Then 126 days add up to $126 \times \frac{0.2^2}{252} = 0.2^2 \times 0.5$, if the days are independent (module 5.5 makes this precise).

Median, mean and the chance of ending above a level

Take case study 2.6.5's stock: $S_0 = 100$, $T = 0.5$, $r = 0.03$ and $\sigma = 0.2$. Then $\ln S_T$ has mean $\ln 100 + 0.005 \approx 4.610170$ and standard deviation $0.2\sqrt{0.5} \approx 0.141421$.

• The median is $e^{4.610170} \approx 100.5013$: half the futures end above it.
• The mean is $100\,e^{0.03 \times 0.5} \approx 101.5113$, as case study 2.6.5 found. The $-\frac{1}{2}\sigma^2$ in the exponent is what puts the median below it.
• $P(S_T > 100) = N(0.035355) \approx 0.514102$. This is $N(d_2)$ of case study 2.6.5, the chance that the call ends in the money.

The 5% and 95% quantiles, $e^{4.610170 \mp 1.644854 \times 0.141421}$, are 79.64 and 126.82. The 90% range reaches 20.86 below the median and 26.32 above it: on the price scale the lognormal leans to the right.

Which drift? Pricing against forecasting

The drift $r$ is the one the pricing model of case study 2.6.5 uses, so $N(d_2)$ is a pricing probability, not a forecast. A forecast puts in the stock's own expected return $\mu$ instead. With $\mu = 8\%$, an assumption for illustration, $P(S_T > 100) \approx 0.584$ and $P(S_T > 120) \approx 0.1407$.

Always say which drift a probability uses. The volatility can be estimated from data far better than $\mu$ can; case study 5.8.5 shows how uncertain a mean return is.

Longer horizons: the mean runs away from the median

With $\mu = 8\%$ and $\sigma = 20\%$, the mean grows at the rate $\mu$ but the median only at $\mu - \frac{1}{2}\sigma^2 = 6\%$. The gap widens with time, and fewer and fewer futures end above the mean.

$P(S_T > \text{mean}) = N(-\frac{1}{2}\sigma\sqrt{T})$, from lesson 5.4.4's $N(-\sigma/2)$ with $\sigma\sqrt{T}$ in place of $\sigma$. Over 30 years the mean is 1,102 but fewer than three futures in ten reach it. A loss over the whole period becomes rarer, from 41.6% to 5.0%.

The last row explains why prices are not modelled as normal. Give a normal price the same mean and standard deviation as the lognormal. Over 10 years it is negative 7.7% of the time, and over 30 years 26%.

Daily returns have fatter tails than the normal

Real daily stock returns usually have more extreme days than a normal distribution allows. Level 5 uses an invented series that behaves the same way, introduced in case study 5.3.5.

It holds 1,000 daily log returns, drawn with a fixed seed from a Student t with 4 degrees of freedom. Scaling sets their volatility near 1% a day. They are illustrative, not market data.

For these returns:

• the mean is $-0.0061\%$ a day and the volatility, the standard deviation dividing by $n$ as in case study 5.3.5, is 1.0073%;
• the skewness is $-0.31$ and the excess kurtosis 5.29 (lesson 5.3.3), where a normal has 0 for both;
• the worst day is $-5.47\%$ and the best $+6.87\%$.

Case study 5.3.5 found these moments and read value at risk from the sample. Here the question is which distribution describes the returns, so compare two models with the sample's mean and volatility.

• Normal: $\mathcal{N}(\bar{x}, s^2)$, with $\bar{x}$ the sample mean and $s$ the volatility.
• Student t, 4 degrees of freedom: $\bar{x} + s\,\frac{T}{\sqrt{2}}$ with $T \sim t_4$. Lesson 5.4.4 gives $t_4$ the variance $\frac{4}{2} = 2$, so dividing by $\sqrt{2}$ scales it to variance 1.

The choice $\nu = 4$ is in the range usually fitted to daily stock returns. Case study 5.9.4 estimates $\nu$ from these data instead of choosing it.

Comparing quantiles

A quantile–quantile comparison lines up the sample's quantiles (lesson 5.4.1) against a model's at the same probabilities. If the model fits, the two agree.

In the far left tail the normal is much too mild: its 0.1% quantile is $-3.12\%$ against the sample's $-5.37\%$. The t gives $-5.12\%$. At the 1% level, where case study 5.3.5 read its value at risk, the normal says $-2.35\%$, the sample $-2.66\%$ and the t $-2.68\%$.

In the middle the normal is too wide: its quartiles, $-0.686\%$ and $0.673\%$, sit outside the sample's, $-0.51\%$ and $0.56\%$. The sample has 76.2% of its days within one standard deviation of the mean, against 68.27% for a normal and 76.98% for the scaled t. Case study 5.3.5 saw the same crowded middle in the data; the t has it too.

The normal and the scaled t cross near 1.95 standard deviations. Inside that, the t's quantiles are closer to the mean than the normal's; beyond it they are further out.

So a fat-tailed model is not simply "wider": it moves probability from the shoulders to both the centre and the tails. That is why case study 5.3.5 found the normal value at risk too large at 95% and too small at 99%.

A 4-sigma day

Count the days more than $k$ standard deviations from the mean, and set each count against what each model expects in 1,000 days. For the scaled t, $|X - \bar{x}| > ks$ is $|T| > k\sqrt{2}$.

At 2 standard deviations the two models agree with each other and with the sample. Beyond that they part fast.

A 4-sigma day has probability $6.33 \times 10^{-5}$ under the normal: 0.016 a year of 252 trading days, or once in 62.6 years. Under the t it has probability 0.0048, or 1.21 a year.

The sample's 6 such days in about four years fit the t, not the normal. A risk model that says "once in 63 years" for something that happens yearly will be surprised often. Case study 4.7.5 made the same warning for joint crashes.

import math

from scipy.stats import norm


def test_in_the_money():
    """Case study 2.6.5's stock ends above 100 with probability N(d₂), about 0.514102"""
    expect_close(prob_above(100, 100, 0.5, 0.03, 0.2), 0.514101801652164, what="prob_above(100, 100, 0.5, 0.03, 0.2)", rel=1e-9)


def test_above_120():
    """The same stock ends above 120 with probability about 0.104948"""
    expect_close(prob_above(100, 120, 0.5, 0.03, 0.2), 0.10494776543969564, what="prob_above(100, 120, ...)", rel=1e-8)


def test_real_world_drift():
    """With an 8% drift instead, it ends above 100 with probability about 0.583998"""
    expect_close(prob_above(100, 100, 0.5, 0.08, 0.2), 0.5839979857136824, what="prob_above with mu = 0.08", rel=1e-9)


def test_long_horizon_mean():
    """Over 30 years only about 0.2919 of futures end above the mean, 100e^2.4"""
    expect_close(prob_above(100, 100 * math.exp(0.08 * 30), 30, 0.08, 0.2), norm.cdf(-0.2 * math.sqrt(30) / 2), what="P(S_T > mean) over 30 years", rel=1e-9)


def test_no_time_left():
    """At T = 0 the price is S0: above 90 for certain, above 110 never"""
    expect_equal(prob_above(100, 90, 0, 0.08, 0.2), 1.0, what="prob_above(100, 90, T=0)")
    expect_equal(prob_above(100, 110, 0, 0.08, 0.2), 0.0, what="prob_above(100, 110, T=0)")


def test_no_volatility():
    """With sigma = 0 the price grows to 100e^0.04 ≈ 104.08: above 103 for certain"""
    expect_equal(prob_above(100, 103, 0.5, 0.08, 0.0), 1.0, what="prob_above(100, 103, sigma=0)")
    expect_equal(prob_above(100, 105, 0.5, 0.08, 0.0), 0.0, what="prob_above(100, 105, sigma=0)")


def test_bad_prices():
    """A zero or negative price, a negative T or a negative volatility raises ValueError"""
    for args in [(0, 100, 0.5, 0.03, 0.2), (100, -5, 0.5, 0.03, 0.2), (100, 100, -1, 0.03, 0.2), (100, 100, 0.5, 0.03, -0.2)]:
        try:
            prob_above(*args)
        except ValueError:
            continue
        raise AssertionError(f"expected ValueError for prob_above{args}")


def test_normal_four_sigma():
    """Under the normal, a 4-sigma day has probability about 0.0000633"""
    expect_close(k_sigma_prob(4), 6.334248366623973e-05, what="k_sigma_prob(4)", rel=1e-9)


def test_t_four_sigma():
    """Under a t with 4 degrees of freedom scaled to variance 1, about 0.004813"""
    expect_close(k_sigma_prob(4, 4), 0.00481267833004422, what="k_sigma_prob(4, nu=4)", rel=1e-8)


def test_zero_sigma():
    """Any move at all, k = 0, has probability 1"""
    expect_close(k_sigma_prob(0), 1.0, what="k_sigma_prob(0)")
    expect_close(k_sigma_prob(0, 5), 1.0, what="k_sigma_prob(0, nu=5)")


def test_far_tail():
    """At 9 sigma the normal's answer is about 2.3e-19, not 0"""
    expect_close(k_sigma_prob(9), 2 * norm.sf(9), what="k_sigma_prob(9)", rel=1e-9)


def test_large_nu():
    """With a million degrees of freedom the t is almost the normal: about 0.0027 at 3 sigma"""
    expect_close(k_sigma_prob(3, 1_000_000), 2 * norm.sf(3), what="k_sigma_prob(3, nu=1e6)", rel=1e-3)


def test_bad_tail_inputs():
    """nu = 2 (no variance) or a negative k raises ValueError"""
    for args in [(3, 2), (3, 1.5), (-1, None)]:
        try:
            k_sigma_prob(*args)
        except ValueError:
            continue
        raise AssertionError(f"expected ValueError for k_sigma_prob{args}")