5.4.4 The lognormal, Student t and chi-squared distributions

The lognormal distribution

Take a normal variable $Y \sim \mathcal{N}(\mu, \sigma^2)$ and exponentiate it. $X = e^Y$ is lognormal: its log, $\ln X = Y$ , is normal. Since $e^y > 0$ for every $y$ (lesson 1.4.1), a lognormal variable is always positive.

Rule: Lognormal distribution

\begin{gathered} X = e^Y, \quad Y \sim \mathcal{N}(\mu, \sigma^2) \\[4pt] P(X \le x) = N\left(\frac{\ln x - \mu}{\sigma}\right) \\[4pt] x_p = e^{\mu + \sigma z_p} \end{gathered}

For $x > 0$ . $\mu$ and $\sigma$ are the mean and standard deviation of $\ln X$ , not of $X$ . $\ln$ keeps order, so $X \le x$ exactly when $Y \le \ln x$ , and the quantiles of $X$ are $e$ to the quantiles of $Y$ .

The median is $e^{z_{0.5}}$ with $z_{0.5} = 0$ , so it is $e^\mu$ . Take $\mu = 0$ and $\sigma = 0.5$ : the median is 1, and $P(X > 2) = N\left(-\frac{\ln 2}{0.5}\right) = N(-1.3863) \approx 0.082829$ .

The 0.05- and 0.95-quantiles are $e^{\mp 0.5 \times 1.644854}$ , about 0.439364 and 2.276017. They are not equally far from the median 1, but they are the same factor away: $1 \div 2.276017 \approx 0.439364$ . On the log scale the distribution is symmetric; on the original scale it stretches out to the right.

The mean is above the median

Rule: Lognormal mean and variance

\begin{gathered} \mathbb{E}[X] = e^{\mu + \sigma^2/2} \\[4pt] \operatorname{Var}(X) = \left(e^{\sigma^2} - 1\right)e^{2\mu + \sigma^2} \end{gathered}

Stated here; case study 5.6.5 derives both from the normal's moment generating function. The mean is the median $e^\mu$ times $e^{\sigma^2/2} > 1$ : rare large values pull it up.

$\mathbb{E}[X] = e^{\mu}$ would be the natural guess, but $\mathbb{E}[e^Y] \ne e^{\mathbb{E}[Y]}$ , as lesson 5.3.1 warned for any curved $g$ . With $\mu = 0$ , the gap grows quickly with $\sigma$ :

$\sigma$	0.1	0.5	1	1.5
median $e^\mu$	1	1	1	1
mean $e^{\sigma^2/2}$	1.005	1.1331	1.6487	3.0802
$P(X > \text{mean}) = N(-\sigma/2)$	0.4801	0.4013	0.3085	0.2266

With $\sigma = 1.5$ , the mean is three times the median, and fewer than a quarter of the values exceed it.

Products stay lognormal. If $X_1$ and $X_2$ are independent lognormals, $\ln(X_1X_2) = \ln X_1 + \ln X_2$ is a sum of independent normals, which is normal (lesson 5.4.3). So $X_1X_2$ is lognormal, with the means and the variances of the logs added. Quantities that grow by multiplying, such as prices, are where the lognormal fits; case study 5.4.6 uses it there.

Raise $\sigma$ and watch the dashed mean pull away to the right of the median, $e^\mu$ . Then switch to the Student t and the chi-squared, the other two distributions of this lesson.

Explore

The curve is a density: probabilities are areas under it. Change the parameters, then set an interval and read the shaded area.

The dashed line is the mean.

μ, mean of ln X = 0

σ, standard deviation of ln X = 0.5

Interval from a = 0.61

to b = 1.65

Probability: P(0.61 ≤ X ≤ 1.65) = 0.6827
Mean: 1.1331
Variance: 0.3647, so the standard deviation is 0.6039

= \sigma

import math import numpy as np from scipy.stats import lognorm, norm mu, sigma = 0, 0.5 X = lognorm(s=sigma, scale=math.exp(mu)) print(X.median(), round(X.mean(), 6), round(math.exp(mu + sigma**2 / 2), 6), round(X.var(), 6)) print(round(X.sf(2), 6), round(norm.cdf(-math.log(2) / sigma), 6)) print(np.round(X.ppf([0.05, 0.95]), 6), round(X.pdf(1), 6), round(norm.pdf(0) / (sigma * 1), 6)) values = np.exp(np.random.default_rng(5).normal(mu, sigma, 1_000_000)) print(round(values.mean(), 4), round(np.median(values), 4), round(np.mean(values > 2), 4))

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The Student t distribution

The Student t distribution with $\nu$ (nu) degrees of freedom is a bell centred at 0, like the standard normal. It has more probability far out, because its density is a power, not an exponential:

f(t) = c_\nu \left(1 + \frac{t^2}{\nu}\right)^{-(\nu + 1)/2}

Here $c_\nu$ is the normalising constant of lesson 5.4.1, which depends on $\nu > 0$ . Far out, $f(t)$ shrinks like the power $\frac{1}{|t|^{\nu + 1}}$ , much more slowly than the normal's $e^{-t^2/2}$ . Data with more extreme values than a normal of the same variance are said to have fat tails.

Rule: Student t

\begin{gathered} \mathbb{E}[T] = 0 \text{ for } \nu > 1 \\[4pt] \operatorname{Var}(T) = \frac{\nu}{\nu - 2} \text{ for } \nu > 2 \\[4pt] \nu \to \infty: \ T \to \mathcal{N}(0, 1) \end{gathered}

The variance is above 1, and it is infinite for $\nu \le 2$ . The excess kurtosis is $\frac{6}{\nu - 4}$ for $\nu > 4$ , and infinite for $\nu \le 4$ . With 30 or more degrees of freedom the t is close to the normal.

The tails decide which moments exist, exactly as for the power tails of lesson 2.6.3. With $\nu = 1$ the density falls like $\frac{1}{t^2}$ , so $t\,f(t)$ falls like $\frac{1}{t}$ , whose tail integral diverges.

That t, also called the Cauchy distribution, has no mean. Its half-mean $\int_0^b t\,f(t)\,dt$ reaches 0.7345, 1.4659 and 2.1988 at $b$ = 10, 100 and 1000, and keeps climbing like a log.

$\nu$	1	2	4	10	30	normal
$P(\lvert T\rvert > 3)$	0.20483	0.09547	0.03994	0.01334	0.00539	0.0027
0.975-quantile	12.706	4.303	2.776	2.228	2.042	1.960

A t with 4 degrees of freedom lands beyond 3 about 15 times as often as a normal. Its 0.975-quantile is 2.776 instead of 1.96, which is why intervals from small samples are wider (lesson 5.8.2).

Part of that gap is the larger variance, $\frac{4}{2} = 2$ . Scaled to variance 1, the t still has fatter tails than the normal; case study 5.4.6 makes that comparison on return data.

The standard normal against two Student t distributions: alike near the centre, they part in the tails. On a log scale, at $x = 4$ the t with 4 degrees of freedom has 255 times the normal's tail.

\nu = 4

from scipy.integrate import quad from scipy.stats import norm, t for nu in [1, 2, 4, 10, 30]: T = t(nu) print(nu, round(2 * T.sf(3), 5), round(T.ppf(0.975), 3), T.var(), float(T.stats(moments="k"))) print("normal", round(2 * norm.sf(3), 5), round(norm.ppf(0.975), 3)) print(round(t(4).sf(3) / norm.sf(3), 1), round(t(4).sf(4) / norm.sf(4))) for b in [10, 100, 1000]: print(b, round(quad(lambda x: x * t(1).pdf(x), 0, b)[0], 4))

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The chi-squared distribution

Square $k$ independent standard normals and add them up. The sum $V = Z_1^2 + \cdots + Z_k^2$ has the chi-squared distribution with $k$ degrees of freedom, written $\chi^2_k$ (chi is the Greek letter $\chi$ ).

Rule: Chi-squared

\begin{gathered} V = Z_1^2 + \cdots + Z_k^2 \sim \chi^2_k \\[4pt] \mathbb{E}[V] = k, \quad \operatorname{Var}(V) = 2k \end{gathered}

$V \ge 0$ , and its density is skewed to the right. For large $k$ the shape is close to a normal.

Each $Z_i^2$ has mean $\mathbb{E}[Z^2] = 1$ . Its variance is $\mathbb{E}[Z^4] - 1 = 3 - 1 = 2$ , since the normal's kurtosis is 3 (lesson 5.4.3). For independent terms means and variances add (lesson 5.3.2), giving $k$ and $2k$ .

Two small cases link it to earlier lessons.

One degree of freedom. $Z^2 \le 1$ exactly when $|Z| \le 1$ , so $P(V \le 1) = 0.682689$ : the 68% of lesson 5.4.3.
Two degrees of freedom. $\chi^2_2$ is the exponential of lesson 5.4.2 with mean 2, that is $\lambda = \frac{1}{2}$ . So $P(V > 3) = e^{-1.5} \approx 0.22313$ .

The 0.95-quantiles are 3.841, 5.991, 7.815 and 18.307 for $k$ = 1, 2, 3 and 10. The first is $z_{0.975}^2 = 1.959964^2$ : 95% of the time $|Z| < 1.96$ .

The t from a normal and a chi-squared

The t is built from the other two. Take $Z$ standard normal and $V \sim \chi^2_\nu$ , independent of $Z$ . Then

T = \frac{Z}{\sqrt{V/\nu}}

has the t distribution with $\nu$ degrees of freedom. $V/\nu$ has mean 1, so $T$ is $Z$ divided by a random number near 1. Now and then that number is small, which makes $T$ large: those are the fat tails.

Lesson 5.8.2 meets this ratio when the standard deviation of a sample is estimated rather than known. Module 5.8 also uses $\chi^2$ for the spread of sample variances.

\chi^2_k

import numpy as np from scipy.stats import chi2, norm, t print(round(chi2(1).cdf(1), 6), round(chi2(2).sf(3), 5), round(np.exp(-1.5), 5)) print([round(float(chi2.ppf(0.95, k)), 3) for k in [1, 2, 3, 10]], round(norm.ppf(0.975) ** 2, 3)) rng = np.random.default_rng(6) V = (rng.standard_normal((100_000, 3)) ** 2).sum(axis=1) print(round(V.mean(), 3), round(V.var(), 3), round(np.mean(V > chi2.ppf(0.95, 3)), 4)) rng = np.random.default_rng(7) Z = rng.standard_normal(200_000) W = (rng.standard_normal((200_000, 4)) ** 2).sum(axis=1) T = Z / np.sqrt(W / 4) print(round(np.mean(np.abs(T) > 3), 4), round(2 * t(4).sf(3), 4), round(T.var(), 3))

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Where each is used

distribution	built from	values	used for
lognormal	$e^Y$ , $Y$ normal	positive	prices and other quantities that grow by multiplying
Student t	$Z / \sqrt{V/\nu}$	any real	small samples; fat-tailed data such as daily returns
chi-squared	$Z_1^2 + \cdots + Z_k^2$	0 or more	sample variances and tests

Case study 5.4.6 models prices as lognormal and compares daily returns with the normal and the t. Case study 5.9.4 fits a t to the same returns.

Common mistakes

The lognormal's $\mu$ and $\sigma$ belong to $\ln X$ . The median is $e^\mu$ ; the mean is $e^{\mu + \sigma^2/2}$ , not $e^\mu$ .
Most lognormal values fall below the mean. $P(X > \text{mean}) = N(-\sigma/2)$ is under one half.
SciPy's lognorm takes s $= \sigma$ and scale $= e^\mu$ , not scale $= \mu$ .
A t's variance is $\frac{\nu}{\nu - 2}$ , not 1. Compare tails at the same variance before calling them fat.
Some t's have no variance, or no mean. For $\nu \le 2$ the variance is infinite; for $\nu = 1$ there is no mean.
Chi-squared needs standard normals. Squares of $\mathcal{N}(0, \sigma^2)$ variables add to $\sigma^2$ times a $\chi^2$ .

The lognormal, Student t and chi-squared distributions

0 of 12 correct

1.
$\ln X \sim \mathcal{N}(1, 0.2^2)$ . Find the median of $X$ (type e for $e$ )
2.
$\ln X \sim \mathcal{N}(0, 1)$ . Find $\mathbb{E}[X]$
3.
$\ln X \sim \mathcal{N}(0, \sigma^2)$ for some $\sigma > 0$ . Find $P(X \le 1)$
4.
$\ln X \sim \mathcal{N}(0, 0.5^2)$ . What z-score does $x = 2$ have on the log scale? (type ln)
5.
$\ln X \sim \mathcal{N}(0, 0.5^2)$ and $z_{0.95} \approx 1.645$ . Find the 0.95-quantile of $X$ (type e^(…) or a decimal)
6.
$X_1$ and $X_2$ are independent, each with $\ln X_i \sim \mathcal{N}(0, 0.5^2)$ . Find the variance of $\ln(X_1 X_2)$
7.
Find the variance of a Student t with 6 degrees of freedom
8.
What is the smallest whole number of degrees of freedom for which a Student t has a finite variance?
9.
Find the excess kurtosis of a Student t with 10 degrees of freedom
10.
$V \sim \chi^2_5$ . Find $\operatorname{Var}(V)$
11.
$V \sim \chi^2_2$ . Find $P(V > 4)$ (type e^(…))
12.
$X_1, \ldots, X_4$ are independent $\mathcal{N}(0, 9)$ . Find $\mathbb{E}[X_1^2 + \cdots + X_4^2]$

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

X = e^Y

The lognormal, Student t and chi-squared distributions

The lognormal distribution

Rule: Lognormal distribution

The mean is above the median

Rule: Lognormal mean and variance

The Student t distribution

Rule: Student t

The chi-squared distribution

Rule: Chi-squared

The t from a normal and a chi-squared

Where each is used

Common mistakes

The lognormal, Student t and chi-squared distributions

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