Random variables and distribution transformations

Definition Fix a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. A random variable is a function $X \colon \Omega \to \mathbb{R}$ that is measurable with respect to $\mathcal{F}$, which means that

for every real $x$. Equivalently, the event $\{X \in B\}$ belongs to $\mathcal{F}$ for every Borel set $B$ of the real line. The distribution, or law, of $X$ is the probability measure $\mu_X(B) = \mathbb{P}(X \in B)$ on the Borel sets, and measurability is exactly what makes these probabilities defined. If $g$ is a Borel function, for example a continuous or a monotone one, then $g(X)$ is again a random variable. Different random variables, even on different probability spaces, can have the same law.

Definition The cumulative distribution function (CDF) of a random variable $X$ is the function

of a real threshold $x$. It is nondecreasing and right-continuous, with $F_X(x) \to 0$ as $x \to -\infty$ and $F_X(x) \to 1$ as $x \to \infty$. Conversely, every function with these four properties is the CDF of some random variable, a result stated here without proof. The CDF determines the law, and for $a < b$,

Writing $F_X(x-)$ for the limit from the left, the jump $F_X(x) - F_X(x-)$ equals $\mathbb{P}(X = x)$. A value with $\mathbb{P}(X = x) > 0$ is an atom of the law. The CDF describes discrete, continuous and mixed laws alike.

Definition A random variable $X$, or its law, is absolutely continuous with probability density function $f_X$ if $f_X \ge 0$ and

for every real $x$. Then $f_X$ integrates to $1$, $\mathbb{P}(X \in B) = \int_B f_X(u)\,du$ for every Borel set $B$, and $F_X'(x) = f_X(x)$ at every point where $f_X$ is continuous. A density is determined only up to changes on a set of length zero. It is not a probability. It has the units of one over the units of $X$, it can exceed $1$, and $\mathbb{P}(X = x) = 0$ for every $x$, also where $f_X(x) > 0$. A law with an atom has no density. A continuous CDF is not enough for a density either, since singular continuous laws exist, but they lie outside this lesson.

Definition Let $X$ have density $f_X$ with $\mathbb{P}(X \in I) = 1$ for an open interval $I$. Let $g$ be continuously differentiable on $I$ with $g'(x) \ne 0$ for every $x \in I$. Then $g$ is strictly monotone, maps $I$ onto an open interval $J$, and has a continuously differentiable inverse $h = g^{-1} \colon J \to I$. The random variable $Y = g(X)$ has the density

for $y \in J$, and $f_Y(y) = 0$ for $y \notin J$. The factor $\lvert h'(y) \rvert = 1/\lvert g'(h(y)) \rvert$ is the absolute value of the Jacobian of the inverse map, which in one dimension is its derivative. To derive it for increasing $g$, note that $\{Y \le y\} = \{X \le h(y)\}$, so $F_Y(y) = F_X(h(y))$ on $J$, and the substitution $u = h(v)$ in the integral of $f_X$ up to $h(y)$ produces the density. For decreasing $g$ the inequality reverses, $F_Y(y) = 1 - F_X(h(y))$, and $h' < 0$, which the absolute value corrects. More generally, suppose $I$ splits, apart from finitely many points, into open intervals $I_1, \dots, I_m$ on each of which $g$ is continuously differentiable with $g' \ne 0$, and let $h_i$ be the inverse of $g$ on $I_i$. Then every branch that reaches $y$ contributes,

The left-out points have probability zero, and the value of $f_Y$ at their images does not matter.

Definition Let $Z \sim N(\mu, \sigma^2)$ with $\mu$ real and $\sigma > 0$, and let $S_0 > 0$. The price $S = S_0 e^{Z}$ has a lognormal distribution, which means that $\ln(S/S_0) \sim N(\mu, \sigma^2)$. The exponential is strictly increasing from $\mathbb{R}$ onto $(0, \infty)$, with inverse $z = \ln(s/S_0)$ and $dz/ds = 1/s$. With the standardized log price

the CDF and the density of $S$ are, for $s > 0$,

and both are zero for $s \le 0$. Since $F_S$ is strictly increasing on $(0, \infty)$ and $F_S(S_0 e^{\mu}) = \Phi(0) = \frac12$, the median is $S_0 e^{\mu}$. The mean $S_0 e^{\mu + \sigma^2/2}$, derived in the next card, lies above the median. If $\sigma = 0$, the price is the constant $S_0 e^{\mu}$, an atom with no density, so the formulas need $\sigma > 0$. Lognormality is a model assumption about prices, not a property of market data.

Definition Completing the square rewrites a quadratic plus a linear term as one shifted square plus a constant. Let $Z \sim N(\mu, \sigma^2)$ with $\sigma > 0$ and let $a$ be any real number. The exponent of the integrand in $\mathbb{E}[e^{aZ}]$ satisfies

so the integrand is $e^{a\mu + a^2\sigma^2/2}$ times the density of $N(\mu + a\sigma^2, \sigma^2)$, which integrates to $1$. Hence

for every real $a$, and the identity also holds when $\sigma = 0$. For the lognormal price $S = S_0 e^{Z}$ it gives $\mathbb{E}[S^a] = S_0^a e^{a\mu + a^2\sigma^2/2}$ for every real $a$. The choices $a = 1$ and $a = 2$ give

and therefore

When $\sigma > 0$, all moments of $S$ are finite, yet $\mathbb{E}[e^{tS}] = \infty$ for every $t > 0$, because $e^{tS_0 e^{z}}$ grows faster than the normal density decays. The moment generating function belongs to the normal logarithm $Z$, not to the price.