Normal Random Variable

Univariate

A normal( $\mu,\sigma$ ) random variable is a continuous random variable $X$ with density $f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2}$ . The moment generating funcion $\phi_X(t) = e^{\mu t + \frac{1}{2} \sigma^2 t^2}$ . The mean $\mathbb{E} X = \mu$ and the variance $\mathbb{V} X = \sigma^2$ . The cumulative distribution of the normal(0,1) random variable $X$ is denoted by $N(x) = P[X \le x]$ . Since $f_X$ is even, $\mathbb{E} f(X) = \mathbb{E} f(-X)$

An affine function $Y = aX + b$ of a normal random variable $X$ is also a normal random variable with mean $\mu_Y = a \mu_X + b$ and variance $\sigma_Y^2 = a^2 \sigma_X^2$ . Hence $Y = \frac{1}{\sigma} (X - \mu)$ is a normal(0,1) random variable.

An important result as it relates to option pricing is the Gaussian Shift Lemma (Büchen) which states

lemma: Let $X$ be normal(0,1). Then $\mathbb{E} e^{aX} f(X) = e^{\frac{1}{2} a^2} \mathbb{E} f(X + a)$ .

this follows from the identity $e^{ax} \phi(x) = e^{\frac{1}{2} a^2} \phi(x - a)$ and the invariance of the domain of integration to shifts.

example: If $X(T) = X(t) e^{(r - \frac{1}{2} \sigma^2) \tau + \sigma \tau^{1/2} Z}$ where $Z$ is normal(0,1), then $\mathbb{E} X(T) = X(t) e^{r \tau}$ using the GSL.

Multivariate

More generally, a normal( $\mu,\Sigma$ ) random vector is a continuous random vector $X$ with density $f_X(x) = (\det 2 \pi \Sigma)^{-1/2} e^{-\frac{1}{2} (x-\mu)^T \Sigma^{-1} (x - \mu)}$ where $\mu = \mathbb{E} X$ and $\Sigma = \mathbb{E} (X - \mu) (X - \mu)^T$ .

When the components are mutually independent, the covariance matrix is a diagonal matrix of marginal variances. In this case the random variable $y = \sum_i c_i X_i$ is a normal with mean $\mu_y = \sum_i c_i \mu_i$ and variance $\sigma_y^2 = \sum_i c_i^2 \sigma_i^2$ .

The multivariate version of the GSL is

lemma: Let $X$ be normal(0,1; $R$ ) where $R$ is the correlation matrix. Then $\mathbb{E} e^{c^T X} F(X) = e^{\frac{1}{2} c^T R c} \mathbb{E} F(x + Rc)$

Since

$\mathbb{E} e^{c^T X} F(X) = \prod \int dx e^{c^T x} \det (2\pi R)^{-1/2} e^{-\frac{1}{2} x^T R^{-1} x} F(x) \\ = \prod \int dx \det (2\pi R)^{-1/2} e^{-\frac{1}{2}(x - Rc)^T R^{-1} (x - Rc)} e^{\frac{1}{2} c^T R c} F(x) \\ = \prod \int dx \det (2\pi R)^{-1/2} e^{-\frac{1}{2}x^T R^{-1} x} e^{\frac{1}{2} c^T R c} F(x + Rc) = e^{\frac{1}{2} c^T R c} \mathbb{E} F(x + Rc)$

Bivariate

When the vector has dimension 2, the distribution is called bivariate. Let $\rho$ be the correlation between the components. When the components are normal(0,1) random variables, the bivariate density with correlation $\rho$ is $f(x_1,x_2) = \frac{1}{2\pi (1 - \rho^2)^{1/2}} e^{-\frac{1}{2} \left( \frac{x_1^2 - 2\rho x_1 x_2 + x_2^2}{1 - \rho^2} \right) }$ and the distribution function $N(x_1,x_2,\rho) = P[X_1 \le x_1, X_2 \le x_2]$ .

The conditional density $P_{X_1 | X_2}$ of the bivariate has density $f(x_1 | x_2) = \frac{ f(x_1,x_2) }{ f(x_2)} = \frac{1}{(2\pi(1 - \rho^2))^{1/2}} e^{-\frac{1}{2} \left( \frac{(x_1 - \rho x_2)^2}{1 - \rho^2} \right)}$

The bivariate version of the Gaussian Shift Lemma is

lemma: Let $X,Y$ be normal(0,1) with correlation $\rho$ . Then $\mathbb{E} e^{aX + bY} f(X,Y) = e^{\frac{1}{2} (a^2 + 2 \rho ab + b^2)} \mathbb{E} f(X + a + b\rho, Y + b + a\rho)$ .

Since $e^{ax+by} \phi(x,y;\rho) = e^{\frac{1}{2} (a^2 + 2 \rho ab + b^2)} \phi(x - a - \rho b, y - b - \rho a)$ the conclusion follows.

Solved Problems

This section comprised worked solutions for chapter 2, Büchen.

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