bivariateNormalCdf

Evaluates the bivariate normal distribution function.

Synopsis

bivariateNormalCdf (x, y, rho)

Required Arguments

float x (Input)

The x-coordinate of the point for which the bivariate normal distribution function is to be evaluated.

float y (Input)

The y-coordinate of the point for which the bivariate normal distribution function is to be evaluated.

float rho (Input)

Correlation coefficient.

Return Value

The probability that a bivariate normal random variable with correlation rho takes a value less than or equal to x and less than or equal to y.

Description

Function bivariateNormalCdf evaluates the distribution function F of a bivariate normal distribution with means of zero, variances of one, and correlation of rho; that is, with ρ = rho, and $|\rho|<1$,

$$F(x,y) = \frac{1}{2 \pi \sqrt{1 - \rho^2}} \int_{-\infty}^{x} \int_{-\infty}^{y} \exp \left(- \frac{u^2 - 2 \rho uv + v^2}{2\left(1-\rho^2\right)}\right) \mathit{dudv}$$

To determine the probability that $U\leq u_0$ and $V\leq v_0$, where $(U,V)^T$ is a bivariate normal random variable with mean $\mu= (\mu_U,\mu_V)^T$ and variance-covariance matrix

$$\begin{split}\Sigma = \begin{bmatrix} \sigma_U^2 & \sigma_{UV} \\ \sigma_{UV} & \sigma_V^2 \\ \end{bmatrix}\end{split}$$

transform $(U,V)^T$ to a vector with zero means and unit variances. The input to bivariateNormalCdf would be X = $(u_0-\mu_U)/\sigma_U$, Y = $(v_0-\mu_V)/\sigma_V$, and $\rho=\sigma_{UV}/(\sigma_U \sigma_V)$.

Function bivariateNormalCdf uses the method of Owen (1962, 1965). Computation of Owen’s T-function is based on code by M. Patefield and D. Tandy (2000). For $|\rho|=1$, the distribution function is computed based on the univariate statistic, $Z=\min(x,y)$, and on the normal distribution function normalCdf.

Example

Suppose $(X,Y)$ is a bivariate normal random variable with mean (0, 0) and variance-covariance matrix as follows:

$$\begin{split}\begin{bmatrix} 1.0 & 0.9 \\ 0.9 & 1.0 \\ \end{bmatrix}\end{split}$$

In this example, we find the probability that X is less than −2.0 and Y is less than 0.0.

from __future__ import print_function
from numpy import *
from pyimsl.stat.bivariateNormalCdf import bivariateNormalCdf

x = -2.0
y = 0.0
rho = 0.9
p = bivariateNormalCdf(x, y, rho)
print("The probability that X is less than -2.0")
print("and Y is less than 0.0 is %6.4f" % p)

Output

The probability that X is less than -2.0
and Y is less than 0.0 is 0.0228