7.3. Univariate Distributions#

7.3.1. Gaussian Distribution#

7.3.1.1. Standard Normal Distribution#

This distribution has a mean of 0 and a variance of 1. It is denoted by

$X \sim \NNN(0, 1).$

The PDF is given by

$f_X(x) = \frac{1}{\sqrt{2\pi}} \exp \left ( - \frac{x^2}{2} \right ).$

The CDF is given by

$F_X(x) = \int_{-\infty}^x f_X(t) d t = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} \exp \left ( - \frac{t^2}{2} \right ) d t.$

Symmetry

$f(-x) = f(x). \quad F(-x) + F(x) = 1.$

Some specific values

$F_X(-\infty) = 0, \quad F_X(0) = \frac{1}{2}, \quad F_X(\infty) = 1.$

The Q-function is given as

$Q(x) = \int_{x}^{\infty} f_X(t) d t = \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \exp \left ( - \frac{t^2}{2} \right ) d t.$

We have

$F_X(x) + Q(x) = 1.$

Alternatively

$F_X(x) = 1 - Q(x).$

Further

$Q(x) + Q(-x) = 1.$

This is due to the symmetry of normal distribution. Alternatively

$Q(x) = 1 - Q(-x).$

Probability of $$X$$ falling in a range $$[a,b]$$

$\PP (a \leq X \leq b) = Q(a) - Q(b) = F(b) - F(a).$

The characteristic function is

$\Psi_X(j\omega) = \exp\left ( - \frac{\omega^2}{2}\right ).$

Mean:

$\mu = \EE (X) = 0.$

Mean square value

$\EE (X^2) = 1.$

Variance:

$\sigma^2 = \EE (X^2) - \EE(X)^2 = 1.$

Standard deviation

$\sigma = 1.$

An upper bound on Q-function

$Q(x) \leq \frac{1}{2} \exp \left ( - \frac{x^2}{2} \right ).$

The moment generating function is

$M_X(t) = \exp\left ( \frac{t^2}{2}\right ).$

7.3.1.2. Error Function#

The error function is defined as

$\erf(x) \triangleq \frac{2}{\sqrt{\pi}} \int_0^x \exp\left ( - t^2 \right) d t.$

The complementary error function is defined as

$\erfc(x) = 1 - \erf(x) = \frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp\left ( - t^2 \right) d t.$

Error function is an odd function.

$\erf(-x) = - \erf(x).$

Some specific values of error function.

$\erf(0) = 0, \quad \erf(-\infty) = -1 , \quad \erf (\infty) = 1.$

The relationship with normal CDF.

$F_X(x) = \frac{1}{2} + \frac{1}{2} \erf \left ( \frac{x}{\sqrt{2}}\right) = \frac{1}{2} \erfc \left (- \frac{x}{\sqrt{2}}\right).$

Relationship with Q function.

$Q(x) = \frac{1}{2} \erfc\left (\frac{x}{\sqrt{2}} \right) = \frac{1}{2} - \frac{1}{2} \erf \left ( \frac{x}{\sqrt{2}} \right ).$
$\erfc(x) = 2 Q(\sqrt{2} x).$

We also have some useful results:

$\int_0^{\infty} \exp\left ( - \frac{t^2}{2}\right ) d t = \sqrt{\frac{\pi}{2}}.$

7.3.1.3. General Normal Distribution#

The general Gaussian (or normal) random variable is denoted as

$X \sim \NNN (\mu, \sigma^2).$

Its PDF is

$f_X( x) = \frac{1}{\sqrt{2 \pi} \sigma} \exp \left ( \frac{1}{2} \frac{(x -\mu)^2}{\sigma^2}. \right)$

A simple transformation

$Y = \frac{X - \mu}{\sigma}$

converts it into standard normal random variable.

The mean:

$\EE (X) = \mu.$

The mean square value:

$\EE (X^2) = \sigma^2 + \mu^2.$

The variance:

$\EE (X^2) - \EE (X)^2 = \sigma^2.$

The CDF:

$F_X(x) = \frac{1}{2} + \frac{1}{2} \erf \left ( \frac{x - \mu}{\sigma\sqrt{2}}\right).$

Notice the transformation from $$x$$ to $$(x - \mu) / \sigma$$.

The characteristic function:

$\Psi_X(j\omega) = \exp\left (j \omega \mu - \frac{\omega^2 \sigma^2}{2}\right ).$

Naturally putting $$\mu = 0$$ and $$\sigma = 1$$, it reduces to the CF of the standard normal r.v.

Th MGF:

$M_X(t) = \exp\left (\mu t + \frac{\sigma^2 t^2}{2}\right ).$

Skewness is zero and Kurtosis is zero.