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
\[M_X(t) = \exp\left ( \frac{t^2}{2}\right ).\]
Error function and its properties
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:
\[\int_0^{\infty} \exp\left ( - \frac{t^2}{2}\right ) d t
= \sqrt{\frac{\pi}{2}}.\]
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.