Normal distribution differential equation

The differential equation having a normal distribution as its solution is

 (dy)/(dx)=(y(mu-x))/(sigma^2),
(60)

This dynamic will be dictated by the distance from mean, the slope  will be positive before mean and negative after mean and y will asymptotically reach 0.

since

 (dy)/y=(mu-x)/(sigma^2)dx
(61)
 ln(y/(y_0))=-1/(2sigma^2)(mu-x)^2
(62)
 y=y_0e^(-(x-mu)^2/(2sigma^2)).
(63)

It can be shown that the optimums of change in Y happen at sigma far from the mean. In other words the maximum fall rate for the probability happens at x=+ – sigma

{y}'=y(\mu -x)/\sigma ^{2}

{y}''={y}'(\frac{\mu -x)}{\sigma ^{2}}+y({\frac{\mu -x}{\sigma ^{2}}})'=0

\frac{y(\mu -x)}{\sigma ^{2}}(\frac{(\mu -x)}{\sigma ^{2}})-\frac{y}{\sigma ^{2}}=0

y(\frac{(\mu -x)^{2}-\sigma ^{2}}{\sigma^{4} })=0

(\mu -x)^{2}-\sigma ^{2}}=0

(\mu -x)=\sigma

 

This equation has been generalized to yield more complicated distributions which are named using the so-called Pearson system.

The normal distribution is also a special case of the chi-squared distribution, since making the substitution

 1/2z=((x-mu)^2)/(2sigma^2)
(64)

gives

d(1/2z) = ((x-mu))/(sigma^2)dx
(65)
= (sqrt(z))/sigmadx.
(66)

Now, the real line x in (-infty,infty) is mapped onto the half-infinite interval z in [0,infty) by this transformation, so an extra factor of 2 must be added to d(z/2), transforming P(x)dx into

P(z)dz = 1/(sigmasqrt(2pi))e^(-z/2)sigma/(sqrt(z))2(1/2dz)
(67)
= (e^(-z/2)z^(-1/2))/(2^(1/2)Gamma(1/2))dz