How is the function expression of normal distribution derived?

How much is the landlord?

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The diversity of & ltbr & gt is as follows: (There is one yuan in the resources)

The definition of & ltbr & gt multivariate normal distribution and the derivation of its density function

& ltbr & gt multivariate normal distribution is defined as follows:

& ltbr & gt assume that u 1, u2, ... up are all independent and obey the distribution of N(0, 1), let U=[u 1, u2, ... up]', a is a nonsingular matrix of order p, and x and μ are column vectors with dimension p.

& ltbr & gt Then the distribution that X=AU+μ obeys is the P-dimensional normal distribution, which is denoted as N(μ, AA').

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& ltbr & gt Thus, the original definition of covariance matrix in multivariate normal distribution is not covariance matrix, but the product of linear transformation.

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& ltbr & gt Let's derive the density function of multivariate normal distribution.

& ltbr & gt suppose that the p-ary random vector X~N(μ, ∑), then the density function of x is

& ltbr & gt 1

& ltbr & gt—————————————exp[(x-μ)'∑^(- 1)(x-μ)]

& ltbr & gt(2*pi)^(p/2)*|∑|^( 1/2)

& ltbr & gt proves that:

& ltbr & gt let ∑=AA' then X=AU+μ.

& ltbr & gt→ U=A^(- 1)(X-μ)

& ltbr & gt because u 1, u2, ... up are all independent and all obey the distribution of N(0, 1), the joint distribution of u is

& ltbr & gt 1

& ltbr & gtp(U)=————————exp[U'*U]