∣det(∂xT∂f(x))∣∣ logpX(x)=log(pZ(f(x)))+log(∣∣det(∂xT∂f(x))∣∣) Let (a,b)=x and (u,v)=z
Contributed by Shuo-Hui Li
u,v∬pZ(u,v)dAZ=a,b∬pX(a,b)dA dA=dadb=(∂u∂a∂v∂b−∂v∂a∂u∂b)dudv=det∣∣∂u∂a∂u∂b∂v∂a∂v∂b∣∣dudv Here, the matrix inside the determinant is the Jacobian, which can be extended to higher dimensions.
u,v∬pZ(u,v)dAZ=a,b∬pX(a,b)det∣Jacobian∣dudv pZ(u,v)=pX(a,b)det∣Jacobian∣ z1,z2∼U(0,1) are two uniform random variables
{x1=−2lnz1cos(2πz2),x2=−2lnz1sin(2πz2). Then, x1,x2 satisfies normal distribution.
Try: get the inverse mapping and prove the result.
A chain of Bijectors, mapping any distribution to normal distribution.
H:(u,v)→(a,b){ab=m0(u,v)u+m1(u,v)=m2(u,v)v+m3(u,v)