Re:Random Variable Algebra
by wvlia5 on Sunday July 11, 2021 @12:14AM
(#27798419)
Visualize it.
You are integrating a function P(X) along a thin surface defined by z<f(X)<z+dz.
What is volume dV for a given dz? It is a surface dS times dX: the amount that X changes for a given dz. Since z=f(x), then dx=dz/|∇f(x)|. So dV=dS*dz/|∇f(x)|
You are integrating a function P(X) along a thin surface defined by z<f(X)<z+dz.
What is volume dV for a given dz? It is a surface dS times dX: the amount that X changes for a given dz. Since z=f(x), then dx=dz/|∇f(x)|. So dV=dS*dz/|∇f(x)|