How Do You Derive the PDF of a Linearly Transformed Vector?

alpines4
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Homework Statement


Define X to be an n-vector of jointly continuous random variables X1, ..., Xn with joint pdf f(x) mapping R^n to R. Let A be an invertible nxn matrix and set Y=AX+b. I want to derive the pdf of f(y) in terms of f(x), the original pdf.



Homework Equations






The Attempt at a Solution



Given a random variable and its PDF f(x), the transformation of Y=g(X) is (given that g is one to one and thus has an inverse) f(g^{-1}(y)) * g'(y). I don't know how to generalize this to a matrix, however. I assume it will be kind of similar... Any help is appreciated. I just need some tips to get started. Thank you!
 
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alpines4 said:

Homework Statement


Define X to be an n-vector of jointly continuous random variables X1, ..., Xn with joint pdf f(x) mapping R^n to R. Let A be an invertible nxn matrix and set Y=AX+b. I want to derive the pdf of f(y) in terms of f(x), the original pdf.



Homework Equations






The Attempt at a Solution



Given a random variable and its PDF f(x), the transformation of Y=g(X) is (given that g is one to one and thus has an inverse) f(g^{-1}(y)) * g'(y). I don't know how to generalize this to a matrix, however. I assume it will be kind of similar... Any help is appreciated. I just need some tips to get started. Thank you!


Intuitively the pdf of Y=g(X) is given by
f_X(x)|dx|=f_Y(y)|dy|
In the case of vector x and y, dx and dy is related through the determinant of the Jacobian matrix between x and y
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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