I have a question, and I am guessing that the question arises due to my lack of good understanding in the change of variable technique.

I would like to evaluate $f_X(x)$. When $f_Y(y)$ exists, I can write $f_X(x)$ as: $$ f_X(x) = \int f_{X|Y}(x|y) f_Y(y) dy $$ In the scenario that I encountered, $f_Y(y)$ is a multivariate normal with mean $\mu$ and variance $V$. Then $Y=\mu + A^T Z$ where $Z$ is a random variable following a standard multivariate normal distribution, and $A^T A =V$.

For some reasons, I would like to evaluate $f_X(x)$ in the way using $Z$ rather than $Y$. That is, I would like to evaluate:

$$ f_X(x) = \int f_{X|Z}(x|z) f(z) dz $$

In this case, what is the form of $f(x|z)$ ?


I tried to solve this problem in this way

Since $Z= A^{-T}[Y-\mu]$, using the change of variable technique, we have $f_Y(y)=f_Z(A^{-T}[y-\mu])|A^{-1}|$. Therefore, I should be able to write $f(x)$ as:

$$ f_X(x) = \int f_{X|Y}(x|y) f_Z(A^{-T}[y-\mu])|A^{-1}|dy $$

However, this calculation did not take me anywhere....


Ok. I just figured out that $$ f_{X|Z}(x|z) = f_{X|Y}(x|A^Tz+\mu). $$

To see this, first, the change of variable technique shows that:

$$ f_{X,Z}(x,z) = f_{X,Y}(x,A^Tz+\mu) |A|. $$

The change of variable technique also shows that:

$$ f_{Z}(z) = f_{Y}(A^Tz+\mu) |A|. $$ (You can get this result by simply integrating out the $f(X,Z)$ above with respect to $X$).


$$ f_{X|Z}(x|z) = \frac{f_{X,Z}(x,z)}{f_Z(z)} = \frac{f_{X,Y}(x,A^Tz+\mu) |A|}{f_{Y}(A^Tz+\mu) |A|}= \frac{f_{X|Y}(x|A^Tz+\mu) f_Y(A^Tz+\mu)|A|}{f_{Y}(A^Tz+\mu) |A|}=f_{X|Y}(x|A^Tz+\mu). $$


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