Vector $\mathbf{y} = (y_1, y_2, ..., y_N)$ and vector $\mathbf{x} = (x_1, x_2, ..., x_N)$ are related according to equation:
$$p(\mathbf{y}) = \frac{1}{|\mathbf{A}|}q(\mathbf{x})$$
And linear equation:
$$\mathbf{y}=\mathbf{A}\mathbf{x}$$
Now I need to change variables for an integral below from $\mathbf{y}$ to $\mathbf{x}$.
$$g(\mathbf{y}) = \int p(\mathbf{y})~ d\mathbf{y}$$
Textbook says its like this:
$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ \bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|d\mathbf{x}$$
$$\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \frac{\partial }{\partial x} Ax = A$$|
$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ d\mathbf{x}$$
I'm a little bit confused by the use of the Jacobian $\bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|$ to change variables in this case... mainly because most of the examples on the internet assume you have two integrals when using Jacobian to change variables...
Questions as follows:
(1) How does the Jacobian works in the case of a single integral where the variables x and y are vectors.
(2) I can see that $dy$ in numerator cancels with $\partial{y}$ in denominator... I'm not understanding how the determinate gets removed and why you can cancel a partial differential with a regular differential...
(3) why can't I just use the variable change method they teach in "calculus I" instead of using a Jacobin when applied to a vector integral?