Computing conditional PDF of multivariate function of random variables.

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I came across the following problem while reading a paper:

Consider two independent random variables $Y$ and $Z$ with respective marginal pdf's $p_{Z}(z)$ and $p_{Y}(y)$. Given an invertible mapping $g: \mathbb{R}^2 \mapsto \mathbb{R}^2: (y, z) \mapsto (g_1(y, z), g_2(y, z))$, let me define the random variables $X_1$ and $X_2$ as:

$$X_1 = g_1(Y, Z),$$ $$X_2 = g_2(Y, Z).$$

I am interested in computing (i.e., finding a formula for) the conditional joint pdf of $X_1$ and $X_2$ given $Y$, that is:

$$p_{X_1, X_2 \,|\, Y} (x_1, x_2 \,|\, y) := \frac{p_{X_1, X_2 , Y} (x_1, x_2, y)}{p_Y(y)}.$$

I know that the joint pdf of $X_1$ and $X_2$ can be computed as:

$$p_{X_1, X_2}(x_1, x_2) = \frac{p_{Y}(y)p_{Z}(z)}{\left| \det \frac{\partial g}{\partial(y,z)} \right|} \bigg\rvert_{(y,z)=g^{-1}(x_1,x_2)},$$

where $\frac{\partial g}{\partial(y,z)}$ is the Jacobian matrix of $g$ and I have used the fact that $Y$ and $Z$ are independent.

But how about $p_{X_1, X_2 \,|\, Y}$? In the paper, they claim (without proof) that this is given by:

$$p_{X_1, X_2 \,|\, Y}(x_1, x_2 \,|\, y) = \frac{p_{Z}(z)}{\left| \det \frac{\partial g}{\partial(y,z)} \right|} \bigg\rvert_{z=f_Z(x_1,x_2)},$$

where $f_Z$ is the inverse mapping from $X_1$ and $X_2$ to $Z$.

However, this result is unintuitive for me: if $Y$ is given, then we should be able to treat it as a constant $Y=y$ from the beginning. Therefore, the result should only depend on the partial derivative $\frac{\partial g}{\partial z}$ and not on the full Jacobian matrix.

Any help would be much appreciated.

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Well, an answer would be that we may write the joint pdf $p_{X_1, X_2, Y} (x_1, x_2, y)$ as:

$$p_{X_1, X_2, Y} (x_1, x_2, y) = p_{X_1, X_2}(x_1, x_2)\, \delta(y-f_Y(x_1, x_2)),$$

where $\delta$ is the Dirac function and $f_Y$ is the inverse mapping from $X_1$ and $X_2$ to $Y$.

Then, using the formula for $p_{X_1, X_2}(x_1, x_2)$ and the definition of conditional probability, the result follows. However, I still do not have an intuition for why the result depends on the derivative of $g$ with respect to $y$. And this is really puzzling me.