Let $\textbf{y}=AX\textbf{w}+\textbf{z}$ where $\textbf{w}$ and $\textbf{z}$ are i.i.d. vectors, why does $p(\textbf{z})=p(\textbf{y}|X,\textbf{w})$? I am a student at the undergraduate level and I have been reading a paper in the information theory/compressed sensing literature which derives a likelihood function that I do not understand.
In this scenario, we have a signal that is corrupted by multiplicative and additive noise. This is modeled by
$$\textbf{y} = AX\textbf{w} + \textbf{z}$$
Where $A$ is a constant matrix of dimensions $M \times N$, $X$ is the original signal matrix with dimensions $N\times N$, $\textbf{w}$ is a random i.i.d. $N \times 1$ vector, $\textbf{w} \sim \mathcal{N}\left(0_n, \frac{1}{\sigma_w^2}I_n\right)$, and $\textbf{z}$ is a random i.i.d. $M \times 1$ vector, $\textbf{z} \sim \mathcal{N}\left(0_m, \frac{1}{\sigma_z^2}I_m\right)$.
Now, the authors have
$$p(\textbf{y} \mid X;A) = \int p(\textbf{y} | X, \textbf{w};A)p(\textbf{w})d\textbf{w}$$
Now I think I understand this first equation. Because
$$\begin{aligned}p(\textbf{y} | X, \textbf{w})  = \frac{p(\textbf{y} \cap X \cap \textbf{w})}{p(X \cap\textbf{w})} = \frac{p(\textbf{y} \cap X \cap \textbf{w})}{p(X) p(\textbf{w})}\end{aligned}$$
by the definition of conditional probability and the because $X$ and $\textbf{w}$ are independent.
Then
$$p(\textbf{y} | X, \textbf{w})p(\textbf{w}) = \frac{p(\textbf{y} \cap X \cap \textbf{w})}{p(X)}$$
$$\implies \int p(\textbf{y} | X, \textbf{w})p(\textbf{w})d\textbf{w} = \int \frac{p(\textbf{y} \cap X \cap \textbf{w})}{p(X)}d\textbf{w} = \frac{p(\textbf{y} \cap X)}{p(X)} = p(\textbf{y} | X)$$.
However, the authors then have
$$\int p(\textbf{y} | X, \textbf{w};A)p(\textbf{w})d\textbf{w} = \int \text{exp}\left(-\frac{1}{2\sigma_z^2}||\textbf{y} - AX\textbf{w}||^2 - \frac{1}{2\sigma_w^2}||\textbf{w}||^2\right)d\textbf{w}$$
Clearly, they have used
$$p(\textbf{y} | X, \textbf{w}) = p(\textbf{z})$$
As that gives us
$$\begin{aligned}\int p(\textbf{z})p(\textbf{w})d\textbf{w} &= c\int\prod_{i=1}^M\text{exp}\left(-\frac{1}{2\sigma_z^2}z_i^2\right)\prod_{j=1}^N\text{exp}\left(-\frac{1}{2\sigma_w^2}w_i^2\right)d\textbf{w} 
\\&= c\int\text{exp}\left(\sum_{i=1}^M -\frac{1}{2\sigma_z^2}z_i^2\right)\text{exp}\left(\sum_{j=1}^N -\frac{1}{2\sigma_w^2}w_i^2\right)d\textbf{w}
\\&= c\int\text{exp}\left(-\frac{1}{2\sigma_z^2}||\textbf{z}||^2\right)\text{exp}\left(-\frac{1}{2\sigma_w^2}||\textbf{w}||^2\right)d\textbf{w}\end{aligned}$$
But I do not understand why, given $\textbf{y} = AX\textbf{w} + \textbf{z}$, it is true that $p(\textbf{y} | X, \textbf{w}) = p(\textbf{z})$. I appreciate any direction you can provide to help me understand this.
 A: Your derivation for the formula of $p(\mathbf y|\mathbf w)$ is correct (I will often drop the parameters $A$ and $X$ from the formulae).
Next, a minor point: note that in the formula for $\int p(\textbf{y} | X, \textbf{w};A)p(\textbf{w})d\textbf{w}$ you left out the normalization factor (the authors of the paper use the $\propto$ sign here). Also, I will write your equation
$$
p(\textbf{y} | X, \textbf{w}) = p(\textbf{z})
$$
as
$$
\tag{*}\label{goal}
p(\textbf{y} | X, \textbf{w}) = p(\textbf{z} = \mathbf y - AX\mathbf w)
$$
for clarity.
Now, let's compute $p(\mathbf y| \mathbf w)$. First, note that since we have a deterministic dependence of $\mathbf y$ on $\mathbf w$ and $\mathbf z$, the probability density function $p(\mathbf y, \mathbf w,\mathbf z)$ is "singular". We can express this with the delta-distribution formalism:
$$
p(\mathbf y, \mathbf w, \mathbf z) = p(\mathbf w, \mathbf z)\;\delta_{\mathbf y = AX\mathbf w + \mathbf z}.
$$
We use this to compute $p(\mathbf y, \mathbf w)$:
$$
\begin{align}
p(\mathbf y, \mathbf w) &= \int p(\mathbf y, \mathbf w, \mathbf z)\;d\mathbf z\\
    &= \int p(\mathbf w, \mathbf z)\;\delta_{\mathbf y = AX\mathbf w + \mathbf z} \;d\mathbf z\\
    &= \int p(\mathbf w)p(\mathbf z)\;\delta_{\mathbf y = AX\mathbf w + \mathbf z} \;d\mathbf z\\
    &=p(\mathbf w) \int p(\mathbf z)\;\delta_{\mathbf y = AX\mathbf w + \mathbf z} \;d\mathbf z\\
    &= p(\mathbf w) p(\mathbf z = \mathbf y - AX\mathbf w).
\end{align}
$$
Finally, since $p(\mathbf y|\mathbf w) = \frac{p(\mathbf y, \mathbf w)}{p(\mathbf w)}$, we obtain the required result $\eqref{goal}$.
