$(g(Y)+Z)|Y=y \sim g(y)+Z$ I'm self learning probability and I have a trouble with the following problem: let $Y,Z$ be continuous independent real random vectors in $\mathbb{R}^n$, $\mathbb{R}^m$ respectively. Let $g:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be $C^1$. Find the distribution of $X|Y=y$, where $X=g(Y)+Z$.
Context: I'm particularly interested in the case where $Z$ is some gaussian noise, this kind of model is useful in robotic probabilistic state estimation.
Guess: $X|Y=y \sim Z+g(y)$, but how can I prove this?
My try: one should

*

*Prove $X$ is a continuous random vector, i.e., it admits a continuous density

*Supposing point 1, we know that the density of $X|Y=y$ can be expressed as $f_{X|Y=y}(x)=f_{X,Y}(x,y)/f_Y(y)$. So I should compute $\mathbb{P}((X,Y)\in C)=\mathbb{P}((g(Y)+Z,Y)\in C)$. Now I would like to invert $(Y,Z) \mapsto (g(Y)+Z,Y)$, but in general I cannot do this. Does it mean that I canno go on to do this?

Can you please help me out/point me to some reference material?
 A: Your definition of "continuous random variable" isn't the usual one, which merely requires the cumulative distribution function of the random variable to be continuous, not that it have a density.  The latter requires the cumulative distribution function to be absolutely continuous, which is a stronger condition.
If $\ X\ $ and $\ Y\ $ are arbitrary random variables, then the distribution $\ P\big(X\in A\,\big|\,Y=y\big)\ $ of $\ X\ $ given $\ Y=y\ $ is defined to be the function of $\ y\ $ which satisfies the equation
$$
P\big(X\in A,Y\in B\big)=\int_B P\big(X\in A\,\big|\,Y=y\big)\,dF_Y(y)
$$
for all measurable $\ B\subseteq\mathbb{R}^n\ $.  This defines $\ P\big(X\in A\,\big|\,Y=y\big)\ $ uniquely for all $\ y\ $ in the support of $\ F_Y\ $.  If $\ F_Y\ $ has a density $\ f_Y\ $, this equation reduces to
$$
P\big(X\in A,Y\in B\big)=\int_B P\big(X\in A\,\big|\,Y=y\big)f_Y(y)\,dy\ ,
$$
but it is not necessary to assume this.
If $\ Y\ $ and $\ Z\ $ are independent, and $\ X=Z+g(Y)\ $ then
\begin{align}
P\big(X\in A,Y\in B\big)&=P\big(Z+g(Y)\in A, Y\in B\big)\\
&=\int_B\int_{\{z\,|\,z+g(y)\in A\}}\,dF_Z(z)\,dF_Y(y)\\
&= \int_BP\big(Z+g(y)\in A\big)\,dF_Y(y)\ .
\end{align}
Comparing this with the definitional equation of $\ P\big(X\in A\,\big|\,Y=y\big)\ $ above, we see that
$$
P\big(X\in A\,\big|\,Y=y\big)= P\big(Z+g(y)\in A\big)
$$
for all $\ y\ $ in the support of $\ F_Y\ $. If $\ F_Z\ $ has a density $\ f_ Z\ $ then
\begin{align}
P\big(X\in A\,\big|\,Y=y\big)&=\int_{\{z\,|\,z+g(y)\in A\}}f_Z(z)dz\\
&=\int_Af_Z(x-g(y))\,dx\ ,
\end{align}
so in this case the conditional distribution $\ P\big(X\in A\,\big|\,Y=y\big)\ $ has the density $\ f_Z(x-g(y))\ $.
