Given two random variables $X $ and $\epsilon$, where $\epsilon \sim N(0,\sigma)$, I would like to derive a conditional distribution of $\beta_0 + \beta_1 X + \epsilon$ given $X$. If it is necessary to assume more, e.g. that $X$ has a density and/or that $X $ and $\epsilon $ are independent please do so! But if possible I would like to solve the problem without such assumptions.
This is what I can say:
If we assume that $X $, and $\epsilon$ are $\mathbb R^n $-valued, then a regular conditional distribution $\kappa$ of $\beta_0 + \beta_1 X + \epsilon$ given $X $ is known to exists and should be such that:
For any $x$
$$A \mapsto \kappa(x,A)$$
is a measure
and for any $A\in \mathcal{B} (\mathbb R^n ) $, the map
$$x \mapsto \kappa(x,A) $$
equals $P[\{\beta_0 + \beta_1 X + \epsilon \in A \} |X = x]$ ($P_X $ almost surely), which means that for any $B \in \mathcal{B} (\mathbb R^n )$
$$P[\{\beta_0 + \beta_1 X + \epsilon \in A \} \cap \{X \in B \}]= \int _B \kappa(x,A ) P_X(dx) $$
Now If we assume that $X $ has a density $f_X $ and that $X $ and $\epsilon $ are independent then $\beta_0 + \beta_1 X$ has density $f_Z(z)=\frac{1}{|\beta_1|}f_X(\frac{z-\beta_0}{\beta_1})$ and $Y:=\beta_0 + \beta_1 X + \epsilon$
has density $(f_Z * f_\epsilon) (y) := \int f_Z(x)f_\epsilon(y-x)\mu (dx) = \frac{1}{|\beta_1|} \int f_X(\frac{x-\beta_0}{\beta_1})f_\epsilon(x-y) \mu (dx) $
What density has $\omega \mapsto (X(\omega),Y(\omega))$ ?
Then if we may find the joint distribution
$$f_{Y|X } (y|x)=\frac{f_{(X,Y)} (x,y)}{f_X(x)}, \ \text {if } f_X(x)>0 $$
[and defined to be zero if $f_X(x)=0$] is a density for a regular conditional probability of $Y $ given $X $.
Can my attempt be completed to derive the distribution explicitly and show that for every $x$ the distribution is $N(\beta_0 + \beta_1x,\sigma^2 )$? And is it possible to do it without assuming that $X $ has a density and that $X $ and $\epsilon $ are independent?
Much grateful for any help provided!