In linear regression, if the random error is N(0,$\sigma^2$) does this mean Y~N($\alpha + \beta X$, $\sigma^2$) In linear regression, if the random error is normally distributed, does this mean the response is normal as well? In particular if $\epsilon$ ~ N(0,$\sigma^2$) does this mean Y~N($\alpha + \beta  X$, $\sigma^2$). More specifically, I am asking if Y will have a normal distribution. I know the mean will be $\alpha + \beta X$ and variance will be $\sigma^2$ but can the distribution be assumed to be normal just because $\epsilon$ is normal? Why?
 A: As is pointed out in this related question, the normality of the error term in a linear regression is not sufficient to ensure the marginal normality of the response variable.  The latter is also affected by the distribution of the explanatory variable, which is not assumed to be normal in a regression analysis.
Under the linear regression model you have specified, the conditional distribution of $Y$ is:
$$Y|x \sim \text{N}(\alpha + \beta x, \sigma^2).$$
The marginal distribution of $Y$ is:
$$F_Y(y) \equiv \mathbb{P}(Y \leqslant y) = \int \limits_{-\infty}^\infty \Phi \Big( \frac{y - \alpha - \beta x}{\sigma} \Big) f_X(x) dx,$$
where $\Phi$ is the CDF of the standard normal distribution.  In the special case where $X \sim \text{N}$ this leads to a normal distribution, but in the more general case where the explanatory variable has some other distribution, you will often get a marginal distribution for the response variable that is not normal.
A: The answer is a most definitive "no."  Marginal normality of $\epsilon$ does not imply that the conditional distributions of $Y$ are normal.  See here for a counterexample:
https://stats.stackexchange.com/a/486951/102879
A: The distribution at a fixed value of x is normal.  Y is not normal.  Just look at the histogram of the response.  It will not look like a normal distribution.  But if you look at the distribution at a fixed x, then it will look normal.
A: Yes if $\varepsilon \sim N(0, \sigma^2)$ $Y=\alpha+\beta x+ \epsilon$  then we can say that $Y \sim N(\alpha+\beta x,\sigma^2)$. This follows from the result that if a random variable $X \sim N(\mu, \sigma^2)$ then $X+a \sim N(\mu+a, \sigma^2)$, for example if $X\sim N(0, 3^2)$ then $X+2 \sim N(2, 3^2)$ 
