# 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?

• Possible duplicate of Where does the misconception that Y must be normally distributed come from? – Ben - Reinstate Monica Nov 7 '18 at 23:22
• @Ben that link there seems more like "what are all these different sets of assumptions?" while this question seems like "why are affine transformations of normal rvs still normal?" – Taylor Nov 7 '18 at 23:30
• @s5s you can show this with moment generating functions. Have you tried that? – Taylor Nov 7 '18 at 23:32
• It's important to distinguish between the conditional distribution of Y given the value of X and the marginal distribution of Y. – Glen_b Nov 8 '18 at 0:41

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.

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.

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 $$X+2 \sim N(2, 3^2)$$

• The fact that if a random variable $X\sim N(\mu,\sigma^2)$ then $X+a∼N(\mu+a,\sigma^2)$ can be established using the method of moment generating functions. – user22546 Nov 7 '18 at 23:48