# Show that target variable is gaussian in simple linear regression

Given the simple linear regression model $$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$$ where $$\beta_0$$ and $$\beta_1$$ are fixed paramters, $$x_i$$ are nonrandom variables and the errors $$\epsilon_i$$ are gaussian distributed according to $$\epsilon_i \sim N(0,\sigma^2)$$.

I know that in this setting the $$y_i$$ are also gaussian distributed and it holds: $$y_i \sim N(\beta_0 + \beta_1 x_i, \sigma^2)$$.

It is straigt forward to show that $$E[y]=\beta_0 + \beta_1 x_i$$ and that $$Var[y]=\sigma^2$$, but what I do not understand is, how I could show that the $$y_i$$ are indeed gaussian distributed (and not distributed to some other distribution with the mean $$E[y]=\beta_0 + \beta_1 x_i$$ and $$Var[y]=\sigma^2$$).

I mean it makes sense that the responses are also gaussian when all the other terms are nonrandom expect for the gaussian noise but what I try to do is to derive this mathematically, i.e., I tried to plug in a gaussian for $$\epsilon_i$$ and plug this into the uppermost equation but that doesnt add up to a gaussian with the mean as calculated above.

Is it possible to show mathematically that $$y_i$$ are indeed gaussian?

• $Y$ is not Gaussian unless you assume Gaussian noise. Linear regression alone, without this assumption, does not make, or need it to be Gaussian. – Tim Nov 28 '18 at 14:54
• yes, but in this set up I specifically assume gaussian noise. – guest1 Nov 28 '18 at 15:16

Define $$\mu_i = \beta_0 + \beta_1x_i$$. Then, by assumption you know that $$\epsilon_i = y_i-\mu_i \sim N(0,\sigma^2)$$. Since the normal distribution belongs to the location-scale family, it follows that
$$y_i-\mu_i+\mu_i= y_i \sim N(\mu_i,\sigma^2).$$
$$\dfrac{y_i-\mu_i}{\sigma} \sim N(0,1).$$