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?

  • $\begingroup$ $Y$ is not Gaussian unless you assume Gaussian noise. Linear regression alone, without this assumption, does not make, or need it to be Gaussian. $\endgroup$
    – Tim
    Commented Nov 28, 2018 at 14:54
  • $\begingroup$ yes, but in this set up I specifically assume gaussian noise. $\endgroup$
    – guest1
    Commented Nov 28, 2018 at 15:16

1 Answer 1


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

$$y_i-\mu_i+\mu_i= y_i \sim N(\mu_i,\sigma^2).$$

$$\dfrac{y_i-\mu_i}{\sigma} \sim N(0,1).$$




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