# Variance of $Y$ in regression model?

In linear regression : $$y_i = \beta_1 + \beta_2x_i + \epsilon_i$$ I don't understand why Var$(y_i)= \sigma^2$

Because

\begin{align} \text{Var}(Y_i) &= \text{Var}(\beta_1 + \beta_2 x_i + \epsilon_i) \\ &= \beta_2^2 \text{Var}(x_i) + \sigma^2 \end{align}

• $x_i$ is a constant so its variance is 0, yielding the $\sigma^2$ result. Commented Oct 24, 2016 at 15:23
• $var(y_i|x_i)=\sigma^2$. $var(y_i) \neq \sigma^2$ as you are attempting to show. Commented Oct 24, 2016 at 15:47

$$x_i$$ is one single non-random variable, so on itself it has a variance of 0, so the formula you wrote simplifies to just $$\sigma^2$$.

Normally $$y_i$$ is expressed as follows:

$$y_i \sim N(\beta_1 + \beta_2x_i, \;\sigma^2)$$

This way it should be evident how the variance of $$y_i$$ is determined. $$\beta_1 + \beta_2x_i$$ only contributes to the expected value of $$y_i$$.

Let's say you have the regression equation:

$$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$$

Different books, different lectures notes etc... follow two different approaches:

1. Treat $x_i$ are scalars. They're entirely exogenous. They're not random.
2. Treat $x_i$ as a random variable.

The answer of @Jarko Dubbeldam takes approach (1). If $x_i$ is a scalar then simply:

$$\mathrm{Var}(y_i) = \mathrm{Var}(\epsilon_i )$$

In any settings, Approach 1 is excessively restrictive (and it isn't necessary). If you take approach two though, you would need to write:

$$\mathrm{Var}(y_i \mid x_i ) = \mathrm{Var}(\epsilon_i )$$