Variance of $Y$ in regression model?

Question

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. — josliber, 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. — Zachary Blumenfeld, Commented Oct 24, 2016 at 15:47

Frans Rodenburg · Accepted Answer · 2019-10-24 10:06:00Z

7

$x_i$ 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 $y_i$.

Frans Rodenburg

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answered Oct 24, 2016 at 15:24

JAD

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Matthew Gunn · Accepted Answer · 2016-10-25 19:25:52Z

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:

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:

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

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