0
$\begingroup$

We have this information :

We have this regression model $Y = \beta x + e$ with ($e\sim N(0, \sigma^2)$) . We also have the estimate of $\beta$ which is $\hat{\beta} = \dfrac{1}{\sum_{i=1}^n x_i^2}\sum_{i=1}^n x_i y_i$. Consider $\hat{\beta}$ as if = $\beta$.

Why does the $E(y_i) = E(\beta x_i + e_i) = \beta x_i$?

Update: Ok, I get that $E(e_i) = 0$ but then I don't understand why does $E(\beta x_i) = \beta x_i$ ? I thought that the $E(x) = \sum x f(x)$. Where is our $f(x)$ in this case?

$\endgroup$
3
  • $\begingroup$ $E(e_i)=0$ and sum rule for expectation. $\endgroup$
    – herb steinberg
    Dec 12, 2019 at 17:29
  • $\begingroup$ but why is the expectation of $\beta x_i$ equal to itself? $\endgroup$
    – WindBreeze
    Dec 12, 2019 at 17:35
  • $\begingroup$ I am not familiar with the details of the example. However, if these quantities are not random, then the expectations are the quantities. Specifically $\beta x_i$ is not random. $\endgroup$
    – herb steinberg
    Dec 12, 2019 at 19:13

1 Answer 1

Reset to default
1
$\begingroup$

I don't understand why does $E(\beta x_i) = \beta x_i$ ?

Because $\beta$ and $x_i$ are deterministic quantities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.