1
$\begingroup$

In a statistics textbook I saw that the linear simple regression model is defined as \begin{equation} Y = \alpha + \beta x + e \end{equation} where $x$ is a value of the independent variable, $Y$ is the response, $\alpha$ and $\beta$ are the parameters, and $e$ is a random error with a mean of 0.

Then this textbook talks about estimators. It says $A$ is an estimator of $\alpha$, and $B$ is an estimator of $\beta$. Also, $A+Bx$ is an estimator of $\alpha+\beta x$. However, according to the textbook, $A+Bx$ is an estimator of $Y$. I don't understand that $A+Bx$ can be an estimator of $Y$, since $Y$ is a random variable but not a parameter. Am I wrong?

To me it makes sense only to say that $A+Bx$ is an estimator of the 'mean' of the random variable $Y$. Then, given that the mean of $e$ is zero, $A+Bx$ is an estimator of the mean of $\alpha + \beta x + e$.

$\endgroup$
2
$\begingroup$

You are right*, we are estimating parameters but not (realizations of) random variables. We would predict those instead.

*Except for the last bit where you say

Then, given that the mean of $e$ is zero, $A+Bx$ is an estimator of $\alpha + \beta x + e$.

Note that $\alpha+\beta x+e$ is a random variable and it equals $Y$. If you oppose to calling $A+Bx$ and estimator of $Y$, you should oppose calling it an estimator of $\alpha+\beta x+e$ for the sake of internal consistency.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer. You're right about my last sentence. I edited my original post. $\endgroup$ – Nownuri Dec 19 '20 at 10:02
0
$\begingroup$

When we say $A$ is an estimator of $\alpha$ it means that $\mathbb{E}(A)=\alpha$. Since $A+Bx$ is a random variable(because of $\alpha$ and $\beta$) we can write: $$\mathbb{E}(A+Bx)=\mathbb{E}(A)+\mathbb{E}(Bx)+0=\mathbb{E}(A)+\mathbb{E}(B)x+\mathbb{E}(e)=\mathbb{E}(Y)$$ Hope it helps.

$\endgroup$
0

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.