3
$\begingroup$

I can't understand how this works:

$e$ is the error term and $x$ is the explanatory variable.

$$Var(e|x) = E(e^2|x) - [E(e|x)]^2$$

I know that $[E(e|x)]^2$ = 0 because $E(e|x) = 0$, and squaring 0 is still 0.

So that leaves $Var(e|x) = E(e^2|x)$

I am confused on this part.

This may be a clearer idea of what I am after:

I am interested in understanding how $E(e^2|x) - [E(e|x)]^2$ = $Var(e|x)$ How do I get $E(e^2|x) - [E(e|x)]^2$? It is just given as a fact in the text, without an after thought.

$\endgroup$
7
  • 1
    $\begingroup$ What exactly is confusing your here? $\endgroup$
    – mpiktas
    Commented Apr 16, 2012 at 6:20
  • 1
    $\begingroup$ And how is your question related to homoscedasticity? $\endgroup$
    – mpiktas
    Commented Apr 16, 2012 at 6:22
  • $\begingroup$ I am studying up some econometrics and looking at the assumptions. The section on constant variance assumption has this expression, but without any explanation of how or why it is. So I would have no Idea how to show that the variance of e given x is equal to the expected value of e squared given x. Is there a way to show this or is it just one of those it is what it is things. $\endgroup$
    – Travis
    Commented Apr 16, 2012 at 6:26
  • 2
    $\begingroup$ In general the definition is $Var(Y)=E[(Y-E[Y])^2]$ if $E[Y]$ exists and is finite. That can then also be written as $E[Y^2] - (E[Y])^2$ by expanding the square and simplifying. You just have a special case of this. $\endgroup$
    – Henry
    Commented Apr 16, 2012 at 7:58
  • 1
    $\begingroup$ So you are confused how the last expression was calculated? The derivation above is yours or comes from the book? Your math is ok, so it is not clear for me, whether you want confirmation whether it is ok, or is this acceptable way of deriving the last expression. $\endgroup$
    – mpiktas
    Commented Apr 16, 2012 at 8:42

2 Answers 2

4
$\begingroup$

I assume you are refering to linear regression. Thus we have $$y=x^T\beta+e$$ Now the homoscedasticity assumption means that the variance does not depend on $x$. so we have $$var[e|x]=var[e]$$ This means each observation is equally important for estimating the mean square error.

$\endgroup$
3
  • 2
    $\begingroup$ This answer is good, but based on the comment stream it's not clear that it actually responds to the doubts of the OP. (Some further clarification from the OP would help.) $\endgroup$
    – cardinal
    Commented Apr 16, 2012 at 14:31
  • $\begingroup$ I am interested in understanding how $E(e^2|x) - [E(e|x)]^2$ = $Var(e|x)$ How do I get $E(e^2|x) - [E(e|x)]^2$? It is just given as a fact in the text, without an after thought. $\endgroup$
    – Travis
    Commented Apr 16, 2012 at 23:02
  • 1
    $\begingroup$ @Travis, what definition for $Var(e|x)$ is given in your book? The formula $Var(e|x)=E(e^2|x)-[E(e|x)]^2$ can be thought as definition. Formal definition is $Var(e|x)=E([e-E(e|x)]^2|x)$. But $E([e-E(e|x)]^2|x)=E(e^2|x)-2E([eE(e|x)]|x)-E(E(e|x)^2|x)=E(e^2|x)-[E(e|x)]^2$, where we employ the properties of conditional expectation. Since this holds for any $e$ and any $x$ (assuming expression on which we operate exists) you can define conditional variance both ways. $\endgroup$
    – mpiktas
    Commented Apr 17, 2012 at 11:16
2
$\begingroup$

I'm dropping the $|x$ for more convenient notation. So read $E(e)$ as $E(e|x)$ if you prefer that...

By definition $Var(e)=E\Big(e-E(e)\Big)^2$. Since $E(e)=0$, we have $Var(e)=E\Big(e-E(e)\Big)^2=E(e-0\Big)^2=E(e^2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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