1
$\begingroup$

I have to show that that if the conditional mean $E( Y\mid X)$ is linear in $X$, and $\operatorname{Var} (Y\mid X)$ is a constant, then

$\operatorname{Var}(Y\mid X) =( \operatorname{Var} Y) [1 - r^2 ] $, where $r$ is the correlation coefficient.

Regression equation is given by

$E(Y\mid X=x) = E(Y) - [ r \operatorname{Var} Y ( x - E(X) ] / \operatorname{Var} X $

Please suggest how to proceed.

  • 1
    Thanks for the effort. Two more hints: Displayed equations (equations on a line by themselves, as opposed to inline) look better and are easier to read when you format them in display style by enclosing them in double dollar signs. And function names such as "Var" get interpreted as a juxtaposition of variables names if you just write them out like that; that causes them to be italicized and to get the wrong spacing. To get proper formatting for those, you can use `\operatorname{Var}` to produce $\operatorname{Var}$. There are also predefined commands for frequently used functions like `\sin`.2012-04-19

1 Answers 1

1

Actually you need $ \mathbb{E}(Y \mid X=x) = \mathbb{E}(Y) + \frac{r\sqrt{\operatorname{var}(Y)}}{\sqrt{\operatorname{var}(X)}}(X - \mathbb{E}(X)). $

So $ \operatorname{var}(\mathbb{E}(Y\mid X)) = \left(\frac{r\sqrt{\operatorname{var}(Y)}}{\sqrt{\operatorname{var}(X)}}\right)^2 \operatorname{var}(X-\mathbb{X}) = \frac{r^2\operatorname{var}(Y)}{\operatorname{var}(X)} \operatorname{var}(X) = r^2\operatorname{var}(Y). $

In other words, the square of the correlation is what fraction of the variability of $Y$ is "explained" by the variability of $X$.

The "law of total variance" partitions $\operatorname{var}(Y)$ into "explained" and "unexplained" parts: $ \operatorname{var}(Y) = \operatorname{var}(\mathbb{E}(Y\mid X)) + \mathbb{E}(\operatorname{var}(Y\mid X)). $ In the special case you're considering, the second term---the "unexplained" part---is the expected value of a constant $\operatorname{var}(Y\mid X)=(1-r^2)\operatorname{var}(Y)$.