1
$\begingroup$

Wikipedia states

[Multiple correlation] is the correlation between the variable's values and the best predictions that can be computed linearly from the predictive variables.

But lecture notes state that the square of multiple correlation between $y$ and a vector of variables $X$ is equal to

$$\frac {Y^TX(X^TX)^{-1}X^TY}{Y^TY}$$

Which is equal to $$\frac {Y^T \hat Y}{Y^T Y}$$

(Where I assume that the first column of $X$ contains only $1$’s, so that there is a constant term in the model). But if the multiple correlation is really equal to the correlation between $Y$ and $\hat Y$, then the its square should rather be equal to $\frac {(Y^T \hat Y)^2}{(Y^T Y)\cdot (\hat Y^T \hat Y)}$.

I don't see that these are the same. Is wikipedia wrong, are my lecture notes wrong, or are these equal?

$\endgroup$
3
  • $\begingroup$ I'm not sure I believe any of these formulas are generally true. Are you perhaps implicitly assuming all variables have been centered? $\endgroup$
    – whuber
    Commented May 22, 2018 at 13:27
  • $\begingroup$ @whuber, no I am not, but I am assuming that $X$ contains a vector of 1’s and should have stated this. Does that solve your concern? $\endgroup$
    – user56834
    Commented May 22, 2018 at 15:05
  • $\begingroup$ Not entirely, because I still don't recognize valid formulas for correlation coefficients: they should have denominators proportional to the standard deviations of $Y$ and $\hat Y$ rather than to their norms. $\endgroup$
    – whuber
    Commented May 22, 2018 at 15:23

1 Answer 1

2
$\begingroup$

So, if we define the $n$ $\times$ $k$ matrix of variables $X$ (respectively, number of observations and number of independent variables) as including the constant, let say $X$ = $($$1_n$ $X^*$$)$, then the $R^2$ (in this case called "centered $R^2$") coincides with the squared simple correlation between $Y$ and $\hat{Y}$.

Indeed, defining:

$M_{[1]}$ =$I_{n}$ $-$ $1_n$$($$1'_n$$1_n$$)^{-1}$$1'_n$

where $I_{n}$ is the $n$ $\times$ $n$ matrix with all 1 on its main diagonal and 0 elsewhere, while $1_n$ is the $n$ $\times$ $1$ vector of all 1, then we have that:

$R^2$ = $\hat{Y}$$'$$M_{[1]}$$\hat{Y}$ $/$ $Y'$$M_{[1]}$$Y$

while the squared simple correlation between $Y$ and $\hat{Y}$ is given by:

$($$r^2_{Y, \hat{Y}}$$)^{2}$ = $($$\hat{Y}$$'$$M_{[1]}$$Y$$)^2$ $/$ $\hat{Y}$$'$$M_{[1]}$$\hat{Y}$$Y'$$M_{[1]}$$Y$

The proof of this fact follows from the fact that $\hat{Y}$$'$$M_{[1]}$$Y$ = $\hat{Y}$$'$$M_{[1]}$$\hat{Y}$ which, in turn relies on the fact that $Y$ = $\hat{Y}$ $+$ $E$, where $E$ is the residual vector.

The problem with your equations is that the first one represents the so called "Uncentered $R^2$", that is the $R^2$ for regression in which the $1_n$ vector is $not$ included in $X$. Under a statistical point of view, this means that we are omitting the constant from the regression.

In this case, the equivalence between the $R^2$ and the squared simple correlation between $Y$ and $\hat{Y}$ does not hold, unless all the variables have $0$ sample mean.

Hope the explanation was clear.

Best Regards,

Niccolo'.

$\endgroup$

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.