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From this source, the estimation of the coefficient of correlation is

$$r = \frac{\Sigma (X_i-E[X])(Y_i-E[Y])}{\sqrt { \Sigma (X_i-E[X])^2 \Sigma (Y_i - E[Y])^2}}$$

If the coefficient correlation is null ($\space\rho = 0\space$), then its estimator $r$ is normally distributed with mean equal 0.

Is this valid whatever is the distribution of $X$ and $Y$? Why?

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    $\begingroup$ The normality/Gaussianity will be approximate: The correlation cannot be less than -1 or more than 1, but a normal/Gaussian sampling distribution won't respect those bounds. $\endgroup$ – Maarten Buis Jun 25 '14 at 9:41
  • $\begingroup$ @Maarten Although I suspect you didn't mean this, you seem to be saying that when $(X_i)$ and $(Y_i)$ are samples of Normal distributions (and the expectations are replaced by means) then $r$ could lie outside the interval $[-1,1]$. That is not possible, no matter what the underlying distribution of $(X,Y)$ might be. Moreover, these are the best possible bounds, because $r$ computed from a sample of a bivariate Normal distribution can have any value in this interval. That's why I don't understand what you mean by "won't respect these bounds." $\endgroup$ – whuber Jun 25 '14 at 15:14
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    $\begingroup$ @whuber I was talking about the sampling distribution of $r$ not about the distribution of either $X_i$ or $Y_i$. If the sampling distribution of $r$ was normal/Gaussian, then in principle it could take any number between $-\infty$ and $+\infty$, which does not respect the known bounds for a correlation $[-1,1]$. So the normal/Gaussian sampling distribution for $r$ cannot be strictly true but only an approximation. $\endgroup$ – Maarten Buis Jun 25 '14 at 18:47
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Oy veh iz mir...

The source is wrong. In fact, the source acknowledges that it is wrong immediately below the claim when it states that under $H_o:\rho=0$ and when $(X, Y)$ are jointly Normal,

$T = \frac{r}{\sqrt{1-r^2}}\sqrt{n-2} \sim T_{n-2}$ .

Consider the source: the slides are from a BIOLOGY course. Many biologists are literate statisticians. Many statisticians are literate biologists. Very few biologists are experts in statistics, and especially mathematical statistics. Very few statisticians are experts in biology.

The proof of the fact (that properly scaled, $R$ follows a $T_{n-2}$ distribution when $\rho = 0$) relies on the underlying Normal. I'm not sure how distributionally robust that result is. What is also true (and is pretty distributionally robust) is that

$Z = \frac{1}{2} \ln \frac{1+r}{1-r} \sim N(\zeta, \frac{1}{n-3})$,

where $\zeta = \tanh^{-1}\rho$. (Incidentally, $Z = \tanh^{-1} r$.)

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On slide 2, they assume that $X,Y$ being only normally distributed.

Then if $\rho=0$, which is exactly their correlation parameter, their estimated correlation would be $r=0$ on average naturally, so its mean $0$ aswell.

Why $r$ would also be normally distributed usually requires a longer proof. It may be related to the central limit theorem, that the mean is normally distributed for $n\rightarrow \infty$.

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One possibility:

You can divide both numerator and denominator by $n$, so the numerator become an average, suggesting we consider the CLT.

So you might try to apply CLT to that scaled numerator (i.e. the covariance), which application requires some additional manipulation.

If you can apply the CLT to the numerator and invoke Slutsky's theorem, you should be able to show* that asymptotically the sample correlation is normal.

*(if the necessary conditions for both those theorems hold)

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    $\begingroup$ To whoever downvoted: It might help to improve my answer if you clarified what you thought the problem was. I'd appreciate an opportunity to give a better answer. $\endgroup$ – Glen_b Jun 25 '14 at 21:37

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