## Correlation – I (6/365)

Let me follow up on the theme – talked about here – of distributing an operator over another. To recap, we saw two ways of distributing an expectation $\displaystyle E(\theta + \phi) = E(\theta) + E(\phi) \\ (E \theta \phi)^2 = E \theta^2 E \phi^2 \\ E\theta \phi = E\theta E\phi \text{ if } \theta, \phi \text{ are independent}$

Can we say anything like this about variance? Consider the variance of the sum of random variables (recall that $V\theta = E(\theta - E\theta)^2$). $\displaystyle V(\theta + \phi) = V\theta + V\phi + 2E(\theta - E\theta)(\phi - E\phi)$

Looks like we have an extra term. Before trying to see what this is, can we think of a situation when this term could be $0$ and thus give us $V(\theta + \phi) = V\theta + V\phi$? The answer is yes because if $\theta$ and $\phi$ are independent, then $2E(\theta - E\theta)(\phi - E\phi) = 2E\theta\phi - 2E\theta E\phi = 2E\theta \phi - 2E\theta \phi = 0$.

In general, though, this extra term will be non-zero. How can we make sense of this extra term? If you go back to the Cauchy-Bunyakovskii inequality we looked at, we can write this term as $\displaystyle (E(\theta - E\theta)(\phi - E\phi))^2 \le E(\theta - E\theta)^2 E(\phi - E\phi)^2 \\ (E(\theta - E\theta)(\phi - E\phi))^2 \le V\theta V\phi$

So it looks as if we can relate the extra term to the variance of the random variables. More specifically, we see that $\displaystyle | \rho(\theta,\phi) | = | \frac{E(\theta - E\theta)(\phi - E\phi)}{\sqrt{V\theta V\phi}} | \le 1$

And this gets to be called the correlation coefficient. Why is this interesting you ask? First, note that $\displaystyle \rho(\theta,\phi) = 0 \implies V(\theta + \phi) = V\theta + V\phi$

You’ve often heard how a lack of correlation does not imply independence of random variables. Well, the above identity is the conclusive proof. Because $\displaystyle \text{if } E(\phi - E\phi)(\theta - E\theta) = E\theta\phi - E\theta E\phi = 0 \\ \text{then } E\theta\phi = E\theta E\phi \text{ does not imply } \theta,\phi \text{ independent}$

Will furnish with an example of when this happens. It seems tricky to come up with an example.

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