Correlation – II (7/365)

In the last post, I said I’d give an example of two random variables \theta, \phi whose expectations are independent E\theta \phi = E\theta E\phi but are not themselves independent.

For a first attempt, this turned out not to be an easy task because when coming up with an example I had to control two things: 1) making sure I can get P(\theta = x | \phi = y) \ne p(\theta = x) for some x, y and 2) making sure I can get E\theta \phi = E\theta E\phi.

I realized that to make like easier I should try to get E\theta \phi = 0 which means I only have to make sure one of the expectations E\theta = 0 and I can use \phi to freely adjust to try and make P(\theta = x | \phi = y) \ne p(\theta = x).

Consider a fair six-sided die. Let

\displaystyle  \theta(2) = \theta(4) = \theta(6) = 1 \text{ i.e. when even} \\ \theta(1) = \theta(3) = \theta(5) = -1 \text{ i.e. when odd} \\ \text{therefore } E\theta = 0

Now, after playing around with \phi, I came up with this

\displaystyle  \phi(1) = 1 \\ \phi(2) = \phi(4) = \phi(6) = 0 \\ \phi(3) = \phi(5) = -0.5 \\ \text{therefore } E\theta\phi = \frac{1}{6} ((-1)(1) + (1)(0) + (-1)(-0.5) + (1)(0) + (-1)(-0.5) + (1)(0) ) = 0

Yay! Now, check for non-independence

\displaystyle  p(\phi = 1 | \theta=1) = \frac{0}{3} \ne p(\phi = 1) = \frac{1}{6}

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