Conditional Probability – Problem (17/365)

Going with the theme of trying to transplant probability laws to expectations consider the following. if A,B are independent events we know that P(A|B) = P(A). Now, if \theta,\phi are independent random variables then

\displaystyle  \text{for all x } E(\theta | \phi = x) \\ = \sum_\omega \theta(\omega) P(\omega | \phi = x) \\ = \sum_\omega \theta(\omega) P(\omega) \\ = E\theta

A question now asks to give an example of random variables \theta,\phi which are not independent but for which E(\theta | \phi) = E\theta.

\displaystyle  \text{Let } P(\omega) = \frac{1}{6} \\ \text{Let } \phi(1)=\phi(2)=\phi(3)=1 \text{ and } \phi(4)=\phi(5)=\phi(6)=0 \\ \text{Let } \theta(1)=1, \theta(4)=2, \theta(5)=-1 \text{ and rest is } 0 \\ \text{Note } P(\theta=0 | \phi=1) = \frac{2}{3} \ne P(\theta=0) = \frac{3}{6} \text{ so not independent} \\ \text{But } E(\theta | \phi=0) = E\theta = \frac{1}{3} = E(\theta | \phi=1)

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