Conditional Probability – Problem (17/365)

Posted on September 28, 2016 by Naren Sundar

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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