Chapter 3: Q.3.26 (page 109)

Prove the equivalence of Equations (5.11) and (5.12).

Short Answer

Expert verified

P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F) \Rightarrow P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)

Both directions are proven so the equivalence is correct.

Step by step solution

Given Information

Prove:

P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F) \Leftrightarrow P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)

Both of those are the defining conditions of conditional independence.

Explanation

Suppose that for $E_{1}, E_{2}, F$ the right equation ${}^{*}: P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)$ is true.

$P (E_{1} ∣ E_{2} F) = \frac{P (E_{1} E_{2} F)}{P (E_{2} F)} = \frac{\frac{P (E_{1} E_{2} F)}{P (F)}}{\frac{P (E_{2} F)}{P (F)}} = \frac{P (E_{1} E_{2} ∣ F)}{P (E_{2} ∣ F)} \overset{*}{=} P (E_{1} ∣ F)$

Taking the leftmost and the rightmost expression this proves the left equation in the hypothesis, i.e.:

P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F) \Leftarrow P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)

Explanation

For the other direction, suppose that for $E_{1}, E_{2}, F$ the right equation ${}^{* *}: P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F)$ is true.

$P (E_{1} E_{2} ∣ F) = \frac{P (E_{1} E_{2} F)}{P (F)} = \frac{P (E_{1} ∣ E_{2} F) P (E_{2} ∣ F) P (F)}{P (F)} = P (E_{1} ∣ E_{2} F) P (E_{2} ∣ F)$

$\overset{* *}{=} P (E_{1} ∣ F) P (E_{2} ∣ F)$

This proves the right equation in the hypothesis, i.e.:

P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F) \Rightarrow P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)

Both directions are proven so the equivalence is correct.

Final Answer

$P (E_{1} ∣ E_{2} F) = P (E_{1} ∣ F) \Rightarrow P (E_{1} E_{2} ∣ F) = P (E_{1} ∣ F) P (E_{2} ∣ F)$

Both directions are proven so the equivalence is correct.

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Prove the equivalence of Equations (5.11) and (5.12).

Short Answer

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