Q.3.26 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

$P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right) \Rightarrow P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

Both directions are proven so the equivalence is correct.

Step 1: Given Information

Prove:

$P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right) \iff P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

Both of those are the defining conditions of conditional independence.

Step 2: Explanation

Suppose that for $E_1,E_2,F$ the right equation ${}^{*}:P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$is true.

$P\left(E_1\mid E_{2}F\right)=\frac{P\left(E_1{E}_{2}F\right)}{P\left(E_{2}F\right)}=\frac{\frac{P\left(E_1{E}_{2}F\right)}{P\left(F\right)}}{\frac{P\left(E_{2}F\right)}{P\left(F\right)}}=\frac{P\left(E_1{E}_2\mid F\right)}{P\left(E_2\mid F\right)}\stackrel{*}{=}P\left(E_1\mid F\right)$

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

$P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right)\Leftarrow P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

Step 3: Explanation

For the other direction, suppose that for $E_1,E_2,F$ the right equation ${}^{**}:P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right)$ is true.

$P\left(E_1{E}_2\mid F\right)=\frac{P\left(E_1{E}_{2}F\right)}{P\left(F\right)}=\frac{P\left(E_1\mid E_{2}F\right)P\left(E_2\mid F\right)P\left(F\right)}{P\left(F\right)}=P\left(E_1\mid E_{2}F\right)P\left(E_2\mid F\right)$

$\stackrel{**}{=}P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

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

$P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right) \Rightarrow P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

Both directions are proven so the equivalence is correct.

Step 4: Final Answer

$P\left(E_1\mid E_{2}F\right)=P\left(E_1\mid F\right) \Rightarrow P\left(E_1{E}_2\mid F\right)=P\left(E_1\mid F\right)P\left(E_2\mid F\right)$

Both directions are proven so the equivalence is correct.