Probability Proof.

Question

Write a proof to show that $\mathbb{P}(X_1 \mid X_3) + \mathbb{P}(X_2 \mid X_3) - \mathbb{P}(C_1 \cap X_2 \mid X_3) = \mathbb{P}(X_1 \cup X_2\mid X_3)$ labeling theorems used for each step.

My attempt:

$$\begin{array}{rcll} \mathbb{P}(X_1 \cup X_2\mid X_3)&=& \frac{\mathbb{P}((X_1 \cup X_2)\cap X_3)}{\mathbb{P}(X_3)} \\ &=&\mathbb{P}((X_1 \mid X_3) \cap (X_2 \mid X_3))\\ &=& \mathbb{P}(X_1 \mid X_3)+\mathbb{P}(X_2 \mid X_3)-\mathbb{P}(X_1 \cap X_2 \mid X_3) \end{array}$$

Answer 1

Hint: Recall that $\Pr(W|X_3)=\dfrac{\Pr(W\cap X_3)}{\Pr(X_3)}$.

Then let $Y_1=X_1\cap X_3$, and $Y_2=X_2\cap X_3$.

Note that $(X_1\cap X_2)\cap X_3=Y_1\cap Y_2$ and $(X_1\cup X_2)\cap X_3=Y_1\cup Y_2$.

Then the thing you are wanting to prove basically comes down to $$\Pr(Y_1)+\Pr(Y_2)-\Pr(Y_1\cap Y_2)=\Pr(Y_1\cup Y_2),$$ which is a familiar fact. Just divide by $\Pr(X_3)$, and use the facts about $Y_1$ and $Y_2$ listed above.

Answer 2

I would use Bayes' rule everywhere on the left-hand side: \begin{align} \mathbb{P}(X_1 \mid X_3) + \mathbb{P}(X_2 \mid X_3) - \mathbb{P}(X_1 \cap X_2 \mid X_3)&= \frac{\mathbb{P}(X_1 \cap X_3)+\mathbb{P}(X_2 \cap X_3)- \mathbb{P}(X_1 \cap X_2 \cap X_3)}{\mathbb{P}(X_3)}\\ &=\frac{\mathbb{P}(X_1\cup X_2\cap X_3)}{\mathbb{P}(X_3)}\\ &=\mathbb{P}(X_1\cup X_2\mid X_3) \end{align}