# show that $P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$ if $X$ and $Y$ are independent,$X$ and $Z$ are independent, and $P(Y\cap Z) = 0$

Show that $$P(X\cap Y)\cup P(X\cap Z)=P(X\cap Y)+P(X\cap Z)$$ if $$X$$ and $$Y$$ are independent, $$X$$ and $$Z$$ are independent, and $$P(Y\cap Z) = 0$$

This makes sense intuitively if one draws a venn diagram. But how can one justify that $$X\cap Y$$ and $$X\cap Z$$ are mutually exclusive with probability rules ?

• Welcome to MSE! It is a little bit unclear to me what you mean by $P(X\cap Y)\cup P(X\cap Z)$. Can you please explain further? Mar 24, 2021 at 7:04
• I'm trying to prove that events $X$ and $Y\cup Z$ are independent, so I'm trying to prove that the Probability of the intersection of two events are simply $P(X)P(Y\cup Z)$ Mar 24, 2021 at 7:07
• @user1337 $P((X \cap Y) \cup (X \cap Z))$ i guess?
– BCLC
Mar 24, 2021 at 7:14
• @JohnSmithKyon I think so Mar 24, 2021 at 7:20
• @user1337 posted answer.
– BCLC
Mar 24, 2021 at 7:20

Part 1. For the venn diagram, what you're doing is thinking of disjoint. Instead, think of almost disjoint.

$$Y$$ and $$Z$$ aren't disjoint but they're almost disjoint (defined as that probability of their intersection is zero). In this case, we're not sure that $$X \cap Y$$ and $$X \cap Z$$ are disjoint but we're sure that they're also almost disjoint. Then $$P((X \cap Y) \cup (X \cap Z)) = P(X \cap Y) + P(X \cap Z)$$.

Remark: Here, I do not use any independence assumption, I think.

Part 2. Oh you said in comments you wanted to show $$X$$ and $$Y \cup Z$$ are independent. Well you can use $$\sigma$$-algebras or...(wait I think you actually figured out the following but anyway)

$$P(X \cap Y) = P(X)P(Y)$$

$$P(X \cap Z) = P(X)P(Z)$$

Then

$$P(X \cap (Y \cup Z)) = P(X)(P(Y)+P(Z))$$

Again use almost disjoint to say

$$= P(X) P(Y \cup Z)$$

$$P(A\cup B)=P(A)+P(B)$$ is true if $$P(A\cap B)=0$$. This is because we always have $$P(A\cup B)+P(A\cap B)=P(A)+P(B)$$.

Now just take $$A=X\cap Y$$ and $$B=X \cap Z$$. Note that $$A\cap B \subset Y\cap Z$$ so $$P(A\cap B)=0$$.