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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 ?

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    $\begingroup$ 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? $\endgroup$
    – user0102
    Mar 24, 2021 at 7:04
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    $\begingroup$ 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 $P(X)P(Y\cup Z)$ $\endgroup$
    – userj
    Mar 24, 2021 at 7:07
  • $\begingroup$ @user1337 $P((X \cap Y) \cup (X \cap Z))$ i guess? $\endgroup$
    – BCLC
    Mar 24, 2021 at 7:14
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    $\begingroup$ @JohnSmithKyon I think so $\endgroup$
    – user0102
    Mar 24, 2021 at 7:20
  • $\begingroup$ @user1337 posted answer. $\endgroup$
    – BCLC
    Mar 24, 2021 at 7:20

2 Answers 2

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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

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

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$P(A\cup B)=P(A)+P(B)$ 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 $P(A\cap B)=0$.

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