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In this question, why is it that

$\mathbb P(X\in B)-\mathbb P(Y\in B)=\mathbb P(X\in B \setminus Y\in B)$

?

I think it should be $\mathbb P(X\in B \setminus Y\in B) = \mathbb P(Y\in B \setminus X\in B) = 0$

$\to \mathbb P(X\in B) - P(X\in B \cap Y\in B) = 0$ and $\mathbb P(Y\in B) - P(X\in B \cap Y\in B) = 0$

$\to P(X\in B) = P(X\in B \cap Y\in B) = \mathbb P(Y\in B)$

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1 Answers1

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It is not true that $P(A \setminus C) = P(A) - P(C)$.

$P(A \setminus C) = P(A \cap C^C) = P(A) - P(A \cap C)$ since probability is finitely additive since it is countably additive

A sufficient condition for $P(A \setminus C) = P(A) - P(C)$ is $C \subseteq A$ since

$P(A \setminus C) = P(A) - P(A \cap C) = P(A) - P(C)$

Now $\mathbb P(X\in B \setminus Y\in B) = \mathbb P(X\in B) - \mathbb P(X\in B \cap Y\in B)$ is true with $A = (X\in B)$ and $C = (Y\in B)$ in $P(A \setminus C) = P(A \cap C^C) = P(A) - P(A \cap C)$

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