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I understand how to prove the following using Venn diagrams (asked and answered in this question) but would also like to be able to show it using the general properties of distributions. The following comes from chapter 6 of Ross's First Course In Probability.

Let $P\{X\leq a \, \wedge Y\leq b\} = F(a,b)$ be the probability that the $X\leq a$ and $Y\leq b$. Prove the following claim:

$P\{a_1 \lt X \leq a_2 \,\, \wedge \,\, b_1 \lt Y \leq b_2 \} = F(a_2,b_2) + F(a_1,b_1) - F(a_1,b_2) - F(a_2,b_1)$

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AnonSubmitter85
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The way to show this is to express $A\cup B$ as the equivalent union of disjoint sets $A \cup A^c B$.

We will use the following notation:

$$P(X\leq a) = F_X (a)$$

$$P(Y\leq b) = F_Y (b)$$

$$P(X\leq a \cap Y \leq b) = F(a,b)$$

We will also use the following identity in the below solution:

$$ \begin{array}{rcl} P\{X > a \cap Y > b\} &=& 1 - P\{ X \leq a \cup Y \leq b\} \\ &=& 1 - P\{X \leq a\} - P\{Y \leq b\} + P\{ X \leq a \cap Y \leq b\} \\ &=& 1 - F_X (a) - F_Y (b) + F(a,b) \end{array} $$

We rewrite the original problem as

$$ \begin{array}{rcl} P(a_1 < X \cap X \leq a_2 \cap b_1 < Y \cap Y \leq b_2) &=& 1 - P(X\leq a_1 \cup X>a_2 \cup Y \leq b_1 \cup Y > b_2 ) \\ &=& 1 -P[ ( X\leq a_1 \cup Y \leq b_1) \cup ( X>a_2 \cup Y > b_2 ) ] \end{array} $$

Using $P(A \cup B) = P(A) + P(A^cB)$ and $P(A\cup B)=P(A) + P(B) - P(AB)$, we have

$$ \\ 1 - \left[ \left(F_X (a_1) + F_Y (b_1) - F(a_1,b_1) \right) + \left( P( (X > a_1 \cap Y > b_1 ) \cap (X > a_2 \cup Y > b_2 ) ) \right) \right] \\ $$

Since $\{X > a_1\} \supset \{X > a_2\}$ and $\{Y>b_1\} \supset \{Y>b_2\}$, the last term in the above equation is equal to

$$ \\ \begin{array}{rcl} P( (X>a_2 \cap Y>b_1) \cup (X>a_1 \cap Y>b_2) ) &=& P(X>a2 \cap Y>b1) + P(X>a_1 \cap Y > b_2)\\ && - P(X>a_2 \cap Y > b_2) \end{array} $$

Using the above identity for $P(X>a \cap Y>b)$ and combining terms yields:

$$ P(a_1 < X \leq a_2 \cap b_1 < Y \leq b_2 ) = F(a_1,b_1)+F(a_2,b_2) - F(a_1,b_2) - F(a_2,b_1) $$

AnonSubmitter85
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