Showing that $P(a<X_1\leq b,c<X_2\leq d)=F(b,d)+F(a,c)-F(a,d)-F(b,c)$

Question

Anyone can help me to prove this condition for rectangols of $\mathbb{R}^2$? I don't really know how to do it.

Thanks for any help!

Answer 1

We know that $F(x,y) = P(X_1 \leq x, X_2 \leq y)$.

By the definition of a distribution, we can say that:

$$ P(a < X_1 \leq b, X_2 \leq d) = P(X_1 \leq b, X_2 \leq d) - P(X_1 \leq a, X_2 \leq d) = F(b,d)-F(a,d). $$

Similarily $P(a < X_1 \leq b, X_2 \leq c) = F(b,c) - F(a,c)$.

Therefore:

$$ P(a < X_1 \leq b, c < X_2 \leq d) =P(a < X_1 \leq b, X_2 \leq d) - P(a < X_1 \leq b, X_2 \leq c) = F(b,d) + F(a,c) - F(a,d) - F(b,c) $$

To think about why this must be true, try drawing the rectangle out in $\mathbb{R}^2$, and then consider what does $F(b,d) + F(a,c)$ represent, and what you must subtract to get only the rectangle.