For $a_1\le b_1$ and $a_2\le b_2$, show that $$P\{ a_1<X_1\le b_1,a_2<X_2\le b_2 \}= F(a_1,a_2)+F(b_1,b_2)-F(b_1,a_2)-F(a_1,b_2)$$ Here $F(a,b)=P\{X_1\le a,\ X_2\le b\}$, the cumulative distribution function.

I know that I need to write down the probability of rectangle as a combination of four probabilities which then I would write down as distribution functions of F. But I don't know how.

  • $\begingroup$ Figured out myself: $P(X_1<=b_1,X_2<=b_2)+P(X_1<=a_1,X_2<=a_2)-P(X_1<=a_1,X_2<=b_2)-P(X_1<=b_1,X_2<=a_2)$ =$F(a1,a2)+F(b1,b2)−F(b1,a2)−F(a1,b2)$ $\endgroup$
    – Jonas
    Oct 20, 2015 at 2:15
  • $\begingroup$ Use \le and \ge to get $\le$ and $\ge$ $\endgroup$
    – user147263
    Oct 20, 2015 at 13:15

1 Answer 1


Figured out myself: $P\{ a_1<X_1\le b_1,a_2<X_2\le b_2 \}$ can be written as $$P(X_1\le b_1,X_2\le b_2)+P(X_1\le a_1,X_2\le a_2)-P(X_1\le a_1,X_2\le b_2)-P(X_1\le b_1,X_2\le a_2)$$ which is equal to $$ F(a_1,a_2)+F(b_1,b_2)−F(b_1,a_2)−F(a_1,b_2)$$ -- Justin


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