I'm reading Shiryaev's Probability and I'm in the section where he introduces the probability measure in $\left(\mathbb{R}^n, \mathbb{B}(\mathbb{R}^n)\right)$. Now he uses a difference operator $$ \Delta_{a_j,b_j} F(x_1, \ldots, x_n) = F(x_1, x_2, \ldots, x_{j-1}, b_j, \ldots x_n) – F(x_1, x_2, \ldots, x_{j-1}, a_j, \ldots x_n), $$ where $F(x_1, \ldots, x_n)$ is a distribution function, meaning that for $\Delta_{a_j,b_j}$ you're working with coordinates $x_j$ only while the rest is fixed.
He states the following theorem without proof:
$$ P(a, b] = \Delta_{a_1,b_2} \ldots \Delta_{a_n,b_n} F(x_1, \ldots, x_n),$$ for $(a,b]= (a_1, b_1] \times \ldots \times(a_n, b_n] $.
I'm trying to prove it for the bidimensional case, meaning that:
$$P(a,b] = F(b_1, b_2) – F(a_1, b_2) + F(a_1, a_2) – F(b_1, a_2)$$
But I'm clueless on where to start this and I'd like some advice or reference for this proof. Thanks!!
Best Answer
The 2-dimensional identity should read $$ P(a,b] = F(b_1,b_2)-F(a_1,b_2)-F(b_1,a_2)+F(a_1,a_2). $$ Let $Q(c,d):=\{(x,y): x\le c, y\le d\}$. I'd start by drawing a figure showing how the rectangle $(a,b]$ can be represented as $$ [Q(b_1,b_2)\setminus Q(a_1,b_2)]\cup[Q(b_1,a_2)\setminus Q(a_1,a_2)]. $$ (This is a disjoint union.)