How to find cumulative distribution function from probability measure?

Question

We define $P$ as the following: $$P(B) = \begin{cases} 0 & (1,1), (-1, -1) \notin B \\ \frac{1}{2} & (1,1) \in B, (-1, -1) \notin B \\ \frac{1}{2} & (1,1) \notin B, (-1, -1) \in B \\ 1 & (1,1) \in B, (-1, -1) \in B \\ \end{cases} $$

Determine cdf for $P$.

Since this is in $R^2$ I know that I can use the cdf formula $F(x_{1}, x_{2}) = P((-\infty, x_{1}]\times(-\infty, x_{2}]).$

I also can see that you can plug in for the points $(1,1), (-1, -1)$ in the formula and break it up into disjoint sets. However, I don't see how this is the answer for the cdf.

$$F(x_{1},x_{2}) = \begin{cases} 0 & x_{1} < -1, x_{2} < -1\\ \frac{1}{2} & -1\leq x_{1} < -1, -1\leq x_{2} < \infty \\ \frac{1}{2} & -1\leq x_{1} < \infty, -1\leq x_{2} < 1 \\ 1 & x_{1} \geq 1,x_{2} \geq 1 \\ \end{cases} $$

Answer 1

$$\def\<{\langle}\def\>{\rangle} P(B) = \begin{cases} 0 & \<1,1\>\notin B, \<-1, -1\> \notin B \\ \frac{1}{2} & \<1,1\> \in B, \<-1, -1\> \notin B \\ \frac{1}{2} & \<1,1\> \notin B, \<-1, -1\> \in B \\ 1 & \<1,1\> \in B, \<-1, -1\> \in B \\ \end{cases}$$

So you wish to define $F(x_1,x_2) := P((-\infty..x_1]\times(-\infty..x_2])$.

For example, $F(0,0)=P((-\infty..0]\times (-\infty..0])$, thus $F(0,0)=1/2$ because $\<-1,-1\>$ is in the interval, but $\<1,1\>$ is not.

Let us study the cases:

• $\<-1,-1\>$ will be in the interval exactly when both $x_1,x_2$ are at least $-1$.
• $\<1,1\>$ will be in the interval when both $x_1,x_2$ are at least $1$.
  • So, if both $x_1,x_2$ are at least $-1$ but at least one among them is less than $1$, then $\<-1,-1\>$ will be in the interval, but $\<1,1\>$ will not. This occurs when $(-1\leq x_1< 1\land -1\leq x_2)\lor(1\leq x_1\land -1\leq x_2< 1)$.

Hence we have:

$$F(x_{1},x_{2}) = \begin{cases} 0 &:& \qquad ~~x_1<-1 &\lor& \qquad ~~x_{2} <-1\\ \frac{1}{2} &:& -1\leq x_1< 1&,& -1\leq x_{2} \\ \frac{1}{2} &:& ~~~1\leq x_{1} &,& -1\leq x_{2} < 1 \\ 1 &:& ~~~1\leq x_{1}&,& ~~~1\leq x_{2} \end{cases} $$