I need to prove that F is not joint cdf of random vector (X,Y) $$F(x,y)=\begin{cases} 0, & x<1 \; \lor \; y<1 \\ 0.5, & x,y\in[1,2) \\ 1, & \text{elsewhere} \end{cases}$$ It's clear to me that it shouldn't be from the very graph of this function, but how to prove it in formal way? Thank you in advance.
Prove that F is not joint cdf of random vector (X,Y)
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probability-theory
random-variables
1 Answers
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Ok, found the answer after 5 minutes after posting, so decided to share it. Notice, that $F(1,3)=P(X\leq 1, Y\leq 3) = 1$ and same goes for $F(3,1)$ according to the formula. So both of these "events" are certain, thus their intersection has probability equal to $1$ as well. But this intersection is $\{\omega \in \Omega :X(\omega),Y(\omega)\leq 1\}$ and $F(1,1)=\frac{1}{2}\neq 1$. Consequently it is not a joint cdf.