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I know $F_{XY}(x,y)$=$P[X \le x,Y \le y]$,but is $F_{XY}(-\infty,2)$ equal to zero?I wonder can it be calculated and be written as the form of $f_{XY}(x,y)$

This question also let me confused,is $F_{XY}(-\infty,\infty)$=$1$?Well, i know the sum of joint CDF must be equal to $1$,but according to $F_{XY}(x,y)$=$P[X \le x,Y \le y]$,it become

$F_{XY}(-\infty,\infty)$=$P[X \le -\infty ,Y \le \infty]$,so this shouldn't be equal to 1,so what is $F_{XY}(-\infty,\infty)$ equal to?

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    Inputting $\infty$ or $-\infty$ into a function defined on the real plane is a little bit sketchy, but under most sensible interpretations, yes, $F(-\infty,2) = F(-\infty,\infty) = 0$. – 2017-02-23
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    but why is F(−∞,2)=F(−∞,∞)=0? – 2017-02-23
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    Note that $\mathbb{P}(X \leq -\infty, Y \leq \infty) = \mathbb{P}(X \leq - \infty) = 0$ for any real-valued random variable $X: \Omega \to \mathbb{R}$ (since $\{X \leq -\infty\} = \{X=-\infty\} = \emptyset$). – 2017-02-23
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    I understand,thankyou – 2017-02-23

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