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I am trying to find an example of two random variables with joint CDF: $$F(x, y) = \mathcal{P} ( X\leq x, Y \leq y) = I_{(x+y\geq 1)}.$$

I can say that there is one such vector because given CDF is actually a CDF i.e. it satisfies all required conditions. And there is a theorem that states that in that case a random vector with given CDF exists. Is my statement correct?

It is obvious that $X$ and $Y$ are dependent and this CDF seems so simple yet I can't think of any example.

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It's not obvious that cumulative distribution function is even a meaningful concept for a random variable whose range does not have an intrinsic total order. Do you have an explicit definition of CDF that applies in this case? – Henning Makholm Oct 13 '12 at 18:51
Sorry, I don't understand what do you mean by that. You think that such $X$ and $Y$ don't exist? Why are you assuming that $X$ and $Y$ have a range that 'does not have an intrinsic total order'? – grozhd Oct 13 '12 at 19:28
x @grozhd: No, I think it doesn't make sense to talk about CDFs for random variables that are not real-valued. In particular it doesn't make sense to me to speak about CDFs for random variables whose values are vectors. – Henning Makholm Oct 13 '12 at 19:30
Maybe I have formulated my question incorrectly. I meant that $X$ and $Y$ are real-valued random variables. And $F(x, y)$ is their joint CDF. – grozhd Oct 13 '12 at 19:34
x @grozhd: Each of $X$ and $Y$ can well be real valued and have a CDF, but their combination $(X,Y)$ is not real-valued. Its possible values are pairs or reals, not reals. What does a CDF with two arguments even mean to you? For example, how would you express $F(3,5)$ as a probability? – Henning Makholm Oct 13 '12 at 19:37
up vote 2 down vote accepted

The "joint CDF" you propose is not possible. It would give us

$$P(X\le 2,Y\le 0) = P(X\le 0,Y\le 2) = 1$$

so therefore $(X,Y)$ would almost certainly satisfy each of $Y\le 0$ and $X\le 0$. But the same joint CDF claims that

$$P(X\le 0, Y\le 0)=0$$

which is a contradiction.

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