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Let X~U([0,1]) and Y=X, find the CDF $ $ $F_{(X,Y)}(x,y)$. Here is my attempt:

$$F_{(X,Y)}(x,y)=P({{X\le x} \cap {Y\le y}})$$ $Since \; \mathbf {Y=X} \; we \; have$ $$P({{X\le x} \cap {X\le y}})=$$ $Let \; \mathbf {m = min\{x,y}\}, \; thus$ $$P({{X\le m} \cap {X\le m}})= P({X\le m})=m$$ $While \; 0 \le x,y \le 1$

I know this is wrong as I have found the (possibly) correct (partial) solution elsewhere, but I don't understand why that is the case.

Thanks in advance

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    What is the other solution? – 2017-01-17
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    ${min\{x,y}\}^2 \; while \; 0 \le x,y \le 1\\$ In the other solution the person never get's rid of Y – 2017-01-17
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    Then they are wrong. What is the source? – 2017-01-17
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    So my solution is actually correct, it's simply ${min\{x,y}\}$, without the squared? The solution is from a classmate, so not very credible. However, he seems to have a better understanding of the material than me, so I assumed I must be wrong. (have been banging my head to the wall for over an hour not understanding what I'm doing wrong) – 2017-01-17

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If the problem is $Y = X$, I believe your solution is right.

If the problem is $X,Y$ are i.i.d., then the solution should be $P(X \leq x, Y \leq y) = xy$.