$X, Y$ are just random variables. I do know what $\mathcal{P}(X=Y)$ is but if I try to think about $\mathcal{P}(Y)$ i dont have idea, so the conditional $\mathcal{P}(X=Y|Y)$ is even more confusing to me...
Thanks
$X, Y$ are just random variables. I do know what $\mathcal{P}(X=Y)$ is but if I try to think about $\mathcal{P}(Y)$ i dont have idea, so the conditional $\mathcal{P}(X=Y|Y)$ is even more confusing to me...
Thanks
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I think that the Brian Tung's comment well points out a core of the problem; It depends on the context, the expression $P(X=Y)$ is ambiguous if both X and Y are all random variable.
Let me give you an example. Suppose that the X is discrete r.v. such that True if and only if the 6-sided dice roll's result is even(2, 4 or 6). And Y is also discrete r.v. such that True if and only if dice roll's result is 3 or 6.
Then what the expression $P(X=Y)$ means? it depends on context, but in this case we can think it's the probability of the result of dice roll is both multiplier of 2 and 3, or both not. The cases satisfy the condition is: 1, 5, 6. So $P(X=Y)=\frac{1}{2}$.
But, the expression $P(X=Y|Y)$ is ambiguous. the restricting condition - $Y$ - does not indicate any additional information about $Y$. So if you want to give some meaningful condition, you may set $Y$'s more condition. For example, $P(X=Y|Y=\text{True})$ is $\frac{1}{3}$, while $P(X=Y|Y=\text{False})$ is $\frac{2}{3}$.