I have two RV $X$ and $Y$, and I am asked to find the probability of $P(X=Y)$. To tackle to problem, I've conditioned on $X$ and used the law of total probability to write :
$ P(X = Y) = \displaystyle \sum_{x \in S_X} P(Y = X | X = x) P(X = x) $
But now I'm wondering if I can write that $P(Y = X | X = x) = P(Y = x | X = x) = P(Y = x)$ without any information about dependence/independence between $X$ and $Y$, can someone tell me if the above equality holds, and if yes why ?
Here the whole problem :
Let $X$ be a r.v such that $G_X(z) = \dfrac{1}{(2-z)^5}$. Let $Y$ be a geometric r.v of parameter $p$. Find $P(X = Y)$. Hint : Do not try to compute $P(X=n)$.
Thanks !
You can write $P(Y=X \mid X=x)$ as $P(Y=x \mid X=x)$ but then you cannot simplify further without resolving the dependence structure of $X$ and $Y$.
This way of proceeding to begin with it is also assuming $X$ and $Y$ are discrete, which maybe they are.