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Let $X$ and $Y$ are both indicator variables that can take values $0$ or $1$. $P(X=1)=0.5$.

If $Y$ is defined only if $X=1$, then are $X$ and $Y$ correlated?

It is mentioned that

If $X=1$, we generate $Y$ from Bernoulli$(0.5)$ distribution.

Does it mean $P(Y=1\mid X=1)=P(Y=0\mid X=1)=0.5$?

But if $P(Y=1\mid X=1)=P(Y=0\mid X=1)$, aren't $X$ and $Y$ uncorrelated?

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    If $Y$ is defined only for part of the sample space it is *not* a random variable, so it is improper to talk about its correlation with random variable $X$. But informally, we can see that $Y$ is in some way dependent on $X$ is it is generated differently for different values. – 2017-02-03
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    @GrahamKemp $Y$ has a distribution. Doesn't if a variable has distribution, then it is a random variable? – 2017-02-03
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    The probability distribution must equal one when measured over the entire sample space. How do you do that when it is not defined everywhere over the sample space? – 2017-02-03
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    I need to know the value of $P(Y=1|X=1).$ All I have the information that: "$X$ is a indicator variable that can take values $0$ or $1$; $P(X=1)=0.5$; $Y$ can also take values $0$ or $1$; $Y$ is defined only if $X=1$; If $X=1$, we generate $Y$ from Bernoulli$(0.5)$ distribution." Can I get the value of $P(Y=1|X=1)$ from this information? – 2017-02-03
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    Ah, if that is *all* you wish to know then the statement is just that $Y$ has a *conditionally* Bernouli distribution *when* $X=1$, $$Y\mid X=1 ~\sim \mathcal{Ber}(0.5)$$ .. so indeed $\mathsf P(Y=1\mid X=1)= \mathsf P(Y=0\mid X=1) =0.5$, and hopefully $Y$ *does* have its distribution defined elsewhere for the condition of $X=0$ . – 2017-02-03
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    @GrahamKemp Thanks. But for $X=0$, we don't need to generate $Y$ at all. That is, if $X=0$, the algorithm stops there. – 2017-02-03
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    Algorithm? .... – 2017-02-03

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