Conditioning on a function of a random variable

Question

Let $X, Y$ denote random variables. Let $f(\cdot)$ denote a real-value function. I know that $$P(X|Y, f(Y))$$ is equivalent to $$P(X|Y).$$

Now suppose I have

$$P(X|a \leq Y \leq b, f(Y))$$

where $a, b$ are constants. Is this equal to

$$P(X|a \leq Y \leq b)?$$

Answer 1

No.

As an example, suppose

• $Y=-1$, $0$, $1$ with equal probabilities $\frac13$,
• $X=Y$
• $f(Y)=Y$
• $a=0$ and $b=1$.

Then

• $P(X=x \mid 0≤Y≤1,f(Y)) = \mathbf{1}_{[Y=x]}$, i.e. $1$ when $Y=x \in \{0,1\}$ and $0$ otherwise
• $P(X=x \mid 0≤Y≤1) = \frac12$ for $x \in \{0,1\}$ and $0$ otherwise

and these are not the same