# On the meaning of the equal sign when used to define the event of a r.v. taking some value.

Let $X$ and $Y$ be random variable and $k$ be some fixed constant.

For a long time I thought that $P(X+Y=k)=P(X=k-Y)$, but recently I read something like $$P(X - Y = k) = E_{Y} \Big( P(X-Y = k) \Big) = E_{Y} \Big( P(X = k+Y) \Big) = \sum_{y=0}^{\infty} P(Y=y)P(X = k+y)$$ that makes me wonder whether it is common to consider as random variables probabilities involving random variables being equal to a quantity that involves a random variable itself, like $P(X=k-Y)$.

I have no background in Probability Theory from a measure-theoretic point of view, but from my naïve understanding an explanation could be that $$P(X=k-Y)=P(\{\omega:\omega\in X^{-1}(k-Y)\})\neq P(\{\omega:\omega\in {(X+Y)}^{-1}(k)\})=P(X+Y=k)$$ where $-1$ obviously indicates the preimage of the argument respect to the function.

In other words, in one case "$P(\cdot)$" is used as a measure of the preimage of $k$ via $X+Y$, in the other case it's used as a function of $Y$ therefore becoming itself a r.v.

Then the confusion rise because of the equal sign being improperly used in place of the set-theoretic way to write down the probability that a r.v. takes some value.

Right?

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The notation $\{\omega : \omega\in X^{-1}(k-Y)\}$ is overly complicated, since the same set can be written simply as $X^{-1}(k-Y)$. One could reasonably write $\{\omega : X(\omega)=k-Y\}$ in order to define the set $X^{-1}(k-Y)$. –  Michael Hardy Dec 2 '12 at 2:30
@MichaelHardy sorry, I rewrote $\{\omega : X(\omega)=k-Y\}$ without noticing to become redundant. –  Emanuele Natale Dec 2 '12 at 2:42

Your understanding that $\Pr(X+Y=k)=\Pr(X=k-Y)$ is correct. Formally, $$\Pr(X+Y=k)=\Pr \{\omega: X(\omega) + Y(\omega) = k)\} = \Pr(X=k-Y).$$
Possibly, what you read was the following (for independent $X$ and $Y$): $$\Pr(X+Y=k)=\mathbb{E}_Y[\Pr_X(X + Y = k)].$$ Here, $\Pr_X(X + Y = k)$ is a random variable (which is a function of $Y$), and $\mathbb{E}_Y[\Pr_X(X + Y = k)]$ is its expectation.