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If $X,Y$ are random variables with $Y$ discrete, and I know that $Y \sim p(\cdot | X)$,

is it true that:

$$ P(X \leq x,Y=y) = P(Y=y|X \leq x)P(X \leq x) $$ ?

In other words, does it make sense to have a $\ \leq \ $ sign in the conditional?

How can I ultimately show that:

$$ P(X \leq x,Y=y) = \int_{-\infty}^{x} p(y|v)p(v)dv $$

where $p(y|v)$ is the conditional of $Y|X$ and $p(v)$ is the density of $X$?

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    Under the ordinary definition $P(A\mid B) = P(A\cap B)/P(B)$, the answer is yes as long as $P(X\leq x)>0$. For the second one, I am not sure which definition of conditional probability you are working with, but it may be understood as the defining property of regular conditional probability applied to this particular case. – 2017-02-27
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    The first result follows directly from the elementary definition of conditional probability. The second result follows from the (continuous) law of total probability / law of total expectation. – 2017-02-27

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In the probability expression, $X\leq x$ is an event.   It represents the set of outcomes whose $X$ measure is less than $x$.

It is quite permissable to condition on an event. $\mathsf P(Y=y\mid X\leq x)$ is a valid expression, and indeed it is true that:

$$\mathsf P(X\leq x, Y=y) = \mathsf P(Y=y\mid X\leq x)~\mathsf P(Y=y)$$

However, you do not want to use that.

$$\begin{align}\mathsf P(X\leq x, Y=y) ~ & = ~ \mathsf P(Y=y)\cdot\mathsf P(X\leq x\mid Y=y) \\[1ex] & = ~ p_Y(y)\cdot \int_{-\infty}^x f_{X\mid Y}(s\mid y)\operatorname d s \\[1ex] & = ~ \int_{-\infty}^x p_Y(y)\,f_{X\mid Y}(s\mid y)\operatorname d s \\[1ex] & = \int_{-\infty}^x f_{X,Y}(s, y)\operatorname d s \\[1ex] & = ~ \int_{-\infty}^x p_{Y\mid X}(y\mid s)\, f_X(s)\operatorname d s \end{align}$$

Although that's not so much a proof of anything, as a definition of how those measures must behave.

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    Is it valid to write down $\{Y=y|X \leq x\}$ as an event itself, like the event $\{Y=y\}$? – 2017-02-27
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    No, that is not valid. The condition separator is not a set operator, it is part of the measure function. The condition constrains where the probability is being measured. – 2017-02-27
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    For the conditional, is there a set analogue by way of indicator functions or does an analogue exist at all? thanks! – 2017-02-27