jointly discrete continuous & discrete probability - conditioning on a range Given discrete and continuous random variables, $X$ and $Y$, respectively, the following conditional probability can be computed:
\begin{equation}
P(Y \leq y_1 | X =x) = \int_{-\infty}^{y_1} f_{Y|X}(y|x)dy
\end{equation}
But say you wanted to compute $P(X=x| Y \leq y_1)$, where you're now conditioning over a range, how would you compute it using the above approach? Typically, I would compute it using Bayes rule, but I wanted to try to derive an expression analogous to the above for $P(X=x| Y \leq y_1)$ (one that involves integrating perhaps the pdf of $Y$ or a conditional pmf of $X$), but I can't think of how this can be done nor have I seen it in any examples (all the examples use Bayes rule). How would one do this? 
 A: Another, maybe a bit more transparent, way to deal with this probability is through the definition
$$P(X=x|Y\leq y_1)=\frac{P(X=x, Y\leq y_1)}{P(Y\leq y_1)}=\frac{\int_{-\infty}^{y_1}dy f_{X,Y}(x,y)}{\int_{-\infty}^{\infty}dx\int_{-\infty}^{y_1}dy f_{X,Y}(x,y)}$$
This expression is inequivalent to $\int_{-\infty}^{y_1}dy f_{X|Y}(x|y)$, since the latter doesn't necessarily represent a probability measure with respect to the variable $Y$. However it is possible to represent this quantity, somewhat expectedly, in terms of it's Bayes' rule equivalent:
$$P(X=x|Y\leq y_1)=\frac{f_X(x)\int_{-\infty}^{y_1}dy f_{Y|X}(y|x)}{\int_{-\infty}^{\infty}dx f_X(x)\int_{-\infty}^{y_1}dy f_{Y|X}(y|x)}$$
EDIT:
If $X$ represents a discrete variable, then replacing $\int dx\to\sum_{x}$ we have that
$$P(X=x|Y\leq y_1)=\frac{P(X=x)\int_{-\infty}^{y_1}dy f_{Y|X}(y|x)}{\sum_{x} P(X=x)\int_{-\infty}^{y_1}dy f_{Y|X}(y|x)}$$
where $P(X=x)=\int_{-\infty}^{\infty}dyf_{X,Y}(x,y)$.
Also, if one has to write an expression with respect to only conditional probability distributions of $X$ and the distribution $Y$then one can write the expression:
$$P(X=x|Y\leq y_1)=\frac{\int_{-\infty}^{y_1}dy f_{X|Y}(x|y)f_Y(y)}{\sum_x\int_{-\infty}^{y_1}dy f_{X|Y}(x|y)f_Y(y)}$$
where $f_Y(y)=\sum_xf_{X,Y}(x,y)$
A: To approach this problem, we start with
\begin{align}
P\left\lbrace X = x | Y \leq y\right\rbrace &= \int_{-\infty}^{\infty} P\left\lbrace X = x | Y \leq y \,\cap Y = v \right\rbrace P\left\lbrace Y = v | Y \leq y \right\rbrace dv
\end{align}
Notice that it is clear that $P\left\lbrace Y = v | Y \leq y \right\rbrace = 0$ if $v > y$. Thus, when $-\infty < v \leq y$, we also know that $P\left\lbrace X = x | Y \leq y \,\cap Y = v \right\rbrace = P\left\lbrace X = x | Y = v \right\rbrace $. This implies that we must instead have that
\begin{align}
P\left\lbrace X = x | Y \leq y\right\rbrace &= \int_{-\infty}^{\infty} P\left\lbrace X = x | Y \leq y \,\cap Y = v \right\rbrace P\left\lbrace Y = v | Y \leq y \right\rbrace dv \\
&= \int_{-\infty}^{y} P\left\lbrace X = x | Y = v \right\rbrace P\left\lbrace Y = v | Y \leq y \right\rbrace dv \\
&= \int_{-\infty}^{y} P\left\lbrace X = x | Y = v \right\rbrace f_{Y|Y\leq y}(v)  dv
\end{align}
It is with this final result that one can make progress in further working out any specific result given knowledge of $P\left\lbrace X = x | Y = v \right\rbrace$ and $f_{Y|Y\leq y}(v)$.
A: Look at the case for a fixed $y$, then integrate over the interval to account for all such $y$ values.
$$P(X = x | Y \leq y_1 ) = \int_{-\infty}^{y_1} P(X=x | Y = y) dy = \int_{-\infty}^{y_1} f_{X|Y = y}(x | y)dy$$
