Conditional probability given conditional pdf Let $X$ and $Y$ be two nonindependent continuous random variables whose support is $[0,1]$. I know $f_{X\mid Y=y}(x)$, i.e., the conditional probability density function of $X$ given the ocurrence of the value $y$ of $Y$, so I can calculate $P(X\le x\mid Y=y)$ as $$\int_0^xf_{X\mid Y=y}(u)du$$
My question is: how can I calculate probabilities like $P(X\le x\mid Y\le y)$? Perhaps $$\int_0^y\int_0^xf_{X\mid Y=v}(u)dudv$$
 A: The probability you ask for requires the use of the "law of total probability" (which has various other names). To understand how to get there, consider first a simpler but similar discrete problem: $Y$ can only assume three mutually exclusive values $y_1$, $y_2$, $y_3$, so we have
\begin{gather}
\mathrm{P}(Y\!=\!y_1 \, \lor\, Y\!=\!y_2 \, \lor\, Y\!=\!y_3) = 1 \ ,
\\
\mathrm{P}(Y\!=\!y_1 \, \land\, Y\!=\!y_2) = 0 \ ,\quad
\mathrm{P}(Y\!=\!y_1 \, \land\, Y\!=\!y_3) = 0 \ ,\quad
\mathrm{P}(Y\!=\!y_2 \, \land\, Y\!=\!y_3) = 0 \ ,
\end{gather}
where "$\land$" is logical "and", and "$\lor$" is logical "or" (if you wonder what I mean by this, check the reference at the end).
Now suppose we want to calculate $\mathrm{P}(X\!=\!x \mid 
Y\!=\!y_1 \, \lor\, Y\!=\!y_2)$. Using the rule for conditional probabilities (which is a consequence of the product rule) we have
$$
\mathrm{P}(X\!=\!x \mid Y\!=\!y_1 \, \lor\, Y\!=\!y_2) =
\frac{\mathrm{P}[X\!=\!x \,\land\, (Y\!=\!y_1 \, \lor\, Y\!=\!y_2) ] }{
\mathrm{P}(Y\!=\!y_1 \, \lor\, Y\!=\!y_2)} \ .
$$
The denominator is easy in view of our initial equations for $Y$:
$$
\mathrm{P}(Y\!=\!y_1 \, \lor\, Y\!=\!y_2) = 
\mathrm{P}(Y\!=\!y_1 ) +
\mathrm{P}(Y\!=\!y_2) \ .
$$
For the numerator we use the usual logical rules for "or" and "and", and the sum rule:
$$\begin{aligned}
\mathrm{P}[X\!=\!x \,\land\, (Y\!=\!y_1 \, \lor\, Y\!=\!y_2) ]
&= \mathrm{P}[(X\!=\!x \,\land\, Y\!=\!y_1) \, \lor\, (X\!=\!x \,\land\,Y\!=\!y_2) ]
\\&= 
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_1) +
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_2) -
\mathrm{P}[(X\!=\!x \,\land\, Y\!=\!y_1) \, \land\, (X\!=\!x \,\land\,Y\!=\!y_2) ]
\\&= 
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_1) +
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_2)
\end{aligned}
$$
the term with the minus sign being zero since $Y\!=\!y_1 \, \land\, Y\!=\!y_2$ is false (has zero probability). Now we apply the product rule twice:
\begin{multline}
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_1) +
\mathrm{P}(X\!=\!x \,\land\, Y\!=\!y_2) = {}\\
\mathrm{P}(X\!=\!x \mid Y\!=\!y_1)\ \mathrm{P}(Y\!=\!y_1) +
\mathrm{P}(X\!=\!x \mid Y\!=\!y_2)\ \mathrm{P}(Y\!=\!y_2)
\ .\end{multline}
Combining all calculations above we find
\begin{multline}
\mathrm{P}(X\!=\!x \mid Y\!=\!y_1 \, \lor\, Y\!=\!y_2) ={}\\
\frac{\mathrm{P}(X\!=\!x \mid Y\!=\!y_1)\ \mathrm{P}(Y\!=\!y_1) +
\mathrm{P}(X\!=\!x \mid Y\!=\!y_2)\ \mathrm{P}(Y\!=\!y_2)}{
\mathrm{P}(Y\!=\!y_1 ) +
\mathrm{P}(Y\!=\!y_2)} \ .
\end{multline}
This is an instance of the law of total probability; as you see it's just a consequence of the three basic laws of probability.
Considering several discrete values $y_i$ we find
$$
\mathrm{P}(X\!=\!x \mid \bigvee_i Y\!=\!y_i ) =
\frac{\sum_i\mathrm{P}(X\!=\!x \mid Y\!=\!y_i)\ \mathrm{P}(Y\!=\!y_i) }{
\sum_i\mathrm{P}(Y\!=\!y_i )} \ .
$$
and note that if we wanted to condition on $y_i$ values less that a given $y^*$ value we could write
$$
\mathrm{P}(X\!=\!x \mid Y\!<\!y^* ) =
\frac{\sum_{y_i<y^*}\mathrm{P}(X\!=\!x \mid Y\!=\!y_i)\ \mathrm{P}(Y\!=\!y_i) }{
\sum_{y_i<y^*}\mathrm{P}(Y\!=\!y_i )} \ .
$$

In your case, conditioning on $Y<y^*$ means conditioning on an infinite (continuum) logical sum: so to speak "$Y\!=\!y_1 \,\lor\,Y\!=\!y_2\,\lor\,Y\!=\!y_3\,\lor\,\dotsb$" for all $y_i<y^*$. Considering a continuum limit (see § 4.5 of the reference below) we find, using probability densities,
$$
\mathrm{p}(x \mid Y<y^*)  = 
\frac{\int_{-\infty}^{y^*} \mathrm{p}(x \mid y)\ \mathrm{p}(y)\ \mathrm{d}y}{
\int_{-\infty}^{y^*}\mathrm{p}(y)\ \mathrm{d}y} \ .
$$
As you see, the calculation requires the density function $\mathrm{p}(y)$, which cannot be obtained from the conditional density $\mathrm{p}(x \mid y)$.
From the formula above you already know how to find the conditional probability for $X<x^*$.

A great reference to reason through this kind of problems in a principled way:

*

*Jaynes: Probability Theory: The Logic of Science (also here and here)

see especially chapters 1–4.
