Confused about conditional probability with continuous random variables Suppose $X$ and $Y$ are continuous with joint PDF $f_{X,Y}$, so the probability of taking on any individual real value is 0. If I want to compute the probability $P(X > 1.5 \mid Y = \frac{1}{2}) = P(X > 1.5 \cap Y = .5) P(Y = .5) = 0$. But shouldn't I be able to compute the probability I get $X = 1.5$ in the event that I first observe $Y = .5$? Why should it have to be 0?
 A: In essence, you are computing the probability
$$
P(X > 1.5 \mid 0.5-\varepsilon < Y < 0.5+\varepsilon)
$$
in the limit as $\varepsilon > 0$.  With that in mind, the probability you're looking for is given by
$$
P(X > 1.5 \mid Y = 0.5) = \frac{\int_{x = 1.5}^{\infty} f_{X, Y}(x, 0.5) \, dx}
                               {\int_{x = -\infty}^{\infty} f_{X, Y}(x, 0.5) \, dx}
$$
ETA: The limiting probability above (my first expression) is not zero.  Since $P(A \mid B) = P(A, B)/P(B)$, we have
$$
P(X > 1.5 \mid 0.5-\varepsilon < Y < 0.5+\varepsilon)
    = \frac{P(X > 1.5, 0.5-\varepsilon < Y < 0.5+\varepsilon)}
           {P(0.5-\varepsilon < Y < 0.5+\varepsilon)}
$$
We can think of this fraction as the ratio between two volumes: the numerator is the volume under the two-variable PDF over the region between $y = 0.5-\varepsilon$ and $y = 0.5+\varepsilon$, with $x > 1.5$, and the denominator is the volume under that PDF over the region between $y = 0.5-\varepsilon$ and $y = 0.5+\varepsilon$, for any $x$.  In the limit as $\varepsilon \to 0$, those volumes go to zero, but their ratio does not, and is given by the expression above.
Also, I forgot to normalize my integral previously; I've fixed that.
