Integrals of probability density functions and their inter-relationships

I am a little bit confused about the relationship between marginal probability density functions (pdfs), joint pdfs (jpdfs), and conditional pdfs (cpdfs), and their integrals. Let me define the following pdfs: $f_X\left(x\right), f_Y\left(y\right)$. Their jpdf is denoted $f_{X,Y}\left(x,y\right)$. And their cpdf is denoted $f_{X|Y}\left(x|y\right)$. I know the following relationship holds:

$$f_{X,Y}\left(x,y\right)=f_X\left(x\right)f_{Y|X}\left(y|x\right)=f_Y\left(y\right)f_{X|Y}\left(x|y\right)$$

What I'm unsure about is what the integrals of these functions are. Intuitively I would expect:

$$\int f_{X|Y}\left(x|y\right)dy=f_X\left(x\right)$$

$$\int f_{X,Y}\left(x,y\right)dy=?$$

or

$$\int f_{X|Y}\left(x|y\right)dx=?$$

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Your first integral is incorrect; it is the second that has value $f_X(x)$ while the third evaluates to $1$. As to why the first intuition is invalid, note that $f_{X,Y}(x,y) = f_{X|Y}\left(x|y\right)f_Y\left(y\right)$ and thus $$\int f_{X,Y}\left(x,y\right)dy = \int f_{X|Y}\left(x|y\right)f_Y\left(y\right) dy$$ is just the continuous version of the law of total probability $$P(A) = \sum_i P(A\mid B_j)P(B_j)$$ while what you write would be akin to $$P(A) = \sum_i P(A\mid B_j).$$ You cannot add probabilities conditioned on different events.... – Dilip Sarwate Oct 28 '13 at 23:52

The first integral you have is not correct. If I plug in the (correct) definition you give for the conditional PDF, you can see the integral is not so simple:

$$\int f_{X|Y}(x|y) dy = \int \frac{f_{X,Y}(x,y)}{f_{Y}(y)} dy.$$

The right hand side does not reduce to the marginal PDF for $X$. Your second integral, however, is precisely the marginal PDF for $X$:

$$\int f_{X,Y}(x,y) dy = f_X(x).$$

The third integral evaluates to one, $$\int f_{X|Y}(x|y) dx = 1,$$ because a conditional PDF is still a probability density function and must have this normalization. By normalization, I mean the result of integrating over the entire support of $X$.

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So there is no simple identity for the first integral? – okj Oct 28 '13 at 23:58
I don't know of one. Like Dilip said, adding probabilities conditioned on different events is just a bunch of "or"'s. Could be anything. – Ian Oct 29 '13 at 0:22