Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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

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

But, what about:

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


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

share|cite|improve this question
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
up vote 1 down vote accepted

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 $X$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.