What does the following equation say about the conditional distribution $P(ab|xy)$? What does the equation below say about the conditional distribution $P(ab|xy)$?
$$ P(ab|xy) = \int{ d\lambda \ \rho(\lambda)\ P(a|x\lambda)\ P(b|y\lambda) }$$
What does $\lambda$ denote here? Could you point me to some resources where I could learn about the semantics of the expression on the right hand side? 
I can provide more context if needed.
 A: It sounds like you have random variables $A$, $B$, $X$, $Y$, $\Lambda$ for which
$$\Pr(A=a, B=b\mid X=x, Y=y) = \mathbb E_\lambda\left[\Pr(A=a\mid X=x, \Lambda=\lambda)\cdot\Pr(B=b\mid Y=y, \Lambda=\lambda) \right]$$
Here $\mathbb E$ is expectation and $\Pr$ is probability (I'm assuming discrete random variables $A$, $B$, $X$, $Y$ here but it would be similar for continuous ones.)
In words, it is not quite true that


*

*$A$ and $B$ are independent given $X$ and $Y$,

*$A$ only depends on $X$, and

*$B$ only depends on $Y$.


But it is true when given $\Lambda$.
For instance let $A$=daughter's hairstyle, $B$=son's hairstyle, $X$=mom's hairstyle, $Y$=dad's hairstyle, and $\Lambda$=the age difference between son and daughter. Something like that.
A: What you wrote is known as marginalization.
Here's the idea:
Suppose I have a red die and a blue die.
What is the probability to roll a 6 on the red die?
$P(r=6) = P(r=6|b=1) * P(b=1) + P(r=6|b=2) * P(b=2)+ ... + P(r=6|b=6) * P(b=6)$
Basically the probability to roll a 6 on the red die is the probability red is 6 given that blue is 1, times the prior probability to get 1 on the blue, plus probability red is 6 given blue is 2 times prior probability blue is 2 and so on...
Basically you are covering every other option of the blue die to get the probability of the red one.
This is what your $\lambda$ does.
Generally speaking: $P(x) = \int P(x|y)P(y)dy$
