Question: 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.

Answer: 
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.
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: $P(x) = \int P(x|y)P(y)dy$