# Marginalising over Dependent Random Variables

Suppose I have two RVs, $$A$$, and $$B$$.

Every place I have looked thus far suggests the following for marginalisation, which for me is fine:

$$f_A(a) = \int_{-\infty}^{\infty} f_{A,B}(a,b)db$$.

However, has it been implicitly assumed anywhere that $$A$$ and $$B$$ are independent random variables?

What is confusing is the following scenario:

Consider RVs where $$B = A^2$$. Is the following true, (with perhaps abuse of notation, I'm not sure how to write this out correctly):

$$f_A(a) = \int_{-\infty}^{\infty} f_{A,B}(a,b)db = \int_{-\infty}^{\infty} f_{A,A^2}(a,a^2)da^2$$.

It just feels strange for me, that you can integrate some "a" and still have a function of "a".

Is it possible to have clarify?

The answer for your first question is no, $$\textbf{A}$$ and $$\textbf{B}$$ don't have to be independent 1. About your second question, note that your integration region is no longer from $$-\infty$$ to $$\infty$$, but from zero to $$\infty$$.
• Thanks :). But regardless of the new limits, the overlap of the variables still trips me out (a and a^2 both in the integral yet some f(a) still remains after integrating). Maybe my mind is too stuck in a high school view of integration? I just cant convince myself why it is ok for some f(a) to remain since it feels like the integral did consider "a" implicitly through "$a^2$". – pche8701 May 16 at 13:25
• I think what's bothering you is to substitute $b$ for $a^2$. Because if $\textbf{B}=\textbf{A}^2$, you have to make a change of variable, and the integral will not be $da^2$, so the last equality is incorrect at this point. – Ga13 May 16 at 17:59