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 $\infty$.

  • $\begingroup$ 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 "$a^2$". $\endgroup$ – pche8701 May 16 at 13:25
  • $\begingroup$ 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. $\endgroup$ – Ga13 May 16 at 17:59
  • $\begingroup$ Thanks. I independently came to a somewhat similar conclusion just recently, so it is nice to know that we both agree. I will mark you as the final answer to this Q and maybe from here look to manually integrate out some dependent RV to enforce some of the principles into my mind $\endgroup$ – pche8701 May 17 at 5:42

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