# Independence of Functions of Multiple Independent Random Variables

Is there a standard method to finding out if functions of multiple random variables are independent? For example, if I have random variables $X$ and $Y$ that are independent exponential RVs with parameter $\lambda$ and $\mu$ respectively, the joint PDF of the two is:

$$f_{XY}(X,Y) = e^{(-\lambda x-\mu y)}$$ when $x$ and $y$ are both greater than 0 and the PDF equals zero otherwise .

Now if I have two functions of $X$ and $Y$ that are

$$A = X+Y$$ $$B = X-Y$$

How could I go about showing whether or not $A$ and $B$ are independent? I know how to solve the marginal PDFs of $X$ and $Y$ as well as the CDF/PDF of $A$ and $B$ (using Law of Total Probability and Substitution Law), but I'm unsure how to determine if $A$ and $B$ are independent.

And beyond the example, is a there a general way to go about determining if functions of the same random variables are independent, regardless of the type of variables that are function inputs?

• One way is to compute the joint density function of $A$ and $B$, then see if you can factor this joint density into something like $f_{A,B}(a,b) = f_A(a) \cdot f_B(b)$. If so, then they are independent. Mar 22, 2018 at 3:54
• How would one go about finding the joint density of two functions of RVs, I think that may be where I'm a bit lost?
– agar
Mar 22, 2018 at 4:01
• In some cases, like the one you are considering, it is easier to use characteristic functions instead of density functions. $A$ and $B$ are independent if and only if $E^{i(tX+sY)}=Ee^{itx} Ee^{isy}$ for all $t,s$. Mar 22, 2018 at 7:53

There is a standard way to do a change of variables transformation from which you should be able to obtain $$f_{A,B}(a,b) = \frac{1}{2}e^{-\frac{1}{2}((\lambda+\mu)a+(\lambda-\mu)b)}$$ for $a>0,$ $|b|<a.$ Even though the density superficially seems to factor, notice that the bounds are dependent on one another (i.e. the support is not a rectangle) which means that they are dependent.
In fact, we could have seen this dependence without doing the change of variables at all. Notice we have $|B|<A,$ so the value of $A$ changes the support of $B$'s conditional distribution.