Using join probability distribution Say I'm given a probability distribution of two random variables $A$ and $B$. What does it mean to calculate the join probability distribution of $3^{(A-B)}$?
The distribution is in fact discrete.
 A: The problem does not specify whether $A$ and $B$ have discrete distribution, continuous distribution, or perhaps something more complicated. We deal with the continuous case. 
Let $W=3^{A-B}$.  To find the density function of $W$, one approach is to find the cumulative distribution function $F_W(w)$ of $W$, and then differentiate. It is clear that $F_W(w)=0$ if $w\le 0$. So assume $w\gt 0$. We have
$$F_W(w)=\Pr(W\le w)=\Pr(3^{A-B}\le w)= \Pr\left(A-B\le \frac{\log w}{\log 3}\right).$$
At this point we cannot provide a detailed solution without some specific information about the joint distribution function of $A$ and $B$. But the calculation will involve an integration. 
We will be integrating the joint density function $f(x,y)$ of $A$ and $B$ over the region where $x-y \le \frac{\log w}{\log 3}$. This is the region above a certain line with slope $1$, namely $y=x-\frac{\log w}{\log 3}$.  
A: If $A$ and $B$ are integer-valued, then $C=3^{A-B}$ has values in the set $\{3^n\mid n\in\mathbb Z\}$ and, for every $n$ in $\mathbb Z$,
$$
\mathbb P(C=3^n)=\mathbb P(A-B=n)=\sum_{k\in\mathbb Z}p_{A,B}(n+k,k),
$$
where $p_{A,B}$ describes the joint distribution of $(A,B)$, that is, for every integers $i$ and $j$,
$$
p_{A,B}(i,j)=\mathbb P(A=i,B=j).
$$
Edit: Assume for example that $A$ and $B$ are independent Bernoulli random variables with respective parameters $a$ and $b$. This means that $\mathbb P(A=1)=a$, $\mathbb P(A=0)=1-a$, $\mathbb P(B=1)=b$, $\mathbb P(B=0)=1-b$, and finally that $\mathbb P(A=i,B=j)=\mathbb P(A=i)\mathbb P(B=j)$ for every $i$ and $j$. 
Then $A-B$ takes values in $\{-1,0,1\}$ hence $C$ takes values in $\{\frac13,1,3\}$. To compute the distribution of $C$, first note that $\mathbb P(C=\frac13)=\mathbb P(A-B=-1)=\mathbb P(A=0,B=1)=(1-a)b$. Likewise,  $\mathbb P(C=3)=\mathbb P(A-B=1)=\mathbb P(A=1,B=0)=a(1-b)$. Finally,  $\mathbb P(C=1)=\mathbb P(A-B=0)=\mathbb P(A=1,B=1)+\mathbb P(A=0,B=0)=ab+(1-a)(1-b)$.
Edit: Continuing on the example,
$$
\mathbb E(C)=\tfrac13\cdot\mathbb P(C=\tfrac13)+1\cdot\mathbb P(C=1)+3\cdot\mathbb P(C=3),
$$
hence
$$
\mathbb E(C)=\tfrac13(1-a)b+ab+(1-a)(1-b)+3a(1-b)=(1+2a)(1-\tfrac23b).
$$
