Consider a random variable $X$ with distribution function F (write, $X \sim F$) and a random variable $Y$ with distribution function $G$ (write, $Y \sim G$).
Now, we can find the characteristic function of $W = XY$ by
\begin{equation} \phi_{W}(t)=E[\exp (\mathrm{i} t W)]=E[\exp (\mathrm{i} t X Y)]=E\{E[\exp (\mathrm{i} t X Y) \mid Y]\} \end{equation}
My question is then: does the characteristic function $\phi_{W}(t)$ uniquely define the distribution of $W$ similarly to how the characteristic function of X, (say, $\phi_{X}(t)$) would uniquely define the distribution of the random variable $X$?
Any insights would be appreciated.
I am going to explain using densities, the general case has the same gist. This was meant to be a comment but too long.
First, I suppose you understand that the c.f. of a random variable determines the distribution of said r.v. (in other words, if $X_1$ and $X_2$ have the same distribution then $\phi_{X_1} = \phi_{X_2}$). You are probably confused as what the distribution of $W$ is since many different random variables $X$ and $Y$ have the prespecified distributions $f_X$ and $f_Y.$ To understand what the distribution of $W$ is you need to first give the distribution for the joint vector $(X, Y)$ and then find the distributions of $W = XY$ using the joint distribution. When $X$ and $Y$ are independent, then the joint distribution must be $f_{X, Y}(x,y) = f_X(x) f_Y(y)$ and you are done. Otherwise you need to specify $f_{X, Y}.$ Notice that specifying $f_{X,Y}$ gives the distributions $f_X$ and $f_Y$ (by integrating against the other variable) but the reverse is never true besides the independent case. I hope this helps.