I have recently come across characteristic functions.
Let $X,Y$ be random variables on $(\Omega, \mathcal{F}, P)$
Let $\widehat{P_{X}}$ and $\widehat{P_{Y}}$ denote the respective characteristic function.
In our notes, I have written down:
$X,Y$ are independent $\iff$ $\widehat{P_{(X,Y)}}(x,y)=\widehat{P_{X}}(x)\times \widehat{P_{Y}}(y)$
But it is also clear that: if $X,Y$ are independent
$\widehat{P_{X+Y}}(x,y)=\widehat{P_{X}(x)}\times \widehat{P_{Y}(y)}$
So that would then mean $\widehat{P_{(X,Y)}}(x,y)=\widehat{P_{X+Y}}(x,y)$
and since every characteristic function uniquely determines the respective distribution $\implies$ $P_{(X,Y)}=P_{X+Y}$, surely this cannot be correct?
I mean $(X,Y)$ induces a prob. space on $(\mathbb R^{2},\mathcal{B}^{2})$ while $X+Y$ induces a prob. space on $(\mathbb R,\mathcal{B})$
$\hat {P_{X+Y}}(x,y)$ does not even make sense. Note that $\hat {P_Z}$ is a function of one variable for any random variable $Z$ whereas $\hat {P_{X,Y}}$ is a function of two variables.
Here are the definitions: $\hat {P_{X,Y}} (x,y)=Ee^{i(xX+yY)}, \hat {P_{X+Y}} (t)=Ee^{it(X+Y)}$. Independence of $X$ and $Y$ is equivalent to $\hat {P_{X,Y}} (x,y)=\hat {P_X} (x) \hat {P_Y} (y)$ for all $x,y \in \mathbb R$.