# confusion on characteristic wrt convolution and product measure

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 $$(\mathbb R,\mathcal{B})$$

• If $X$ and $Y$ are independent, then $\widehat{P_{X+Y}}(t)=\widehat{P_X}(t)\times \widehat{P_Y}(t)$. – Mike Earnest Feb 12 at 0:21

$$\hat {P_{X+Y}}(x,y)$$ 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 $$x,y \in \mathbb R$$.