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

  • $\begingroup$ If $X$ and $Y$ are independent, then $\widehat{P_{X+Y}}(t)=\widehat{P_X}(t)\times \widehat{P_Y}(t)$. $\endgroup$ – 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 $x,y \in \mathbb R$.


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