1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.