(1) Let $X_1,\ldots,X_n$ be independent randiom variables with $\mathbb{P}(\left\{|X_i|<\infty\right\})=1, i=1,\ldots,n$. Then for the characteristical function $\varphi$ of $Y_n:=\sum_{i=1}^{n}X_i$ it is $$ \varphi_{Y_n}=\prod_{i=1}^{n}\varphi_{X_i}. $$ (2) Let $\mu_1,\mu_2$ be probability measures on $\mathbb{R}$, then for the characteristical function of the convolution it is $$ \varphi_{\mu_1\star\mu_2}=\varphi_{\mu_1}\varphi_{\mu_2}. $$

(1) $$ \varphi_{Y_n}=\mathbb{E}(\exp(itY_n))=\mathbb{E}(\prod_{i=1}^{n}\exp(itX_i)) $$ and because the $X_i$ are independent and the function $h_t\colon\mathbb{R}\to\mathbb{C}, x\longmapsto\exp(itx)$ is measurable, the compositions $h_t\circ X_i$ are independent, too, so that $$ \mathbb{E}(\prod_{i=1}^{n}\exp(itX_i))=\prod_{i=1}^{n}E(\exp(itX_i))=\prod_{i=1}^{n}\varphi_{X_i}. $$

(2) We defined the convolution as follows:

Let $\mathcal{M}$ be the set of all probability measures on $(\mathbb{R},\mathcal{B})$ and consider $s\colon\mathbb{R}^2\to\mathbb{R}, (x,y)\longmapsto x+y$. Then for $\mu,\nu\in\mathcal{M}$ we call $$ \mu\star\nu:=(\mu\otimes\nu)\circ s^{-1}\in\mathcal{M} $$ the convolution of $\mu$ and $\nu$.

Using this, I get $$ \varphi_{\mu_1\star\mu_2}(t)=\int e^{itx}\, d(\mu_1\star\mu_2)=\int e^{itx}\, d((\mu_1\otimes\mu_2)\circ s^{-1})=\int e^{itx}\circ s\, d(\mu_1\otimes\mu_2)\\=\int e^{it(x+y)}\, d(\mu_1\otimes\mu_2)=\int e^{itx}\cdot e^{ity}\, d(\mu_1\otimes\mu_2)=\int e^{itx}\int e^{ity}\, d\mu_2(y)\, d\mu_1(x)=\varphi_{\mu_1}\varphi_{\mu_2} $$

By the way: Why can I apply Fubini?

I think because of Tonelly, because $$ \int\int\lvert e^{itx}\cdot e^{ity}\rvert\, d\mu_2(y)\, d\mu_1(x)=1<\infty. $$

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    $\begingroup$ I'm not really an expert on probability theory, but what does not add up is the equality $e^{itx}\circ s = e^{its}$. You should get something like $e^{it(x_1+x_2)}$ and, using Fubini and the fact that $e^{it(x_1+x_2)}=e^{itx_1}\cdot e^{itx_2}$ you should get to $\varphi_{\mu_1}(t)\varphi{\mu_2}(t)$. $\endgroup$ – dinosaur Feb 22 '14 at 16:37
  • $\begingroup$ Thank you. I corrected it. Ís it right that I can Fubini because of Tonelli? $\endgroup$ – mathfemi Feb 22 '14 at 16:49
  • $\begingroup$ yes, I believe that is correct. $\endgroup$ – dinosaur Feb 22 '14 at 16:56
  • $\begingroup$ And to apply Fubini I have to be sure, that the integrand $e^{it(x+y)}$ is $\mathcal{B}^2-\mathcal{B}^2$ measurable, right? This is the case, because $(x,y)\mapsto e^{it(x+y)}$ is continious? $\endgroup$ – mathfemi Feb 22 '14 at 16:57
  • $\begingroup$ The function $(x,y) \mapsto e^{it(x+y)}$ is measurable and bounded, hence integrable and so Fubini applies. $\endgroup$ – copper.hat Feb 22 '14 at 22:02

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