Characteristic function of independent Poisson random variables

Question

Let $X_{i}$ be independent Poisson distributed random variables with parameter $\lambda_{i} > 0$ for $i = 1,\ldots,n$. Now the joint distribution is given by \begin{equation*} \mathbb{P}\left(X_{1} = k_1,\ldots,X_n = k_n\right) = \prod_{j=1}^{n}e^{-\lambda_j}\frac{\lambda_j^{k_j}}{k_j!}. \end{equation*} The characteristic function $\varphi_{X_{i}}(t_{i})$ of $X_{i}$ is given by $$ \varphi_{X_{i}}(t_{i}) = e^{\lambda_{i}\left(e^{it_{i}} - 1 \right)}. $$ Is it right that the characteristic funtion of $(X_{1},\ldots,X_n)$ is given by $$ \varphi_{X_{1}}(t_{1})\cdot\ldots\varphi_{X_{n}}(t_{n}) = e^{\sum_{i=1}^{n}\lambda_{i}\left(e^{it_{i}} -1 \right)}? $$

Answer 1

Yes. The joint characteristic function of the $X_j$ is $\Bbb E\exp it\cdot X=\Bbb E\prod_j\exp it_jX_j$. By independence, $\Bbb E$ commutes with $\prod_j$.