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 $\prod_j$.