Vector's characteristic function.

Question

Let $\phi_x(t)=E(e^{it^Tx})$ be the characteristic function of the random vector x and $\phi_z(\alpha)=E(e^{i\alpha z})$ the characteristic function of the random variable $z$. Suppose $z=t^Tx$. Could I conclude that if $\alpha=1$ then the characteristic function of de random variable (regarded as a function of $t$) is equal to the characteristic function of the random vector? Can the characteristic funtion be equal for a variable and random vector?

Answer 1

They are equal in the sense that the former is a vector valued function whereas the latter is not. They can be understood as equal in the following sense: define $\mathbf{t}=\alpha \mathbf{d'}$ where $\alpha\in\mathbb{R}$ and $\mathbf{d}\in\mathbb{R}^{n}$ and let $z=\mathbf{d}'\mathbf{x}$ then $$ \varphi_{x}(\alpha\mathbf{d})=\mathbb{E}\left[e^{i\alpha \mathbf{d}'\mathbf{x}}\right]=\mathbb{E}\left[e^{i\alpha z}\right]=\varphi_{z}(\alpha) $$ So the answer to your question is yes, but you must be careful on your definition of "being equal to".

Just to provide an example of this, in a sense you do this in order to prove the Law of Large Numbers: you have a characteristic function of a random vector $\mathbf{x}$ and you would like to find the characteristic function of its sum $z=\mathbf{1}'x$. If we assume that variables are $iid$ then $\varphi_{x}(\alpha \mathbf{1}')=\left(\varphi_{x_{1}}(\alpha)\right)^{n}$.