Characteristic Function of Multivariate Normal

Question

I am reading through the lecture notes of a statistics class, and it describes several steps on how to transfer the characteristic function of a standard normal into the characteristic function of a multivariate normal. Most of the steps I have no problem with, but there is one that seems slightly too big a leap. It says: $$X\in N(0,I) \rightarrow \phi_X(t)=\exp\left(-\frac{1}{2}t^Tt\right)$$ which is fair enough, but then immediately jumps to: $$X\in N(\mu,\Sigma) \rightarrow \phi_X(t)=\exp\left(it^T\mu-\frac{1}{2}t^T\Sigma t\right)$$ I can see how this makes sense given the characteristic function of the normal distribution, but would this step not have to be substantiated a little bit more? Or is there some property that I am missing, that allows for this step to be easily made?

Answer 1

Let $\sqrt{\Sigma}$ denote a positive-definite symmetric matrix that squares to $\Sigma$, so $\sqrt{\Sigma}^T\sqrt{\Sigma}=\Sigma$. (To prove one exists, consider a basis that diagonalises $\Sigma$, then square-root the eigenvalues.) Write $X=\mu+\sqrt{\Sigma}Z$ so$$\begin{align}\exp it^TX&=\exp(it^T\mu+i(\sqrt{\Sigma}t)^TZ)\\&=\exp(it^T\mu)\phi_Z(\sqrt{\Sigma}t)\\&=\exp\left(it^T\mu-\frac12t^T\sqrt{\Sigma}^T\sqrt{\Sigma}t\right)\\&=\exp\left(it^T\mu-\frac12t^T\Sigma t\right).\end{align}$$

Answer 2

If $X\sim\mathsf{Norm}(\mu,\Sigma)$ then a matrix $A$ exists such that $X=\mu+AU$ where $U\sim\mathsf{Norm}(0,I)$ and $\Sigma=AA^T$.

Then: $$\phi_X(t)=\mathbb Ee^{it^T(\mu+AU)}=e^{it^T\mu}\mathbb Ee^{it^TAU}=e^{it^T\mu}\phi_{U}(A^Tt)=e^{it^T\mu}e^{-\frac12t^TAA^Tt}=e^{it^T\mu-\frac12t^T\Sigma t}$$