# The characteristic function of a multivariate normal distributed random variable

The characteristic function of a random variable $X$ is defined as $\hat{X}(\theta)=\mathbb{E}(e^{i\theta X})$. If $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma\ge 0$, then its characteristic function can be found as follows:

$$\hat{X}(\theta)=\mathbb{E}(e^{i\theta X}) =\int_{-\infty}^{\infty}\frac{e^{i\theta x-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}dx=\ldots=e^{i\mu\theta-\frac{\sigma^2\theta^2}{2}}$$

(to be honest, I have no idea what to put instead of the "$\ldots$"; I've looked here, but that's only for the standard case. Anyway, this is not really my question, even if it is interesting and might be relevant)

Now, if I got it right, a random Gaussian vector $X$ (of dimension $n$) is a vector of the form $X=AY+M$ where $A$ is any real square matrix $n\times n$, $Y$ is a vector of size $n$ in which each coordination is a standard normally distributed random variable, and $M$ is some (constant) vector of size $n$.

I am trying to find the characteristic function of such $X$. The generalization of the formula for characteristic functions to higher dimensions is straight-forward:

$$\hat{X}=\mathbb{E}(e^{i<\theta,X>}),$$ where $<.,.>$ here is an inner product. So I can start with the following:

$$\hat{X}(\theta) = \mathbb{E}(e^{i<\theta,X>}) = \mathbb{E}(e^{i<\theta,AY>}\cdot e^{i<\theta,M>})\\ =e^{i<\theta,M>}\cdot \mathbb{E}(e^{i<\theta,AY>})$$

And I'm left with an expectation of a complex product of random variables. That probably means that the covariance matrix of some random variables should be involved, but that touches the boundaries of my knowledge about probability.

• $\theta$ - what is that?
– dtn
May 15, 2023 at 14:47

You wouldn't want to use the bracket notation for inner product when you're essentially dealing with matrices. Instead, write $\mathbb{E}\left[e^{i\theta^{T}X}\right] = \mathbb{E}\left[e^{i\theta^{T}\left(AY+M\right)}\right] = e^{i\theta^{T}M}\mathbb{E}\left[e^{i\theta^{T}AY}\right]$. You're only left with computing the characteristic function of a multivariate Gaussian distribution. \begin{align*}X &\sim \mathcal{N}\left(\mu, \Sigma\right)\\ \mathbb{E}\left[e^{is^{T}X}\right] &= \exp \left\{i\mu^{T}s - \frac{1}{2}s^{T}\Sigma s \right\} \end{align*} Just find out the mean vector and the covariance matrix of $AY$ since Gaussian variables have the affine property which means they don't change under linear transformation (They're still Gaussian completely defined by the mean vector and covariance matrix). If $Y \sim \mathcal{N}\left(\mu_{Y}, \Sigma_{Y}\right)$, then \begin{align*} \mathbb{E}\left[AY\right] &= A\mu_{Y} \\ \operatorname{Var}\left[AY\right] &= A\Sigma_{Y} A^{T} . \end{align*}
Using the relationship between $X$ and $Y$, \begin{align*} AY &= X-M \\ \mathbb{E}\left[AY\right] &= \mu_{X} - M \\\operatorname{Var}\left[AY\right] &= \Sigma_{X}\\ \mathbb{E}\left[e^{i\theta^{T}AY}\right] &= \exp \left\{i\left(\mu_{X}-M\right)^{T}\theta - \frac{1}{2}\theta^{T}\Sigma_{X} \theta \right\} . \end{align*} This is as far as I can get with the information you gave.
You are basically finished! See, you obtained $$\Psi_X(\theta) = e^{(i\langle\theta,M\rangle)}\mathbb{E}(e^{(i\langle\theta,AY\rangle)})$$ What is left is noticing that A can move to the other side of the inner product $$= e^{(i\langle\theta,M\rangle)}\mathbb{E}(e^{(i\langle A'\theta,Y\rangle)}) = e^{(i\langle\theta,M\rangle)}\Psi_Y(A'\theta)$$ All you have left is plugging in the characteristic function of the multivariate normal distribution.