Consider an $n$ dimensional multivariable gaussain $X$ with $\mu = E[X]$ and $V = var(X)$, the covariance matrix. I wish to derive its distribution $$ f_X (x) = (2\pi)^{-n/2} (\det V)^{-1/2} \exp\left(-\frac{1}{2} (x - \mu, V^{-1}(x - \mu))\right). $$
Note that here my definition of a multivariable gaussian is
Let $X$ be a random variable. We say that $X$ is a Gaussian on $\mathbb{R}^n$ if $\langle u, X\rangle$ is Gaussian on $\mathbb{R}$ for all $u \in \mathbb{R}^n$.
My attempted Proof: Consider the Gaussian vector $\mathbf{Y} = (Y_1, \cdots, Y_n)$, where the $Y_i \sim N(0, 1)$ are independent. Then the density of $\mathbf{Y} $ is $$ f_Y(y) = (2\pi)^{-n/2} e^{-|y|^2/2}. $$ Define $$ \tilde{\mathbf{X} } = V^{1/2} \mathbf Y + \mu. $$ I know that $$ \phi_{\tilde{ \mathbf X}}(u)=e^{i(u, \mu)-(u, V u) / 2} . $$ I want to use
Let $X$ be a random variable on $\mathbb{R}^d$ and let $\phi_X$ be its characteristic function. Then if $\phi_X$ is integrable, then $\mu_X$ has a bounded and continuous density function, given by $$ f_X(x)=(2 \pi)^{-d} \int_{\mathbb{R}^d} \phi_X(u) e^{-i(u, x)} \mathrm{d} u . $$
So \begin{align*} f_{\tilde{\mathbf{X}} } (\mathbf{x} ) & = \frac{1}{(2\pi )^{n}} \int_{\mathbb{R} ^{n}}^{} e^{-i u \cdot \mathbf{x} } \phi_{\tilde{\mathbf{X}} } (u) \, \mathrm{d} u \\ & = \frac{1}{(2\pi )^{n}} \int_{\mathbb{R} ^{n}}^{} e^{-i u \cdot \mathbf{x} } e^{i u \cdot \mu - \frac{1}{2} (u, Vu)} \, \mathrm{d} u \\ & = \frac{1}{(2\pi )^{n}} \int_{\mathbb{R} ^{n}}^{} e^{-i u \cdot \left(\mathbf{x} - \mu \right)} e^{- \frac{1}{2} (u, Vu)} \, \mathrm{d} u \\ & = \left| \frac{1}{\det V} \right| \frac{1}{(2\pi )^{n}} \int_{\mathbb{R} ^{n}}^{} e^{-i V ^{-1}u \cdot \left(\mathbf{x} - \mu \right)} e^{- \frac{1}{2} (u, V^{-1}u)} \, \mathrm{d} u \end{align*}
I am not stuck.
Question How do I proceed? If my approach is at a dead end, how do I show this result given my definitions?