I was asked to show that $\mu$ and $\Sigma$ are respectively the Mean and Covariance of the Multivariate Gaussian Distribution. $X$ is a Random Vector, and $x \in \Re ^n$
$$P(X = x) = \frac{1}{(2\pi)^\frac{n}{2} |\Sigma|^\frac{1}{2}}e^{\frac{-1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)}$$
I started by showing that $$\int_{x \in R^n} P(X = x) \,dx = 1$$
My solution: Since $\Sigma$ is real and symmetric, it has an Eigen Value Decomposition $$\Sigma = Q\Lambda Q^T$$ where $\Lambda$ is diagonal (Since $\Sigma$ is symmetric positive definite, all eigenvalues $\lambda_i > 0$) and $Q$ is orthonormal ($QQ^T = Q^TQ = I_n$) and columns of $Q$ are eigenvectors of $\Sigma$ and form an Orthonormal Basis of $\Re^n$.
Using the transformation (this transformation will be Invertible), $$Q^T(x - \mu) = t$$ The Jacobian will be $Q$ itself, so the integral reduces to: $$\frac{1}{(2\pi)^\frac{n}{2} |\Sigma|^\frac{1}{2}}\int_{t \in R^n} |Q|e^{\frac{-1}{2}t^T\Lambda^{-1}t} dt$$ And the answer now is clearly visible, as $[diag(\lambda_1....\lambda_n)]^{-1} = diag(\frac{1}{\lambda_1}....\frac{1}{\lambda_n})$ and the integral splits into product of $n$ independent integrals (the Gaussians in 1 dimension).
The problem: How does $x \in \Re^n$ lead to $t \in \Re^n$. I believe this has something to do with the transformation being linear invertible and / or $Q$ being orthonormal. Any help will be appreciated, plus for example, why this happens here and a case where this won't be true. (Intuitively, when the transformation is not invertible, the limits of all $t_i$'s shouldnt't be independent.)
So for any $t \in \Re^n$ we wish to "test" if it should be in the limits, we want to see if $\exists x \in \Re^n $ such that $$Q^T(x - \mu) = t$$ $Q^T$ is invertible, guarantees this set of linear equations has a unique solution (in $\Re^n$), which is: $$x = \mu + Qt$$ So every $t \in \Re^n$ should be in the limits of the transformed integral!