Brunei Math Club

 Marginal distributions of a multivariate normal distribution are also normal distributions. Let's prove this. See also : Multivariate normal distribution [Wikipedia] The density function of a multivariate normal distribution is given as $$f(\mathbf{x})=\frac{1}{\sqrt{(2\pi {)}^n|\mathrm{\Sigma }|}}\exp [-\frac{1}{2}(\mathbf{x}-\mathit{\mu }{)}^{\mathrm{\top }}{\mathrm{\Sigma }}^{-1}(\mathbf{x}-\mathit{\mu })]$$ where $\mathbf{x}\in {\mathbb{R}}^n$ is the random vector, $\mathit{\mu }$ is the mean vector and $\mathrm{\Sigma }$ is the covariance matrix. By changing the variables $\mathbf{x}-\mathit{\mu }\mapsto \mathbf{x}$, we can assume the mean is zero without losing generality. So, in the following, we only consider $$f(\mathbf{x})=\frac{1}{\sqrt{(2\pi {)}^n|\mathrm{\Sigma }|}}\exp [-\frac{1}{2}{\mathbf{x}}^{\mathrm{\top }}{\mathrm{\Sigma }}^{-1}\mathbf{x}].$$ We need the following theorems from linear algebra. Theorem 1 Let $A$ be an $n\times n$ regular, $D$ be $m\times m$ regular, \(B\...