# Show that $\bf{X}\sim N_{np}(\mu,\Sigma)$

Let $X_1,\dots,X_n$ be independent random variables such that $X_i\sim N_p(\mu_i,\Sigma_i)$. Show that $\bf{X}\sim N_{np}(\mu,\Sigma)$ where $\bf{\mu}=(\mu_1,\dots,\mu_n)$ and $$\bf{\Sigma}=\begin{bmatrix}\Sigma_1 & 0&\dots& 0\\ 0 & \Sigma_2& \dots & 0 \\ \dots&\dots&\dots & \dots \\ 0 & 0 & 0 & \Sigma_n\end{bmatrix}$$

I know from theorem 2 from here Gaussian that $\bf{X}$ have multivariate normal distribution if $\bf{t'X}$ have normal distribution. So using mgf $$M_\bf{X}(\bf{t})=E[\exp(t'X)]=E[\exp(t_1X_1)]\dots E[\exp(t_nX_n)]$$

$$=\exp(t_1'\mu_1+\frac{1}{2}t_1'\Sigma_1t)\dots\exp(t_n'\mu_n+\frac{1}{2}t_n'\Sigma t_n)=\exp(\bf{t'\mu}+\frac{1}{2}t'\Sigma t)\sim N_{np}(\mu,\Sigma)$$

because each $X_i$ is $N_p$ and we have a vector of length $n$ and the covariance matrix is diag since each $X_i$ are iid.

EDIT: Each $\mu_i$ is a vector and $\Sigma_i$ a matrix.

• @kimchilover In this case each $\mu_i$ is a vector of length $p$, you mean that each component of each vector need to be the same? – Roland Sep 10 '17 at 18:39
• @kimchilover I made a mistake, the $X_i$ are only independent. – Roland Sep 10 '17 at 18:55