In the following, we are going to give three definitions of multivariate normal distribution and show that these definitions are equivalent to each other. The explicit and transparent definition.\\ **Definition 1** A random vector $X$ has a non-degenerate (multivariate) nor­mal distribution if it has a joint density function of the form $$ f_X(x)=\sqrt{\frac{1}{(2\pi)^n \text{det}(Q) }}\text{exp}\big\{-\frac{1}{2} (x-\mu)^t Q^{-1} (x-\mu)\big\} $$ for some real vector $\mu$ and some positive definite matrix $Q$ . The constructive definition.\\ **Definition 2** A random vector $X$ has a (multivariate) normal distribution if it can be expressed in the form $$ X = UY + \mu, $$ for some matrix $U$ and some real vector $\mu$, where $Y$ is a random vector whose components are independent $N(0, 1)$ random variables. The elegant definition.\\ **Definition 3** A random vector $X$ has a (multivariate) normal distribution if for every real vector $a$, the random variable $a' X$ is normal. //Proofs of the equivalence// First, by the definition of density function and a variable change, we also see that **Definition 2** is equivalent to **Definition 1**. It is also easy to see that **Definition 2** implies **Definition 3**, and we thus only need to show **Definition 3** implies **Definition 2**. Let $E(X_i)=\mu_i$ and $\text{Cov}(X_i,X_j)=B_{ij}$, then by direction computation, we find that $$ (1)\hspace{30pt}E(a'X)=a'\mu,\hspace{10pt} \text{Cov}(a'X)=\text{Cov}(a'X, a'X)=a'Ba, $$ where $\mu=(\mu_1,\cdots,\mu_n)'$ and $B=(B_{ij})$ is a symmetric matrix. Since Cov$(a'X)$ is always positive, we see from $(1)$ that $B$ is positive definite. Now suppose that $B=U^{-1}U^{'-1}$ where $U$ is non-singular orthogonal matrix, and let $Y = U(X-\mu)$, then $E(Y)=0$ and $$ (2)\hspace{20pt}\text{Cov}(Y_i,Y_j)=\text{Cov}(U_{il}(X_l-\mu_l), U_{jk}(X_k-\mu_k))=\text{Cov}(U_{il}X_l, U_{jk}X_k)=U_{il}B_{lk}U_{jk}=\delta_{ij}, $$ where implicitly we have summation over $l,k$. From (2), we conclude that $Y$ is a random vector whose components are $N(0,1)$ random variables, and the co-variance of each two components is $0$, and furthermore, $a'Y$ is also normal. We need the following result to complete our proof. **Lemma 1** If $Y_1, Y_2$ are both $N(0,1)$ random variables with co-variance $0$, and $a_1Y_1+a_2Y_2$ is normal for all $a_1,a_2$. Then $Y_1$ and $Y_2$ are independent. By **Lemma 1**, we can conclude that $Y$ is a random vector whose components are independent $N(0, 1)$ random variables. $\square$ **Remark 1** The proof of **Lemma 1** may be achieved by using characteristic function. === References #Fundamentals of Probability - MIT OpenCourseWare, [[http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-436j-fundamentals-of-probability-fall-2008/lecture-notes/MIT6_436JF08_lec15.pdf|MULTIVARIATE NORMAL DISTRIBUTIONS]] #Jin-Ting Zhang, Lecture Notes, [[http://www.stat.nus.edu.sg/~zhangjt/teaching/ST4233/Lecture/ch3.pdf|Linear Models - Multivariate Normal Distribution]]