# Multivariate normal random variables

Prove that $$X\sim N(\mu,\Sigma)$$ is a multivariate random vector if and only if $$\langle X,x\rangle$$ has a normal distribution for each $$x\in\mathbb{R}^n$$.

Hint: use Cramér–Wold theorem.

My idea for the $$(\Leftarrow)$$ part is to use the characteristic function but I do not know how to formalize my reasoning correctly. For the opposite direction I have no ideas on how to use the hint.

Can someone help me?

Edit:

Definition of a standard normal random vector: a random vector $$X$$ has standard normal distribution in $$\mathbb{R^n}$$ if it has density: $$$$f(x)=\frac{1}{(2\pi)^{n/2}}e^{-\frac{\|x\|_2^2}{2}}$$$$ we write $$X\sim N(0,I_n)$$

Definition: $$X$$ random vector has general normal distribution in $$\mathbb{R}^n$$ and we write $$X\sim N(\mu,\Sigma)$$ for $$\mu\in\mathbb{R}^n$$ and $$\Sigma$$ $$n\times n$$ positive semi-definite matrix if and only if $$Z=\Sigma^{-1/2}(X-\mu)\sim N(0,I_n)$$

• The Cramer Wold theorem involves convergence in distribution, but there is no statement about convergence in the problem. Also note that some textbooks define the multivariate distribution by saying that $X$ is multivariate normal if and only if $\langle X, x \rangle$ has a normal distribution for all $x\in \mathbb{R}^n$. If this is your definition, then the problem is trivial, but if your textbook uses another definition, then you should probably add it to the question. Nov 8, 2021 at 11:28
• @LeanderTilstedKristensen I have edited the definition of the textobook Nov 8, 2021 at 12:42
• Actually, after writing my answer, i now realize that the definition given in your textbook is wrong. This definition can only work for positive definite matrices and not for positive semi-definite matrices. The reason is that if $\Sigma$ has eigenvalue $0$, then $\Sigma^{-1/2}$ has a column space of dimension less than $n$ and it is therefore impossible for $\Sigma^{-1/2}(X-\mu)$ to have a standard normal distribution! Nov 8, 2021 at 16:16
• Yes but I forgot to add in the in the definition that the matrix is also required to be invertible. Nov 8, 2021 at 16:22
• Ok. Do note however that the multivariate normal distribution is also defined when $\Sigma$ is not invertible and the result still holds. When $\Sigma$ is not invertible we can define $N(\mu,\Sigma)$ as the distribution of $\Sigma^{1/2}Z + \mu$, where $Z \sim N(0,I_n)$. Nov 8, 2021 at 16:26

Let's first consider the implication $$(\Rightarrow)$$. Note, that if $$X \sim N(0,I_n)$$, then $$X=(X_1,\dots,X_n)^*$$ where $$X_i$$ are independent $$N(0,1)$$ variables for $$i=1,\dots,n$$. In particular we get for any $$x \in \mathbb{R}^n$$, that $$\langle X , x \rangle = x^*X= \sum_{i=1}^n x_iX_i \sim N(0,\sum_{i=1}^{n} x_i^2)=N(0,||x||^2)$$ by properties of the $$N(0,1)$$ distribution. For general $$X \sim N(\mu,\Sigma)$$ note that \begin{align*}(x^*X-x^*\mu) &= x^*(X-\mu) \\ &= x^* \Sigma^{1/2} \Sigma^{-1/2} (X-\mu) \\ &=(\Sigma^{1/2}x)^* \Sigma^{-1/2}(X-\mu) \\ &= \langle Z,\Sigma^{1/2}x\rangle \\ &\sim N(0,||\Sigma^{1/2}x||^2) \\ &= N(0,x^*\Sigma x) \end{align*} were we used the established result for $$Z = \Sigma^{-1/2}(X-\mu) \sim N(0,I_n)$$. This gives us that $$x^* X \sim N(x^*\mu , x^*\Sigma x)$$, which concludes the proof of the first implication.
Now let's show $$(\Leftarrow)$$. Let $$X$$ be a random vector with the property that $$\langle X , x \rangle$$ has a normal distribution for all $$x \in \mathbb{R}$$ and let $$Y \sim N(\mu,\Sigma)$$ with $$\mu = \mathbb{E}[X]$$ and $$\Sigma=\operatorname{Cov}(X)$$. This implies for all $$x \in \mathbb{R}^n$$ that $$\langle X,x\rangle \sim \langle Y,x \rangle \sim N(x^* \mu , x^*\Sigma x)$$ and thus, that $$\mathbb{E}[e^{i\langle X , x\rangle}] = \mathbb{E}[e^{i \langle Y,x \rangle}]$$
for all $$x \in \mathbb{R^n}$$, which means that $$X$$ and $$Y$$ have the same characteristic function and thus the same distribution, which is what we wanted to prove.
• I don't understand why $\langle X,x\rangle$ has the same distribution of $\langle Y,x\rangle$ Nov 8, 2021 at 16:27
• We know that $\langle X , x \rangle$ has a normal distribution, so we only need to determine the parameters. Clearly the mean parameter must be $\mathbb{E}[ \langle X, x \rangle]$ and the variance parameter must be $\operatorname{Var}(\langle X , x \rangle)$. Can you prove, that $\mathbb{E}[x^* X] = x^* \mathbb{E}[X]=x^*\mu$ and $\operatorname{Var}(x^* X) = x^*\operatorname{Cov(X)}x = x^* \Sigma x$ ?? (hint: Linearity of expectation) Nov 8, 2021 at 16:33
• Also the fact that $\langle Y , x \rangle \sim N(x^* \mu, x^* \Sigma x)$ was proven in the proof of the other implication $(\Rightarrow)$. Nov 8, 2021 at 16:38
• $\operatorname{Var}(x^*X)=E[(x^*(X-\mu)^2)]=E[(x^*(X-\mu))(x^*(X-\mu))^*]=E[x^*(X-\mu)(X-\mu)^*x]=x^*\operatorname{Cov}(X)x=x^*\Sigma x$ correct? Nov 8, 2021 at 16:52