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?


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

  • 2
    $\begingroup$ 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. $\endgroup$ Nov 8, 2021 at 11:28
  • $\begingroup$ @LeanderTilstedKristensen I have edited the definition of the textobook $\endgroup$
    – Pefok
    Nov 8, 2021 at 12:42
  • $\begingroup$ 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! $\endgroup$ Nov 8, 2021 at 16:16
  • $\begingroup$ Yes but I forgot to add in the in the definition that the matrix is also required to be invertible. $\endgroup$
    – Pefok
    Nov 8, 2021 at 16:22
  • $\begingroup$ 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 $Z \sim N(0,I_n)$. $\endgroup$ Nov 8, 2021 at 16:26

1 Answer 1


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.

  • $\begingroup$ I don't understand why $\langle X,x\rangle$ has the same distribution of $\langle Y,x\rangle$ $\endgroup$
    – Pefok
    Nov 8, 2021 at 16:27
  • $\begingroup$ 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) $\endgroup$ Nov 8, 2021 at 16:33
  • $\begingroup$ Also the fact that $\langle Y , x \rangle \sim N(x^* \mu, x^* \Sigma x)$ was proven in the proof of the other implication $(\Rightarrow)$. $\endgroup$ Nov 8, 2021 at 16:38
  • $\begingroup$ $\operatorname{Var}(x^*X)=E[(x^*(X-\mu)^2)]=E[(x^*(X-\mu))(x^*(X-\mu))^*]=E[x^*(X-\mu)(X-\mu)^*x]=x^*\operatorname{Cov}(X)x=x^*\Sigma x$ correct? $\endgroup$
    – Pefok
    Nov 8, 2021 at 16:52
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    $\begingroup$ Yes I made a typo. thank you so much $\endgroup$
    – Pefok
    Nov 8, 2021 at 16:57

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