characterization of multivariate normal distribution by its one-dimensional marginals Let $X$ be a random vector in $\mathbb R^n$. we say $X$ has the normal distribution $N(\boldsymbol{\mu},\boldsymbol{V})$ if it has the density function:
$$\frac{1}{(2\pi)^{D/2}|\boldsymbol{V}|^{1/2}} \exp \left( -\frac{1}{2}(\boldsymbol{x}-\boldsymbol{\mu})^T \boldsymbol{V}^{-1}(\boldsymbol{x}-\boldsymbol{\mu}) \right) $$
Using this definition, I wonder how to show that $X$ has a multivariate normal distribution if and only if every one dimensional marginal $\langle X,\theta \rangle$ is normal (in one-dimensional sense).
I have trouble in both directions:
(1) For the forward direction, I am not sure how to find the density of $\langle X,\theta \rangle$ for a fix vector $\theta$ so that I can show the one-dimensional marginals are normal.
(2) For the backword direction, a hint says we may use the Cramer-Wold device, which says the sequence $X_n$ converges to $X_0$ in distribution if $\langle X_n,\theta \rangle$ converges to $\langle X_0,\theta \rangle$ in distribution for all fix vector $\theta$. But I am not sure how to use this theorem in the proof (where is the convergence?).
This is the Exercise 3.3.4 in Vershynin's high dimensional probability book. But I can't find it in some other prbability books, say Durrett.
 A: I don't know Vershynin's book, so don't know what ingredients may be legitimately used in answering his homework exercise.  So I will try to use as little as I can to give a proof that normality of all linear forms in a random vector implies the vector itself is normally distributed.
Zeroth, you need to know the characteristic function of gaussian random variables and gaussian random vectors.
First, the Cramér–Wold theorem tells us that the distribution of $X$ is specified by the distributions of all the linear forms in $X$.  If you believe that the characteristic function $\phi(a)=E[\exp(i\langle a,X\rangle)]$ of $X$ determines the distribution of $X$, this is easy to see, as the characteristic function $\phi_a$ of the scalar random variable $\langle a,X\rangle$ is simply $t\mapsto \phi(ta)$, that is, $\phi$ restricted to the line through $a$. To know all the functions $\phi_a$ is to know $\phi$.
Second. since the $i$-th component of $X$ is a linear form in $X$, and hence gaussian, and hence with finite variance, we know each of $E[X^2_i]<\infty$, and hence (by Cauchy-Schwarz) the covariance matrix $\mathbf V$ is finite.  As is the mean vector, $\mathbf \mu$.  Now we know the expectation and variance of the linear form $\langle a,X\rangle$, namely $\langle a,\mathbf \mu\rangle $ and $a'\mathbf V a$.  And hence the characteristic function of the gaussian rv. $\langle a,X\rangle$, with that mean and variance, namely, $\phi_a(t) =\exp( it\langle a,\mathbf \mu\rangle - t^2 a'\mathbf V a/2)$.
Finally, put these together: the characteristic function of a $N(\mathbf \mu,\mathbf V)$ random vector (call it $\varphi)$ has exactly the same restrictions to lines through $a$:
for all $a$, we have $\varphi_a(t)=\phi_a(t)$, for all $a$ and all $t$.  So $\phi=\varphi$, as desired.
