Even though there are problems here have the same title to my problem but I have different question.
From Jacod-Protter ''Probability Essentials''
Multivariate CLT: Let $(X_j)_{j≥1}$ be i.i.d. $,\mathbb{R}^d$-valued r.v. Let the (vector) $\mu = E\{X_j\}$, and let $Q$ denote the covariance matrix: $Q = (q_{k,l})_{1 \leq k,l \leq d}$, where $q_{k,l} = Cov(X_j^k, X_j^l)$, where $X_j^k$ is the $k^{th}$ component of the $\mathbb{R}^d$-valued r.v. of $X_j$. Then $$\lim_{n \to \infty}\frac{S_n - n\mu}{\sqrt{n}} = Z,$$ where L$(Z) = N(0,Q)$ and where converges is in distribution.
I found some resources that hinted me to use the Cramer-Wold Device:
Let $(X_n)_{n \geq 1}$ and $X$ be a random vectors in $\mathbb{R}^d$. Then $X_n \to X$ if and only if $\zeta \cdot X_n \to \zeta \cdot X$ for all $\zeta \in \mathbb{R}^d$.
Here is my attempt:
Let $\zeta \in \mathbb{R}^d$. Then $(\zeta \cdot X_n)_{n \geq 1}$ are independent and since $(X_n)_{n \geq 1}$ are identically distributed $$\varphi_{\zeta \cdot X_n}(u) = E\{e^{iu\langle \zeta, X_n \rangle}\} = E\{e^{\langle iu \zeta, X_n \rangle}\} = \varphi_{X_n}(u\zeta) = \varphi_{X_1}(u\zeta). $$ Thus $(\zeta \cdot X_n)_{n \geq 1}$ are i.i.d. Note that, $$E\{\zeta \cdot X_n\} = \zeta \cdot E\{X_n\} = \zeta \cdot E\{X_1\},$$ \begin{eqnarray} Var\{\zeta \cdot X_n\} &=&E\{(\zeta \cdot X_n)^2\} - (E\{\zeta \cdot X_n\})^2 \\ &\vdots& \\ &=& \zeta \cdot \left( E\{X_nX_n^T\} - E\{X_n\}E\{E_n\}^T\right)\zeta \\ &=& \zeta \cdot Q\zeta. \end{eqnarray}
Now, using the Basic Central Limit Theorem
Let $(X_n)_{n \geq 1}$ be i.i.d. with $E\{X_n\} = \mu$ and $Var\{X_n\} = \sigma^2$ for all $n$ with $0 < \sigma^2 \infty$. Let $Y_n = \frac{\sum_{j=1}^n X_j - n\mu}{\sigma \sqrt{n}}$. Then $Y_n$ converges in distribution to $Y$, where L$(Y) = $N$(0,1).$
we have $$\frac{1}{\sqrt{\zeta \cdot Q \zeta}} \left( \frac{\sum_{j=1}^n \left(\zeta \cdot X_j - \zeta \cdot E\{X_j\}\right)}{\sqrt{n}} \right) \to Z, \text{ where } L(Z) = N(0,1)$$ \begin{eqnarray} &\implies& \sqrt{\zeta \cdot Q\zeta} \left( \frac{\zeta}{\sqrt{\zeta \cdot Q \zeta}} \right) \left( \frac{\sum_{j=1}^n X_j - n E\{X_1\}}{\sqrt{n}} \right) \to \sqrt{\zeta \cdot Q\zeta}\ Z, \text{ where } L(Z) = N(0,1) \\ &\implies& \zeta \left(\frac{\sum_{j=1}^n X_j - n \mu}{\sqrt{n}} \right) \to \sqrt{\zeta \cdot Q\zeta} \ Z, \text{ where } L(Z) = N(0,1). \\ & \implies& \zeta \left(\frac{\sum_{j=1}^n X_j - n \mu}{\sqrt{n}} \right) \to \zeta \ Z, \text{ where } L(Z) = N(0,Q) \\ & \implies& \left(\frac{\sum_{j=1}^n X_j - n \mu}{\sqrt{n}} \right) \to \ Z, \text{ where } L(Z) = N(0,Q), \end{eqnarray} as required.
My question: Is $\sqrt{\zeta \cdot Q\zeta} Z, \text{ where } L(Z) = N(0,1)$ is the same or equivalent to $ \zeta Z$, where $L(Z) = N(0,Q)$ ?
If so, can you show me why? If not, can you show me where did I go wrong? Thank you.
