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Let $\mathbf{U}^n = (U_1^n,U_2^n,\cdots, U_k^n) $ converge in distribution to $N(\mathbf{0},\Sigma)$. My question is that if $V_{k+1} \sim N(0,1)$ and is independent of $\mathbf{U}^n$ then is it true that $$ (\mathbf{U}^n,V_{k+1}) \stackrel{D}\to N(0,\Sigma') \quad \text{where} \quad \Sigma' = \begin{bmatrix} \Sigma & \mathbf{0} \\ \mathbf{0} & 1 \end{bmatrix}? $$

If so, can I generalize this to $(\mathbf{U}^n,V_{k+1},\cdots, V_{k+m}) \stackrel{D}\to N\left(0,\begin{bmatrix} \Sigma &\mathbf{0} \\ \mathbf{0} & I_m\end{bmatrix}\right)$?

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Yes, it is true. This can be validated by the so-called Cramer-Wold device. Given any $a = (a_1^T, a_2^T)^T \in \mathbb{R}^{k + m}$, denote $V = (V_{k + 1}, \ldots, V_{k + m})^T$. By condition and the property of multivariate normal distribution, we have $$a_2^T V \sim N(0, a_2^Ta_2)$$ and $a_2^TV$ is independent of $U^n$. Therefore, by $a_1^TU^n \Rightarrow N(0, a_1^T\Sigma a_1)$ and the independence between $a_1^T U^n$ and $a_2^TV$, we conclude that \begin{align*} a_1^T U^n + a_2^T V \Rightarrow N(0, a_1^T\Sigma a_1 + a_2^Ta_2) = N(0, a^T\Sigma'a) \end{align*} as $n \to \infty$. By Cramer-Wold device, this means $$\begin{bmatrix}U^n \\ V\end{bmatrix} \Rightarrow N(0, \Sigma').$$

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