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Let $Z=(Z_1,Z_2)$ be a bivariate standard normal vector and $Y_{1,n},Y_{2,n}$ two sequences of real valued random variables with finite variance such that $Y_{1,n}\xrightarrow{d}Z_1$ and $Y_{2,n}\xrightarrow{d}Z_2$ (in distribution) for $n\rightarrow\mathbb{N}$. Assume further that \begin{align*} \operatorname{E}\left[Y_{1,n}\right]=\operatorname{E}\left[Y_{2,n}\right]=\operatorname{E}\left[Z_1\right]=\operatorname{E}\left[Z_2\right]=0 ,\end{align*} \begin{align*} \operatorname{Var}\left(Y_{1,n}\right)\xrightarrow{n\rightarrow\infty}1, \operatorname{Var}\left(Y_{2,n}\right)\xrightarrow{n\rightarrow\infty}1\end{align*} and \begin{align*} \operatorname{Cov}\left(Y_{1,n},Y_{2,n}\right)\xrightarrow{n\rightarrow\infty}\operatorname{Cov}\left(Z_1,Z_2\right).\end{align*} Does now hold that \begin{align*} \left(Y_{1,n},Y_{2,n}\right)\xrightarrow{d}\left(Z_1,Z_2\right) \end{align*}for $n\rightarrow\infty$?

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This is not even true for constant sequences. If $Y_{1,n} := Y_1$ is a Normal(0,1) random variable, and $Y_{2,n} = BY =: Y_2$ where B is a Bernoulli(1/2) random variable (taking values in $\{1,-1\}$), then this sequence satisfies all the convergence hypotheses since the sequences are constant in $n$.

We have that $Cov(Y_1, Y_2) = 0$, but the vector fails to be a bivariate normal distribution. You can see their distribution here. So they can't converge to the Normal$(diag(1,1))$ bivariate normal distribution.

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