Does convergence in distribution with correlation tending to 1 implies convergence in probability? Let be $(X_n)_{n \in\mathbb{N}}$ a sequence of random variables and $Z$ another random variable such that, when $n$ goes to infinity:

*

*$X_n$ converges to $Z$ in distribution : $X_n \overset{\mathcal{D}}{\to} Z$,

*The (Pearson's) correlation of $X_n$ and $Z$ converges to 1 (assuming these correlations are all well defined) : $\mathrm{corr}(X_n, Z) \to 1$.

Would this mean that $X_n$ converges to $Z$ in probability ? If not, do you think there might be another measure of dependence for which the statement could be true ?

Isn't this a reminiscent of the CLT? If the distribution of $X_{n}$ converges to Z, this means that the probability converges to Z as well. After all, the probabilities are obtained from distributions. Pearson coefficient:
\begin{equation}
\rho_{X,Z}=\frac{cov(X,Z)}{\sigma_{X} \sigma_{Z}},
\end{equation}
but in the limit n$\rightarrow$ $\infty$: X$\rightarrow$Z, thus:
\begin{equation}
\rho_{X,Z}=\frac{cov(X,Z)}{\sigma_{X} \sigma_{Z}}=\frac{E[XZ]-E[X]E[Z]}{\sigma^{2}}=\frac{E[Z^{2}]-E[Z]^{2}}{\sigma^{2}}=\frac{\sigma^{2}}{\sigma^{2}}=1,
\end{equation}
where I have defined $\sigma_{X}$=$\sigma_{Z}$=$\sigma$ in the limit n$\rightarrow$ $\infty$.
