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 ?

Convergence in probability holds under additional (weak) conditions
It is possible to establish convergence in probability here if you can first establish the moment convergence $\mathbb{E}(X_n) \rightarrow \mathbb{E}(Z)$ and $\mathbb{V}(X_n) \rightarrow \mathbb{V}(Z)$.  Additional conditions for the moment convergence require that the sequence of moments $\mathbb{E}(X_n)$ and $\mathbb{V}(X_n)$ are bounded for sufficiently large $n$ (see e.g., here).  Here it is worth noting that the antecedent limit condition $\mathbb{Corr}(X_n, Z) \rightarrow 1$ already implies that the correlation exists, so $\mathbb{E}(Z)$ and $\mathbb{V}(Z)$ are finite and the other moments are finite for sufficiently large $n$ so all we need to go further to get the required condition is to assume that they are also bounded for sufficiently large $n$.
Now, assuming you can first establish the required convergence in moments, you then have $\mathbb{E}(X_n - Z) \rightarrow 0$ as an initial property.  Moreover, since $\mathbb{Corr}(X_n, Z) \rightarrow 1$ and $\mathbb{V}(X_n) \rightarrow \mathbb{V}(Z)$ you also have the property:
$$\begin{align}
\mathbb{V}(X_n - Z)
&= \mathbb{V}(X_n) - 2 \cdot \mathbb{Cov}(X_n, Z) + \mathbb{V}(Z) \\[12pt]
&= \mathbb{V}(X_n) - 2 \cdot \mathbb{Corr}(X_n, Z) \sqrt{\mathbb{V}(X_n) \mathbb{V}(Z)} + \mathbb{V}(Z) \\[6pt]
&\rightarrow \mathbb{V}(Z) - 2 \times 1 \times \sqrt{\mathbb{V}(Z) \cdot \mathbb{V}(Z)} + \mathbb{V}(Z) \\[10pt]
&= 2 \mathbb{V}(Z) - 2 \mathbb{V}(Z) \\[12pt]
&= 0. \\[6pt]
\end{align}$$
These two properties establish convergence in mean-square which then implies convergence in probabilty (using Markov's inequality).
