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 ?

Two notes on the convergence in moments assumed for @Ben's answer
It's not easy.
Suppose we had instead that $X_n\stackrel{d}{\to} Z$, and $Z_n\stackrel{p}{\to}Z$ and $\mathbb{Corr}[X_n, Z_n]\to 1$. Convergence in probability of $X_n$ to $Z$ need not hold in this slightly modified problem.
Take

*

*$U\sim U[0,1]$,

*$V\sim N(0,1)$

*$X_n\sim N(0,1)$ if $U>1/n$ and $X_n=2^n$ if $U<1/n$

*$Z_n=V$ if $U>1/n$ and $Z_n=2^n$ if $U<1/n$

*$Z=V$
with all the $N(0,1)$s independent. Then $\mathbb{Corr}[X_n, Z_n]$ exists for every $n$ and converges to 1, but $X_n$ does not converge in probability to $Z$ (it doesn't converge in probability at all).
By using the same $N(0,1)$ for all $n$ in the definition of $X_n$,  you could also arrange for $X_n$ to converge in probability to a $N(0,1)$ that was independent of $Z$
The result is true.
You can't do anything like the construction in part 1, because $Z$ doesn't vary  with $n$.  Heuristically, you need increasingly rare and extreme outliers in the $X_n$ and they can't stay correlated with fixed outliers in $Z$
Proof
Rather than working with the variance, we work with a truncated variance.  Given a finite, positive $M$, write $X^M_n$ for $X_n\{|X_n|<M\}$. We know the variance of $Z$ is finite, and there's only one of it so it's also uniformly bounded and we don't need to truncate it.
Now for any fixed $M$, the truncated  variance is continuous with respect to convergence in distribution, so
$$\mathbb{V}[X^M_n-Z]\to \mathbb{V}[Z^M-Z]$$
and given any $\epsilon>0$ we can choose $M$ so the limit is less than $\epsilon$, by finiteness of $\mathbb{V}[Z]$.
So, given $\epsilon$, we can find $N$ and $M$ such that for $n>N$
$$\mathbb{V}[X^M_n-Z]<2\epsilon$$
and (since $X_n$ converges in distribution)
$$\mathbb{P}[X_n\neq X_n^M]<\epsilon$$
Now for any $\eta$
$$\mathbb{P}[|X_n-Z]>\eta]\leq \mathbb{P}[|X_n-X_n^M]>\eta]+ \mathbb{P}[|Z-X_n^M]>\eta]$$
The first term is bounded by $\epsilon$ and the second (via Chebyshev's inequality) by something like $2\epsilon/\eta$. So we can choose $\epsilon$ to make it small and we are (finally) done.
Check
Why wouldn't this proof work for the modified problem where the result is false? The very first line
$$\mathbb{V}[X^M_n-Z]\to \mathbb{V}[Z^M-Z]$$
fails, since the correlation condition is on $Z_n$ rather than $Z$. It's important to the proof that $Z$ doesn't need truncation.
