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 ?
 A: 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).
A: 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.
A: 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$\rightarrow$ $\infty$.
