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).

• $X_n$ converging in distribution to $Z$ doesn't imply $\mathbb{E}[X_n-Z]\to 0$ or $\mathbb{V}(X_n)\to \mathbb{V}(Z)$, though it may follow from the convergence of the correlations (I don't currently have a counterexample) Commented Nov 3, 2022 at 6:48
• I think it's true, but it wouldn't be true if we had, say, $Z_n\stackrel{p}{\to}Z$ and $\mathrm{corr}[X_n,Z_n]\to 1$. The reason it's true is that $Z$ is fixed. To cause a problem, outliers of $X_n$ have to be more and more extreme and less and less probable as $n\to\infty$, so they can't correlate well with the fixed outliers of $Z$. Commented Nov 3, 2022 at 7:44
• @ThomasLumley: Thanks, well spotted --- I jumped the gun a bit. I've now edited the question to point out the additional step going from convergence in distribution to convergence in moments. Please have a look and let me know if you think there are any remaining gaps.
– Ben
Commented Nov 3, 2022 at 21:30
• Yes, that looks correct. I still need to think about how to prove that the correlation condition establishes convergence of the moments. Commented Nov 3, 2022 at 22:50
• I don't think that works: the moments are finite but that doesn't imply they are bounded Commented Nov 4, 2022 at 0:37

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 $$X_n\{|X_n|. We know the variance of $$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.

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: $$$$\rho_{X,Z}=\frac{cov(X,Z)}{\sigma_{X} \sigma_{Z}},$$$$ but in the limit n$$\rightarrow$$ $$\infty$$: X$$\rightarrow$$Z, thus: $$$$\rho_{X,Z}=\frac{cov(X,Z)}{\sigma_{X} \sigma_{Z}}=\frac{E[XZ]-E[X]E[Z]}{\sigma^{2}}=\frac{E[Z^{2}]-E[Z]^{2}}{\sigma^{2}}=\frac{\sigma^{2}}{\sigma^{2}}=1,$$$$ where I have defined $$\sigma_{X}$$=$$\sigma_{Z}$$=$$\sigma$$ in the limit n$$\rightarrow$$ $$\infty$$.

• Does this show convergence in probability? I don’t see it.
– Eli
Commented Nov 3, 2022 at 2:06
• I'm not sure of what this shows, could you detail ? Plus I don't get why E[XZ] goes to E[Z^2] in the limit.. Commented Nov 7, 2022 at 0:01
• You use $E(XZ) = E(Z^2)$ which is not correct. That $X_n$ converges in distribution to $Z$ does not mean that "X becomes Z" (even in a probabilistic sense), it only means that they ultimately have the same distribution. Commented Nov 17, 2022 at 23:19