# Convergence of diagonal of double sequence of random variables

I have a double sequence of random variables $$X^n_m$$, where $$n, m \in \mathbb{N}$$.

There exist variables $$X^n$$ and $$X$$ such that $$X_m^n \overset{m\to \infty}{\longrightarrow} X^n$$ almost surely and $$X^n \overset{n\to \infty}{\longrightarrow} X$$ in probability where $$X$$ is a constant.

There also exist variables $$X_m$$ such that $$X_m^n \overset{n\to \infty}{\longrightarrow} X_m$$ in probability and $$X_m \overset{m\to \infty}{\longrightarrow} X$$ almost surely.

My question: under what conditions can it be concluded that the 'diagonal' sequence $$X_n^n \overset{n\to \infty}{\longrightarrow} X$$ in probability?

In my setting, the sequence $$X^n$$ are uniformly bounded and at least $$0$$, i.e. $$\exists K : \mathbb{P}(0 \leq X^n \leq K)=1$$ for all $$n$$. But I'm not sure how to use this fact or whether it is helpeful.

Here's what I've tried so far:

To prove that $$X_n^n \overset{n\to \infty}{\longrightarrow} X$$ in probability, it is suffices to prove that the double limit $$X_m^n \overset{n,m\to \infty}{\longrightarrow} X$$ in probability, i.e. that $$\forall \epsilon, \delta>0\ \ \exists C$$ such that $$n, m > C \implies |X - X_m^n| > \epsilon$$ with probability $$\leq \delta$$.

So, let $$\epsilon, \delta > 0$$ and try to prove existence of such a $$C$$. Observe that

$$|X - X_m^n| \leq |X - X^n| + |X^n - X_m^n|$$

and using the fact that almost sure convergence implies convergence in probability, it follows that

$$\exists N(\epsilon, \delta) : n > N(\epsilon, \delta) \implies |X - X^n| > \epsilon$$ with probability $$\leq \delta$$

$$\exists M(\epsilon, \delta, n) : m > M(\epsilon, \delta, n) \implies |X^n - X^n_m| > \epsilon$$ with probability $$\leq \delta$$

The issue here is that since $$M$$ depeneds on $$n$$, it could be that $$M(\epsilon, \delta, N(\epsilon, \delta)) > N(\epsilon, \delta)$$. If this is the case, then (informally) $$m$$ needs to grow faster than $$n$$ meaning that no such $$C$$ can exist. I'm not sure how to resolve this problem.

• My suggestion is the following: forget about random variables and types of convergence for them. Take $X_n^{m}$ to be just real numbers (which means we are looking at the question for constant random variables). Ask the same question now. Can you think of a simple condition under which the diagonal sequence converges? Feb 28, 2019 at 11:45
• Thanks for the helpful comment! In the case of real numbers, a sufficient condition would be that for any $\epsilon > 0$ there exists $M(\epsilon) : m > M(\epsilon) \implies |X^n - X_m^n| < \epsilon$ for any $n$. The generalisation of this to the case of random variables is simply replacing $M(\epsilon, \delta, n)$ with $M(\epsilon, \delta)$ in my attempted proof above. This says that $X_m^n$ converges to $X^n$ uniformly in some sense across $n$. Mar 1, 2019 at 9:48

The assumptions given are not sufficient. Consider for example the constant random variables:

$$X_m^n = 0$$ if $$n=m$$

$$X_m^n = 1$$ if $$n\not=m$$

Then $$X^n_m \overset{n\to\infty}{\longrightarrow} X_m = 1$$ for any fixed $$m$$ and $$X^n_m \overset{m\to\infty}{\longrightarrow}X^n = 1$$ for any fixed $$m$$, so if we take $$X=1$$ then these variables satisfy the assumptions as given.

But in this case $$X_n^n = 0$$ for all $$n$$, so $$X^n_n \overset{n\to\infty}{\longrightarrow} 0 \not= X$$.

Convergence of the diagonal sequence does hold if we assume that $$X_m^n \overset{m\to\infty}{\longrightarrow}X^n$$ in probability uniformly in $$n$$, in the sense that:

$$\forall \epsilon, \delta > 0 \ \ \exists M(\epsilon, \delta) : m > M(\epsilon, \delta) \implies |X^n - X^n_m| > \epsilon \$$ for any $$n$$.

If this holds, the attempted proof in the question statement works, taking $$C = \max\{ N(\epsilon, \delta), M(\epsilon, \delta) \}$$. In fact, it suffices to have convergence of all variables only in probability, and not almost surely.

• I'm happy with this answer, but will wait before marking it as accepted in case someone writes a better answer soon. Mar 1, 2019 at 10:00