Convergence of maximum of sequences of random variables

by Baham91   Last Updated August 14, 2019 00:20 AM

Let $X_1^n,\ldots,X_k^n$ be sequences of random variables, where $k$ is fixed. The sequences are not independent. I'd like to prove the following statement:

Assume that $X_1^n \to x_1, \ldots, X_k^n \to x_n$ in probability, where $x_1,\ldots,x_n$ are constants. Then: $$ \max_{i \in \{1,\ldots,k\}} X_i^n \to \max_{i \in \{1,\ldots,k\}} x_i $$ in probability.

Notice that this setting is different than the usual one, where the maximum is usually taken over $n$.

I'm not entirely sure if this can be seen as a direct application of the continuous mapping theorem, especially if the sequences live in other spaces than $\mathbb{R}$ (for instance on $\mathbb{N}$).

Any insight is very much appreciated!

Answers 2

It is indeed just the ($k$-dimensional) continuous mapping theorem, since $F(x_1, \dots, x_k) := \max(x_1, \dots, x_k)$ is a continuous function from $\mathbb{R}^k$ to $\mathbb{R}$ (easy exercise). You have a sequence of $\mathbb{R}^k$-valued random vectors $\mathbf{X}^n := (X_1^n, \dots, X_k^n)$ converging in probability to a constant vector $\mathbf{x} = (x_1, \dots, x_k)$, and you are claiming that $F(\mathbf{X}^n) \to F(\mathbf{x})$ in probability. That is exactly the continuous mapping theorem.

It makes no difference if your random variables only take values in a subset of $\mathbb{R}$, such as $\mathbb{N}$ or $\mathbb{Z}$ or what have you. You can still view them as real-valued random variables.

(Of course, if your random variables take values in some space completely different from $\mathbb{R}$, then you need to know quite a bit about the order and topological structure of that space to even make sense of the statement.)

Nate Eldredge
Nate Eldredge
August 13, 2019 23:10 PM

It is easy to verify that $|max \{a_1,a_2,...,a_k) - max \{b_1,b_2,...,b_k)|>\epsilon$ implies $|a_i-b_i| >\epsilon$ for some $i$. [Use proof by contradiction]. Hence the result follows immediately from definition of convergence in probability.

Kavi Rama Murthy
Kavi Rama Murthy
August 13, 2019 23:28 PM

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