Notation for Convergence in Probability

Question

I am struggling with intuitively understanding convergence in probability, and I think my difficulty lies in understanding the notation of the definition. I was unable to find other questions/answers that address the notation specifically.

From the following definition:

Let $\{{X_n}\}_{n\geq1}$ be a sequence of RVs. We say that $X_n$ converges in probability to an RV $X$ if:

$\lim_{n\to\infty} P(|X_n - X|<\epsilon)=1$ for every $\epsilon>0$

In this definition, what exactly does $X_n$ refer to? Is it the whole sequence (i.e. the originally defined $\{X_n\}_{n\geq1}$), the $\text{n}^{\text{th}}$ term of the sequence, or something else?

If it is referring to the $\text{n}^{\text{th}}$ term, if each $X_i$ in the sequence is identically distributed, what is special about $X_n$ compared to say $X_3$?

If it is referring to the whole sequence, how can you "subtract" from a whole sequence? For example, saying $X_n - X = X_1 - X, X_2 - X, X_3 - X,...$ is still a sequence.

I understand how sample mean $\frac{1}{n}\sum X_i$ can converge to a constant, but the $X_i$ terms here are not a sequence, but rather a sum of the terms in a sequence.

This seems simple but I am having trouble with wrapping my head around this. Thanks!

Answer 1

Let $\{{X_n}\}_{n\geq1}$ be a sequence of RVs. We say that $X_n$ converges in probability to an RV $X$ if:

$\lim_{n\to\infty} P(|X_n - X|<\epsilon)=1$ for every $\epsilon>0$

In this definition, what exactly does $X_n$ refer to? Is it the whole sequence (i.e. the originally defined $\{X_n\}_{n\geq1}$), the $\text{n}^{\text{th}}$ term of the sequence, or something else?

$X_n$ appears in multiple places in this definition.

• In "$X_n$ converges in probability to $X$," the $X_n$ is shorthand for the full sequence $\{X_n\}_{n \ge 1}$. It makes sense to say that "a sequence of random variables converges...," but it does not make sense to say that "a single random variable converges..."
• In $P(|X_n - X|<\epsilon)$, the $X_n$ simply represents a single random variable, specifically the $n$th term of the sequence. This probability is a real number for each $n$, so the sequence $$P(|X_1-X|<\epsilon),\; P(|X_2-X|<\epsilon),\; P(|X_3-X|<\epsilon), \ldots$$ is simply a sequence of real numbers, and the claim is that its limit is $1$.

Answer 2

Let me try to clarify.

$X_n$ refers to the generic, $n$-th term of the sequence, which is one random variable depending on $n$. For any fixed $\epsilon$, $P(|X_n-X|<\epsilon)$ is a function of $n$, i. e. a sequence of real numbers in $[0,1]$. For convergence in probability, that sequence should have limit $=1$ no matter how small $\epsilon>0$ you choose.

Two r.v. having the same distribution does not make them equal (as functions on the sample space). They can actually be very different, but the only condition needed for many theorems is that they be identically distributed (and sometimes independent).

The whole sequence is denoted $\{X_n\}$, and a generic term is denoted $X_n$. Just like $f(x)$ denotes the value of the function $f$ at $x$, not the function, which is simply $f$.

You are not subtracting from the whole sequence, you are considering the sequence of differences $X_1-X, X_2-X,\dots$. Each term in this sequence is a r.v.