Let $X_n$ be a random variable (say in $\mathbb{R}^p$), and consider this statement:

Claim: "For any sequence of random variables $X_n$ being bounded in probability, there exists a subsequence $X_{n_k}$, and some $X$, such that $X_{n_k}$ converges in probability to $X$."

I wonder if it is true, or if a counter-example can be found?

Here is what I found so far on the topic:

Fact 1: Prokhorov's theorem states that bounded sequences in probability admit a subsequence converging in law. But convergence in law is weaker than convergence in probability. Since this theorem is presented as being very important, I can guess it is sharp.

Fact 2: A sequence converging in probability admits a subsequence converging almost surely (see this other post), which is a stronger mode of convergence than in probability. If my claim is true, we could deduce that any stochastically bounded sequence admits an almost surely converging subsequence. This would make the conclusion of Prokhorov's theorem even weaker.

up vote 1 down vote accepted

No, this is, in general, wrong as the following (counter)example shows.

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of independent real-valued random variables such that $X_n \sim \mu$ for some non-trivial distribution $\mu$. Then $(X_n)_{n \in \mathbb{N}}$ is bounded in probability, but does not admit a subsequence which converges in probability.

Indeed: The boundedness in probability is obvious from the fact that each $X_n$ has the same distribution. If $(X_n)_{n \in \mathbb{N}}$ had a subsequence which converges in probability, say $X_{n_k} \to Y$, then $Y \sim \mu$ and we could choose a further subsequence $X_{n_{k_{j}}}$ such that $X_{n_{k_j}} \to Y$ almost surely. It is well-known that the pointwise limit of independent random variables is almost surely constant (see the lemma below) and therefore $Y$ is trvial; this contradicts our assumption that $Y \sim \mu$ is non-trivial.

Lemma Let $(Y_j)_{j \geq 1}$ be a sequence of independent real-valued random variables such that $Y_j \to Y$ almost surely. Then $Y$ is constant almost surely.

Proof: For any $c \in \mathbb{R}$ the event $\{Y \leq c\}$ is a tail event and therefore it follows from Kolmogorov's 0-1 law that $\mathbb{P}(Y \leq c) \in \{0,1\}$; as $c \in \mathbb{R}$ is arbitrary this implies that $Y$ is constant almost surely.

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