# Is convergence in probability equivalent to "almost surely... something"

Convergence in probability is weaker than almost sure convergence.

I wonder if the property "$X_n$ converges to $0$ in probability" can be expressed as: "For almost all $\omega$, the sequence $X_n(\omega)$ has a certain property (to determine)"

In other words, what are the consequences of convergence in probability that can be observed on each individual realization of the sequence. And do they constitute a characterization?

More formally, is there a property $\phi$ whose argument is a sequence of real numbers (or if you prefer $\phi$ is a function from $\mathbb{R}^\mathbb{N}$ to $\{true,false\}$) such as for all sequences of random variables $(X_n)_{n\in\mathbb{N}}$:

"$(X_n)_{n\in\mathbb{N}}$ converges to $0$ in probability" $\Longleftrightarrow$ "for almost all $\omega$, $\phi((X_n(\omega))_{n\in\mathbb{N}})$ holds"

(and find $\phi$)

A candidate for $\phi$ I'm thinking of would be something like that:

$$\phi((x_n)_{n\in\mathbb{N}}): \forall\epsilon>0\quad\lim_{n\rightarrow+\infty}\frac{\#\{k<n\mid |x_k|>\epsilon\}}{n}=0$$

• "has a certain property" is a bit blurry, everything in the world has some property. Convergence in probability is $\lim_{n\to\infty}\Pr\big(|X_n-X| > \varepsilon\big) = 0$, could you please tell us what exactly are you referring to in the definition?
– Tim
Mar 12, 2018 at 11:46
• I tried to clarify Mar 12, 2018 at 11:55
• I'm afraid your notation is unclear. What does $P((X_n(\omega))_{n\in\mathbb{N}})$ mean? "For each value there is a probability of observing it"? Sure there is, it is in [0, 1]....
– Tim
Mar 12, 2018 at 11:58
• Sorry, using $P$ was a bad idea. I called it $\phi$ and tried more clarification. Mar 12, 2018 at 12:05
• $\phi(x) = I(x \in (-\infty, \infty))$ is a property that is always valid... Sorry for nitpicking, but your question is unclear. Maybe you could tell us why exactly would you assume that convergence in probability leads to discovering some "property"?
– Tim
Mar 12, 2018 at 12:11

The answer is no: there is no such property. Any property of the form "a.s. something" that implies convergence in probability also implies a.s. convergence, hence cannot be equivalent to convergence in probability.

Proof: Write $X=(X_n)_{n\in\mathbb{N}}$. Let's assume all variables $X_n$ are binary (in $\{0;1\}$) for simplicity. Then convergence in probability (implicitly to 0 in all that follows) of $X$ is simply: $$\lim_{n\rightarrow+\infty} P(X_n=1)=0$$ For any property $\psi$ of sequences of 0s and 1s, "a.s. $\psi(X)$" means "for almost all $\omega$, $\psi(X(\omega))$".

For every infinite part $A$ of $\mathbb{N}$ define the property (of sequences of 0s and 1s):

$$\phi_A(x): \exists n\in A\quad x_n=0$$

Clearly, convergence in probability of $X$ implies a.s. $\phi_A(X)$. Now assume convergence in probability of $X$ is implied by a.s. $\phi(X)$ for some property $\phi$.

For any $A$, a.s. $\phi(X)$ implies convergence in probability of $X$ implies a.s. $\phi_A(X)$. Then it is clear that for any sequence $x$ of 0s and 1s, $\phi(x)$ implies $\phi_A(x)$: just use a random sequence $X$ identically equal to $x$. Thus $\phi(x)$ implies $\forall A\space \phi_A(x)$ where $A$ ranges over all infinite parts of $\mathbb{N}$. $\forall A\space \phi_A(x)$ says there is no infinite part of $\mathbb{N}$ where $x$ is constant to 1. In other words it says $x$ has only finitely many 1s. That is $x$ tends to 0. Thus a.s. $\phi(X)$ implies a.s. convergence of $X$.

$$\phi((x_n)_{n\in\mathbb{N}}): \forall\epsilon>0\quad\lim_{n\rightarrow+\infty}\frac{\#\{k<n\mid |x_k|>\epsilon\}}{n}=0$$

"Almost surely $\phi$" is not equivalent to convergence in probability.

First, it is not implied by convergence in probability. A counter example is not so easy to find. The implication holds for independent variables. A counter example can be constructed by creating successive independent groups of 0/1 variables. All variables in the same group are equal: their common value is called the value of the group. And we have:

• the length of each group is large enough (say $2^k$) so that a group having value 1 influences the average sufficiently
• almost surely infinitely many groups have value 1 so that the above happens infinitely many times
• the probability for a group to have value 1 tends to 0 so that convergence in probability (to 0) holds

The last two points can be be obtained by saying that the probability for the $k^{th}$ group to have value 1 is $1/k$ and using the second Borrel-Cantelli lemma.

"Almost surely $\phi$" does not imply convergence in probability either. But it's not far from it. "For all subsequences of $X$, almost surely $\phi$ for this subsequence" implies convergence in probability and this is quite easy to prove.

Going further than this seems to raise key theoretical difficulties: how to define asymptotic empirical frequency in a more general way than this.

• Something might be missing from the limit expression, because "$Y_k=1$" means the random variable $Y_k$ is identically the constant function $1.$ If that's what you intended to write, then the reverse implication obviously is false.
– whuber
Mar 13, 2018 at 14:40
• I meant "$Y_k(\omega)=1$". I re-wrote it like this. Do you see a shorter notation? Mar 13, 2018 at 17:05