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Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Suppose that we have a sequence of random variables ${X_n}$ with the property that

$$ \mathbb{P}\Big( \lim_{n\to\infty} X_n = 0 \Big) = 1. $$

My intuition from this equation is that the event

$$A = \Big\{ \omega \in \Omega | \lim_{n\to\infty} X_n(\omega) = 0 \Big\}$$

is a high probability set. In an example, I encountered that it follows

$$ \mathbb{E}\Big( \lim_{n\to\infty} X_n \Big) = 0. $$

However, I couldn't truely understand why this happens. I do not know much about measure theory and its applications in probability, so it would be nice if you could explain this in terms of basic mathematical analysis or at least a minimal amount of measure theory supported by some analysis and intuition. Another natural question is that what can we say about the cardinality of $A$?

Hosein Rahnama
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    @Sarilass That's an argument in favor not being able to exchange limit and integral. Right now the question is why $E(\lim X_n)=0$. – AlvinL Nov 23 '23 at 10:12
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    @Sarilass Is $n$ a constant in your example, or you meant $\mathbb{P}(X_n=n)=\frac{1}{n}$? In either case, if the $X_n$ are independent then from the second Borel-Cantelli lemma it's not hard to deduce that $X_n$ can't tend to $0$ almost surely. On the other hand, if you want $\mathbb{P}(X_n=n^2)=\frac{1}{n^2}$ then yes, $X_n\to 0$ almost surely by the first Borel-Cantelli lemma. – Mark Nov 23 '23 at 10:18
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    For OP question, I think it is hard to show this in basic mathematical analysis without measure theory. Essentially, what OP asks is the implication $$ \mathbb{P}(X = 0) = 1 \Longrightarrow \mathbb{E}(X) = 0$$ The proof of this will involve the fact that expectation of a random variable over a null set is $0$, i.e a basic fact of measure theory. – Thành Nguyễn Nov 23 '23 at 10:23
  • @ThànhNguyễn: OK, then please use a minimal amount of measure theory supported by some basi analysis. :) – Hosein Rahnama Nov 23 '23 at 10:24
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    The conclusion $\mathbb{E}\Big( \lim_{n\to\infty} X_n \Big) = 0$ is correct. But beware! The alternate conclusion $\lim_{n\to\infty}\mathbb{E}\Big( X_n \Big) = 0$ may fail. – GEdgar Nov 23 '23 at 10:33
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    It seems to me that it is better to state that for every random variable $Y$ on the space we have $\mathbb EY=0$ if $Y$ coincides wth $\lim X_n$ on the set ${(X_n)_n\text{ is convergent}}$. This set has probability $1$ but is not necessarily the whole set $\Omega$ and if it is not then $\lim X_n$ is not formally a random variable (hence has no expectation). – drhab Nov 23 '23 at 11:46
  • @drhab: Hmmm, I agree. Do we have to say that $Y$ should be bounded on the set where $(X_n)$ is not convergent? – Hosein Rahnama Nov 23 '23 at 12:29
  • No, we do not need any further restrictions on random variable $Y$. – drhab Nov 23 '23 at 14:01

3 Answers3

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Denote $\lim X_n = Y$ (1). Since the $X_n$ are measurable, $Y$ is also measurable. By assumption $\mathbb P\{Y=0\}=1$. Then $$ EY = \int _{\{Y=0\}}Yd\mathbb P + \int _{\{Y\neq 0\}}Yd\mathbb P = 0. $$ In the first case, we integrate zero and in the second case we integrate over a set of size zero. In both cases, the result is zero.

You have tagged probability theory, so one cannot feasibly hope to dodge measure theory.

As for cardinality, it depends on $\Omega$. E.g if $\Omega =[0,1]$, which is uncountable, and $\mathbb P$ is the Lebesgue measure, then any event with positive measure would have to be uncountable.

(1) It is fair to point out that $Y(\omega)$ need not be defined because the limit might not exist. For this particular discussion, it does not pose a problem because by assumption, the subset of those $\omega$ for which $Y(\omega)$ is not defined is negligible (i.e contained in a measure zero subset). So we could set $$ Y(\omega) = \begin{cases}\lim X_n(\omega), &\exists \lim X_n(\omega) \\ +\infty, &\text{otherwise} \end{cases} $$

AlvinL
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Here is the continuation of my comment. Essentially, what we want to show is the following: For a random variable $X$, $$ \mathbb{P}(X = 0) = 1 \Longrightarrow \mathbb{E}(X) = 0 $$ To prove this, we split the random variable $X$ into two: $$ X = X \cdot 1\{X = 0\} + X \cdot 1\{X \neq 0\} $$ where $1\{A\}$ is the indicator function of some event $A$. By linearity of expectation, $$ \mathbb{E}(X) = \mathbb{E}(X \cdot 1\{X = 0\}) + \mathbb{E}(X \cdot 1\{X \neq 0\}) $$ Now, consider the first expectation of $X \cdot 1\{X = 0\}$. Since $X(\omega) \cdot 1\{X = 0\}(\omega)$ equals to $0$, no matter what is the value of $\omega \in \Omega$, therefore, $X \cdot 1\{X = 0\} \equiv 0$, and $$ \mathbb{E}(X \cdot 1\{X = 0\}) = \mathbb{E}(0) = 0 $$ So, $$ \mathbb{E}(X) = \mathbb{E}(X \cdot 1\{X \neq 0\}) $$ Since $\mathbb{P}(X = 0) = 1$, we have $\mathbb{P}(X \neq 0) = 0$. Thus, the right hand side is the expectation of a random variable on a null set and this must be $0$ too (the proof of this fact can be found here, which is based on measure theory)

Thành Nguyễn
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This is kind of a fundamental consequence of the definition of the Lebesgue integral, and so one cannot avoid this definition, and hence one cannot avoid measure theory in actually showing this.

Here's the standard development:

  1. A nonnegative simple random variable is (roughly) one that only takes a finite number of nonnegative values, each of which is finite. More precisely, these are maps of the form $$U(\omega) = \sum_{i = 1}^n a_i \mathbf{1}_{E_i}(\omega),$$ where the $E_i$ are disjoint events, $a_i \in [0, \infty),$ and $n < \infty$. The idea is that its easy to define the integral of simple RVs by basically thinking about linearity: $\mathbb{E}[U] = \sum a_i P(E_i).$

  2. For a non-negative random variable $X$, the expectation is defined as $$\mathbb{E}[X] := \sup\{\mathbb{E}[U] : U \le X, U \textrm{ simple, nonnegative}\}.$$ While the above is certainly well defined, it is not obvious that this matches your usual sense of the expectation: e.g., if the RVs have a nice density (what does that even mean, in the measure theoretic sense?), then does this definition yield $\mathbb{E}[X] = \int x f_X(x) \mathrm{d}x$? I'll leave it to you to consult an appropriate textbook for actually understanding this definition.

  3. Proposition: If a random variable $X \ge 0$ satisfies $P(X = 0) = 1,$ then $\mathbb{E}[X] = 0$.

Pf. Let $E = \{\omega: X(\omega) \neq 0\}$. For any simple function $ U = \sum_{i \le n} a_i \mathbf{1}_{E_i},$ if $U \le X,$ then for any $i : a_i > 0, E_i \subset E,$ and so $P(E_i) = 0.$ It thus follows that $\mathbb{E}[U] = \sum_{i } a_i P(E_i) = \sum_{i : a_i \le 0} a_i P(E_i) = 0$. But then $\mathbb{E}[X]$ is a supremum of the singleton set $\{0\},$ and so equals $0$.

There are now two complications in the case of your question:

For general random variables, the definition of integrals needs to consider the positive and negative parts separately. However, when dealing with $P(X = 0) = 1,$ we can simply deal with the nonnegative random variable $|X|$ instead, since $x = 0 \iff |x| = 0$, and since $|\mathbb{E}[X]| \le \mathbb{E}[|X|]$. So, in the following, lets set $Y_n(\omega) = |X_n(\omega)|$. Of course, by continuity, $X_n(\omega) \to x \implies Y_n(\omega) \to |x|$.

A final complication in the situation of the question is that we don't know that $\lim Y_n(\omega)$ is a random variable, since this may not exist for some $\omega$. Here the almost sure convergence lets us avoid this issue: pick any random variable $Z,$ and let us define $$ Y(\omega) = \begin{cases} \lim Y_n(\omega) & \lim Y_n(\omega) \textrm{ exists } \\ Z(\omega) & \lim Y_n(\omega) \textrm{ doesn't exist}\end{cases}.$$ Since $P(\lim Y_n = 0) =1,$ no matter what $Z$ we pick, we also have $P(Y = 0) = 1.$ The technical term is that each such $Y$ is a version of $\lim Y_n$.

But now, for any version, since $Y \ge 0, P(Y = 0) = 1,$ the conclusion that $\mathbb{E}[Y] = 0$ follows from the proposition above. This controls $\mathbb{E}[\lim X_n]$ (or again, any version of this) to have mean $0$ using $0 \le |\mathbb{E}[\lim X_n]|\le \mathbb{E}[Y] = 0.$

stochasticboy321
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  • How did you conclude the inequality in the last line? – Hosein Rahnama Nov 23 '23 at 11:57
  • Can we make the argument more concrete for this sample space which is not really a continuum? – Hosein Rahnama Nov 23 '23 at 12:16
  • The quickest way to see $|\mathbb{E}[X]| \le \mathbb{E}[|X|]$ is via Jensen's inequality. More directly, define $X^+ = \max(X,0), X^- = \max(-X,0),$ both of which are nonnegative. Then $|X| = X^+ + X^-, X = X^+ - X^-.$ So $|\mathbb{E}[X]| = |\mathbb{E}[X^+] - \mathbb{E}[X^-]| \le \max(\mathbb{E}[X^+], \mathbb{E}[X^-]) \le \mathbb{E}[X^+] + \mathbb{E}[X^-] = \mathbb{E}[|X|]$ (where we are using that for positive $a,b, |a-b| \le \max(a,b)$). – stochasticboy321 Nov 23 '23 at 12:58
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    As for your sample space in the linked question, this definitely is a set of size continuum: map $\omega \mapsto \sum_i \mathbf{1}_{{H}}(\omega_i) 2^{-i}$. Then you generate every real number in $[0,1]$ (basically $H$ or $T$ corresponding to $1$ or $0$ in the binary expansion of the number). – stochasticboy321 Nov 23 '23 at 13:01
  • About the inequality, I mean $0 \le |\mathbb{E}[\lim X_n]|\le \mathbb{E}[Y]$. I am not sure that it really makes sense to talk about $\mathbb{E}(\lim X_n)$ as $\lim X_n$ is not really defined over all of $\Omega$ as you also pointed out and so it is not a random variable. – Hosein Rahnama Nov 23 '23 at 13:13
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    Oh, like I said in the parentheses, it's really for any version of this, since the object is not well defined. Strictly speaking, we I guess we should start by defining a version of $\lim X_n$, and extend that to a version of $\lim Y_n$, and then use the above, but I guess I wrote the two bits in the opposite direction and missed this. – stochasticboy321 Nov 23 '23 at 13:14
  • Whenever you got time, please fix the things you mentioned in your last comment. Then, I will accept your answer. Thanks for all the time you put in. :) – Hosein Rahnama Nov 24 '23 at 18:26