Example of $X_n$ which converge a.s. but not in mean

Question

Provide an example of a sequence of random variables which converge a.s. but not in mean.

I know that the random variables $X_n=n\cdot\mathbb{1}_{(0,\frac{1}{n})}$ converge in probability as given any $\varepsilon>0$ \begin{align*} P(|X_n-0|>\varepsilon)=P(X_n>\varepsilon)\le P(X_n>0)=P\big(\big(0,\frac{1}{n}\big)\big)=\frac{1}{n}\to 0 \end{align*} However, they do not converge in mean as

\begin{align*} E|X_n-0|=E\big(n\cdot\mathbb{1}_{(0\frac{1}{n})}\big)=n\cdot P\big(\big(0,\frac{1}{n}\big)\big)=n\cdot\frac{1}{n}=1\,\,\text{for all n} \end{align*}

So, my question here is do these $X_n$ converge to $0$ a.s.? And if so, how does one show this rigorously with an $\varepsilon$ proof? I know we need to find $N\in\mathbb{N}$ such that $X_N<\varepsilon$. To this end, we can make the lengths of the intervals $\big(0,\frac{1}{n}\big)$ arbitrarily small but the multiplication by $n$ stops $X_n$ from being arbitrarily small, so I am sort of thinking that these $X_n$ do not converge a.s. If thats the case, whats an example that would work here?

What is $P((0,\frac1n))$ here? Or rather, what's the underlying probability measure? Uniform on $[0,1]$? — mostsquares, Jul 17 '20 at 18:34
@cwindolf I think he means "the measure of the event $(0, \frac{1}{n})$". Since his universe is $\Omega = (0,1)$, this makes sense. — FiMePr, Jul 17 '20 at 18:38
@FiMePr Yes, that is exactly what I meant, sorry that wasn't more clear. — The Mad Scientist, Jul 17 '20 at 18:39

FiMePr · Accepted Answer · 2020-07-17T18:35:49.770

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You want to prove that the event $\lbrace X_n(\omega) \rightarrow 0 \rbrace$ has probability $1$.

I claim that this event is the universe $\Omega$ itself. In fact, for any $\omega \in (0,1)$, the sequence $X_n(\omega)$ eventually vanishes.

edited Jul 17 '20 at 18:35

answered Jul 17 '20 at 18:27

FiMePr

1,219

Ahhh, indeed. I was not working with the $\omega$'s properly. So, for any $\omega\in\Omega$ there exists an $N\in\mathbb{N}$ such that $\omega\notin\big(0,\frac{1}{N}\big)$ and thus $X_n=0<\varepsilon$ for all $n\ge N$. And so $X_n(\omega)\to 0$ for all $\omega$, thus $P(X_n\to 0)=1$. Is that correct? – The Mad Scientist Jul 17 '20 at 18:47
@SpiderBite Exactly. – FiMePr Jul 17 '20 at 18:50
Awesome, thank you! – The Mad Scientist Jul 17 '20 at 18:50

score 1 · Answer 2 · edited May 15 '25 at 01:56

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Provide an example of a sequence of random variables which converge a.s. but not in mean.

Consider a random variable $U$ which is uniform in $[0;1]$ and let

$$ X_n= \begin{cases} n, & \text{if $U \leq \frac{1}{n}$ } \\ 0, & \text{if $U >\frac{1}{n}$} \end{cases}$$

We have

$$\lim_{n\rightarrow\infty}\mathbb{E}[X_n]=\lim_{n\rightarrow\infty} n\mathbb{P}(U\leq\frac{1}{n})=1$$

On the other hand, for any outcome $\omega$ for which $U(\omega)>0$ (which happens with probability 1), $X(\omega)$ converges to zero.

Thus $X_n \rightarrow 0$ a.s. but $\mathbb{E}[X_n]$ does not.

edited May 15 '25 at 01:56

jII

3,158

answered Jul 17 '20 at 18:42

tommik

33,201
4
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Beautiful, thanks for the alternate example! – The Mad Scientist Jul 17 '20 at 19:02
Isn't it the same example ? – FiMePr Jul 23 '20 at 09:10

score 0 · Answer 3 · answered Jul 17 '20 at 19:54

Payoff in the Gambler's ruin problem with a double until the first win, then stop. If you start with A coins and double the bet every time you lose, assuming you have infinite money, current wealth converges: $$ X_n(\omega) \to_n X(\omega)=A+1 \text{ wp 1} $$ So $E \lim X_n = A+1$. At the same time, $\lim_n EX_n = -3 2^{n+1} \frac{1}{2} + 2^{n+1}\frac{1}{2}$

Example of $X_n$ which converge a.s. but not in mean

3 Answers3