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Suppose a sequence $U_n$ of random variables satisfies the following conditions: For each $n$,

$$P(U_n = 1) = 1/n$$ and $$P(U_n = 0) = 1 - 1/n.$$

Can someone tell me whether this converges almost surely to 0? I know it converges to 0 in $L_1$. I'd say it converges to 0 a.s as well because $P(U_n = 0) = 1- 1/n \to 1$, which I think is consistent with the definition of a.s convergence, but apparently according to some exercise I'm wrong.

Can someone tell me what is correct?

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Consider the functions $\chi_{[0,1]}$, $\chi_{[0,1/2]}$, $\chi_{[1/2,1]}$, $\chi_{[1,1/3]}$, $\chi_{[1/3,2/3]}$, $\,\ldots$. This sequence converges to $0$ in $L_1([0,1])$; but the sequence converges pointwise nowhere. (The sequence can be thought of as a line segment of height $1$ and width tending to $0$ that forever slides across the entire interval $[0,1]$ (if that makes any sense).) A sequence similar to this can be constructed satisfying your conditions (note $\sum(1/n)$ diverges here). – David Mitra Dec 12 '12 at 15:31
up vote 3 down vote accepted

Let $\Omega=[0,1]$ endowed with the Borel $\sigma$-algebra and the uniform distribution. Define inductively two sequences $a_n$ and $b_n$ such that $a_n=0$, $b_1=1$, $a_{n+1}=b_n$ $b_{n+1}=a_{n+1}+1/n\textrm{ mod } 1$. For any $n$, if $a_n<b_b$ let $U_n(\omega)=1$ if $\omega\in[a_n,b_n]$ and $0$ otherwise. If $a_n>b_n$ let $U_n(\omega)=1$ if $\omega\geq a_n$ or $\omega\leq b_n$ and $0$ otherwise. You can check that this satisfies your assumptions, but for all $\omega$, $U_n(\omega)=1$ for infinitely many $n$. The latter folows from the divergence of the harmonic series.

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Ok then What I don't understand is why my original sequence doesn't verify this definition :… (the first formula). It seems to be exactly what I wrote in my question and it converges to 0 – lezebulon Dec 12 '12 at 15:35
@lezebulon You didn't specify any original sequence. In some cases, a sequence of random variables satisfying your assumptions converges to $0$ a.s. My example shows that in some cases, it will not. – Michael Greinecker Dec 12 '12 at 15:37
But what I wrote in my 1st question ( P(Un=0)=1−1/n→1 ) is correct for every sequence that verifies these assumption right? I just don't see where is the incorrect part – lezebulon Dec 12 '12 at 15:52
@lezebulon: You appear to be confusing almost sure convergence with convergence in probability. As an interesting aside: While your random variables here are discrete, their properties imply that they cannot be constructed on a discrete probability space! Exploring this fact may help to clarify where your confusion arises. – cardinal Dec 12 '12 at 16:10
@lezubulon A random variable is a measurable function $f:\Omega\to\mathbb{R}$ defined on some probability space $(\Omega,\Sigma,\mu)$. You cannot discuss almost sure convergence without actually looking at this probability space. – Michael Greinecker Dec 12 '12 at 16:33

This is one typical example for showing that almost-sure convergence is a strong property: this sequence does not converge a.s, even if it converges (to zero) in probability.

To get an informal insight: imagine you have a lot (say, 1000) realizations of this process. Ask yourself how many of this realizations will converge to 0. Recall that, for each fixed realization, what we have is a plain numeric sequence, so we are speaking of the common (non-random) sequence convergence now. But our sequences are composed of 0's and 1's; hence a sequence will converge to zero only if it has finite number of ones; ie. if $\exists n_j,$ $x[n_j]=1; x[n]=0 , n>n_j$. Intution says that is not very probable that this will happen for the majority of our 1000 realizations; actually, on the contrary, it's extremely improbable that it happens ever.

We can compute the probability explicitly (this not necessary, nor even usual, to prove/disprove convergence). Let $A_k$ be the event $x[k]=1; x[n]=0 , n>k$. By telescoping the product we get that each probability (for fixed $k$) is zero:

$$P(A_k)= \frac{1}{k} \prod_{n=k+1}^{\infty} \left(1 - \frac{1}{n} \right)= \frac{1}{k} \frac{k}{k+1}\frac{k+1}{k+2} \cdots = 0$$

The probability of convergence is $\sum_k P(A_k)$ ; but the (countable) union of events of probability zero has probability zero. Hence, the probability is not one (as we should have got, if convergence was almost sure), but, on the contrary, zero.

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