0

Consider the sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ defined by $$\begin{align*} X_1(\omega) &:= 1_{\big(\frac{1}{2},1 \big]}(\omega) \\ X_2(\omega) &:= 1_{\big(0, \frac{1}{2}\big]}(\omega) \\ X_3(\omega) &:= 1_{\big(\frac{3}{4},1 \big]}(\omega) \\ X_4(\omega) &:= 1_{\big(\frac{1}{2},\frac{3}{4} \big]}(\omega)\\ &\vdots \end{align*}$$

I need to show whether this sequence converges in $L^1$ and a.s. but I am not sure about what $X_5$, $X_6$, $X_7$ ...looks like. Can you help me?

LearningProb
  • 498
  • 2
  • 11
  • 1
    This looks like a standard example: I assume the pattern is supposed to be that on each round $k$ we chop the unit interval into $2^k$ equal sized disjoint subintervals. Then each round $k$ has $2^k$ sub-rounds where we individually consider each subinterval associated with round $k$. If you put all the subrounds in order one after the other you can enumerate them, obtaining a sequence of intervals as is done above. – Michael Dec 05 '19 at 23:38
  • 1
    Thank you. So we have $X_5=(1/4,1/2]$,$X_6=(0,1/4]$,$X_7=(7/8,1]$,$X_8=(6/8,7/8]$. Am i right? – LearningProb Dec 06 '19 at 11:31
  • Yes, that is the sequence of intervals. The intervals are shrinking in size to 0-size so $P[X_n=1]\rightarrow 0$ and $P[X_n=0]\rightarrow 1$ (as would be the case for any other sequence of intervals with size that goes to zero). The intervals are arranged like this to ensure prob 1 convergence is impossible. – Michael Dec 06 '19 at 16:48
  • Note that convergence with probability 1 is also called convergence almost surely. This is different from the much weaker type of convergence called convergence in probability. You have a textbook example of a random process ${X_n}_{n=1}^{\infty}$ that converges to 0 in probability, but not with probability 1. – Michael Dec 07 '19 at 18:38

1 Answers1

1

It is impossible to say for sure what $X_5,X_6...$ are, but this looks like a standard example of sequence which convreges in measure but not almost surely. So my guess is $(X_n)$ is the sequence obtained by arranging the functions $I_{[\frac {i-1} {2^{n}},\frac i {2^{n}})}$ in sequence with increasing order of the denominator $2^{n}$ and increasing order of $i$ within each block. In that case $EX_n$ is the sequence $(\frac 1 2,\frac 1 2, \frac 1 4,\frac 1 4 ,\frac 1 4,\frac 1 4,...)$ where $\frac 1 {2^{n}}$ is repeated $2^{n}$ times. Hence $EX_n \to 0$ which means $X_n \to 0$ in $L^{1}$. The sequence does not converge almost surely because, at every point, there are infinitely many $0$'s and $1$'s.

Kavi Rama Murthy
  • 311,013
  • So the idea is that each interval becomes of measure zero as gets larger so that the lim of $E[X_n]= 0$ but the sequence of andom variable keeps depending on N as N gets larger, so that it does not converge a.s to any random variable, not only to 0. Actually $P(w:X_n(w)=0)=0\neq1$. Am I right? – LearningProb Dec 06 '19 at 11:10
  • @LearningProb Yes, what you have written is correct. – Kavi Rama Murthy Dec 06 '19 at 11:43
  • @LearningProb : Actually, in this example we have $$ \lim_{n\rightarrow\infty}P[\omega: X_n(\omega)=0]=1$$ So $X_n$ converges to 0 in probability, and in mean square, and in L1, but not with probability 1. – Michael Dec 06 '19 at 16:39
  • https://math.stackexchange.com/questions/1170559/convergence-types-in-probability-theory-counterexamples/1170661 according to this there is a.s covergence so we can not have that limit to be one – LearningProb Dec 06 '19 at 16:51
  • So, do we have convergence in L1 or not? – LearningProb Dec 06 '19 at 17:58
  • Yes: compute the $L^1$ norm : it clearly goes to zero. So you have convergence in $L^1$ to the variable $0$. – justt Dec 06 '19 at 18:06
  • Yes, I meant to say do we have A.S. convergence? This is not clear since we have two conflicting answers – LearningProb Dec 07 '19 at 10:35
  • @Michael claims that \lim_{n\rightarrow\infty}P[\omega: X_n(\omega)=0]=1 implying $X_n$ converges a.s., while Kabo claims that this is not the case. – LearningProb Dec 07 '19 at 16:29
  • There is no disagreement between Kabo's answer and my comments. We are both saying the same thing (to emphasize that, I have now +1'd Kabo's answer, which I neglected to do before). Kabo incorrectly wrote a comment "yes what you have written is correct" in response to an (incorrect) suggestion that $P[\omega : X_n(\omega)=0]=0$. But that was likely only because Kabo did not read the last sentence fully. – Michael Dec 07 '19 at 18:19
  • The fact that $P[X_n=0]\rightarrow 1$ shows: (i) The LearningProb assertion $P[X_n=0]=0$ for all $n$ is false; (ii) $X_n\rightarrow 0$ in probability. It does not imply that $X_n\rightarrow 0$ with probability 1. For that you need to evaluate $$P\left[\omega : \lim_{n\rightarrow\infty} X_n(\omega) = 0\right]$$ which is not the same thing as $\lim_{n\rightarrow\infty}P[\omega : X_n(\omega)=0]$. – Michael Dec 07 '19 at 18:24
  • Ok thank you for clarifying – LearningProb Dec 07 '19 at 18:58