Timeline for Kolmogorov 0-1 law counter examples for almost independent variables
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1, 2022 at 12:52 | comment | added | Iosif Pinelis | @BCLC : Thank you for the links. | |
| Sep 1, 2022 at 4:51 | comment | added | BCLC | Relevant? Link 1 Link 2 Link 3 | |
| Aug 31, 2022 at 20:33 | comment | added | Petri Kattilakoski | Yes, they are i.i.d. | |
| Aug 31, 2022 at 20:32 | comment | added | Iosif Pinelis | @PetriKattilakoski : The definition in your comment is better than the one in the OP. Still, are the $X_i$'s with $i\ne0\mod k+1$ independent? | |
| Aug 31, 2022 at 20:19 | comment | added | Petri Kattilakoski | In your example the $Y_n$ are not well defined because you access non defined indices (what is $Y_1$?). The natural fixing is $ Y_{n}=\bigoplus_{i=1}^{k}X_{n+i} $, and this way they are indeed totally independent (to the best of my understanding), as you can't infer anything about $Y_{i}$'s value given $Y_{j}$'s value. | |
| Aug 31, 2022 at 20:16 | comment | added | Petri Kattilakoski | $a$ is any integer. To clarify, this is my example for any $i\not\equiv0\pmod{k+1}$, we define $X_{i}\sim U\left(\left\{ 0,1\right\} \right)$, and for $i\equiv0\pmod{k+1}$ we define $X_{i}=\bigoplus_{j=1}^{k}X_{i-j} $ | |
| Aug 31, 2022 at 20:16 | comment | added | Iosif Pinelis | @PetriKattilakoski : I still do not understand the definition in your example. In particular: (i) What is $a$ there? (ii) What is the joint distribution of the $X_i$'s? (iii) What do you mean by "override"? Etc, etc. I also do not understand (a) why you think the $Y_n$'s are not well defined; (b) how you propose to "fix" them; (c) how a fact about a marginal one-dimensional distribution can imply such a statement about a joint distribution as a statement of independence. | |
| Aug 31, 2022 at 19:52 | comment | added | Petri Kattilakoski | In my post I "override" $ X_i $ (so in each k+1 variables, the first k are $\sim U\left(\left\{ 0,1\right\} \right)$, and the one after them is determined from the k $X_i$s before it. In your example in the above comment, the variables $ Y_n $ are not well defined, but the natural fixing of them makes them totally independent, as $X_{i}\oplus X_{j}$ is also $\sim U\left(\left\{ 0,1\right\} \right)$. | |
| Aug 31, 2022 at 19:45 | comment | added | Iosif Pinelis | @PetriKattilakoski : The definition of strong mixing is given in the paper linked in my answer. As for your example, I cannot discern any definition of a sequence of random variables there. If you meant $Y_n:=\bigoplus_{i=1}^k X_{n-i}$, where the $X_i$'s are independent $U(\{0,1\})$, then the sequence $(Y_n)$ is $m$-dependent for $m=k-1$ and therefore must obey the 0-1 law. | |
| Aug 31, 2022 at 19:39 | comment | added | Petri Kattilakoski | Naturally speaking, k-wise independence means that given a variable $X_i$, knowing the values of any (other) k-1 variables, does not give more information about $X_i$. Formally, it means that (for discrete variables) $$ \prod_{i=1}^{k}\mathbb{P}\left(X_{n_{i}}=a_{i}\right)=\mathbb{P}\left(\bigwedge_{i=1}^{k}X_{n_{i}}=a_{i}\right) $$ | |
| Aug 31, 2022 at 19:32 | comment | added | Iosif Pinelis | @ZachHunter : I do not know what $k$-independent would mean. I have added a link to the definition of $m$-dependence. | |
| Aug 31, 2022 at 19:32 | comment | added | Petri Kattilakoski | I am not quite sure I understand the concept of strong mixing, so take this comment with a grain of salt. While my example is indeed a strong mixing, I can think of a sequence of random variables such that ANY (K+1) of them are dependent, for example using linear equations with k variables, am I wrong? | |
| Aug 31, 2022 at 19:31 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 73 characters in body
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| Aug 31, 2022 at 19:26 | comment | added | Zach Hunter | do you mean $k$-independent and any natural $k>1$? | |
| Aug 31, 2022 at 19:21 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |