Timeline for Expected value of the maximum of the periodogram
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2020 at 3:38 | vote | accept | Cm7F7Bb | ||
| Nov 28, 2018 at 15:49 | comment | added | fedja | @Cm7F7Bb It is true that there formally may be atoms but, since only the distribution matters, you can always think that the probability space is $[0,1]$ (or something else like that), so you can choose some part of the level set that has exactly the right measure. | |
| Nov 28, 2018 at 15:19 | comment | added | Cm7F7Bb | I’m sorry to bother you again, but could you explain how you define $s$? If we choose some $p$, the set $F$ such that $P(F)=p$ might not necessarily exist unless the random variable is continuous. Is that correct? Can we set $s=s(p)=\inf\{x>0:P(|X|>x)\le p\}$? Then we would have $P(|X|>s)\le p$ and $P(|X|>s-\delta)>p$ for $\delta>0$. I’m really interested in this proof and I think that there are a lot of great ideas in this proof that I want to understand, but please let me know if I’m bothering you too much with my questions. | |
| Nov 28, 2018 at 15:01 | comment | added | fedja | @Cm7F7Bb You are completely correct about how the set $F$ is chosen: it is a level set of measure $p$. The thing that seems to confuse you is that you denote the level itself by $s$ while my $s$ is merely an upper bound for this level. So your $s^qp\le 1$ is exactly my $s^qp=1$. | |
| Nov 28, 2018 at 10:24 | comment | added | Cm7F7Bb | Could you please explain in more detail how the set $F$ is chosen? We have that $P(F)=p$ and $F=\{|X|>s\}$ since $|X'|\le s$. What is the relation between $s$ and $p$? I can only obtain that $s^qp\le 1$ using Markov's inequality and the assumption that $E|X|^q=1$. However, it is stated in the proof that $s^qp=1$ which is obtained using Jensen's inequality. Could you explain in more detail? I would greatly appreciate that. | |
| Nov 23, 2018 at 11:37 | comment | added | fedja | @Cm7F7Bb You just show that the expectation of the maximum for $X_r$ is at most $C2^{-2\delta r}\log n$ and then use the triangle inequality. | |
| Nov 23, 2018 at 10:49 | comment | added | Cm7F7Bb | I’m struggling to understand why it is enough to treat each $X_r$ separately. We need to show that $$ E\max_{1\le j\le n}\Bigl|\sum_{t=1}^na_{jnt}\sum_{r=0}^{\log_2n}X_{rt}\Bigr|^2=O(\log n), $$ where $|a_{jnt}|\le n^{-1/2}$. I find it a little bit confusing when the notation is changed from $X_r$ to $2^{-\delta r}X$ since $X$ is a random variable that we want to decompose. Could you please explain in more detail how the extra exponential factor $2^{-\delta r}$ works and why it is enough to consider each $X_r$ separately? I’d greatly appreciate that. | |
| Nov 21, 2018 at 16:01 | comment | added | fedja | @Cm7F7Bb What I'm doing is slightly different. I take the set of largest values of $|X|$ of fixed probability and replace $X$ on it by its average to get $X'$. I also start with the largest level, so the first thing I get is $Z$ (so I write $X=Z+X'$). Then I apply this trick to $X'$ to chop off $X_{\log_2 n}$ and so on. | |
| Nov 21, 2018 at 15:49 | comment | added | Cm7F7Bb | I'd like to ask for a clarification about the decomposition of $X$. Define $F_0=\{|X|^q\le 1\}$, $F_r=\{2^{r-1}<|X|^q\le 2^r\}$ for $r\ge1$, $p_r=P(F_r)$ and $G_r=\{|X|^q>2^r\}$ for $r\ge0$. Denote $$ X_r =(X-p_r^{-1}E[X\chi_{F_r}])\chi_{F_r}, \quad Z =X\chi_{G_{\log_2n}}+\sum_{r=0}^{\log_2n}p_r^{-1}E[X\chi_{F_r}]\chi_{F_r} $$ so that $X=Z+\sum_{r=0}^{\log_2n}X_r$. If I use such a decomposition, $Z$ is never equal to $0$. I can modify this decomposition so that $P(Z\ne0)\le n^{-1}$, but then $X_r$'s are not equal to $0$. Could you please give a hint? | |
| Sep 10, 2018 at 1:18 | history | answered | fedja | CC BY-SA 4.0 |