Timeline for A maximal inequality
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| May 15 at 11:51 | comment | added | Iosif Pinelis | @MathRevenge : I guess I misunderstood you "approximate the limit". I thought you were concerned with the rate of convergence of $q_j$ to $q$. Of course, there is no problem finding $q$ with any degree of accuracy. | |
| May 15 at 5:04 | comment | added | MathRevenge | Thank you @Iosif Pinelis for your patience. Regarding the approximation of the limit $q$ I think that math.stackexchange.com/questions/356903/… and math.stackexchange.com/questions/3126681/… may help! Indeed, in this case the minimum is reached at $u = \frac{1}{2}$ and that probability is approximately $q \approx 0.0479435$ | |
| May 14 at 22:08 | comment | added | Iosif Pinelis | @MathRevenge : Concerning your latter comment, we divide $b_{j-1}\sim2^{(j-3)/2}$ by $\sim2^{(j-1)/2}$ and get $\sim1/2$. | |
| May 14 at 22:06 | comment | added | Iosif Pinelis | @MathRevenge : Concerning your penultimate comment, there are somewhat explicit bounds on the rate of convergence in the invariance principle. See e.g. papers by Borovkov. But, no, there is no easy way to derive such bounds. | |
| May 13 at 14:55 | comment | added | MathRevenge | I'm having trouble in understanding this step: $\min_{x\in[-b_{j-1},b_{j-1}]}P(|S_m+x|\le b_j\ \forall m\in\{1,\dots,2^{j-1})\to\min_{u\in[-1/2,1/2]}P(|W_t+u|\le1/\sqrt2\ \forall t\in[0,1])$ . I guess you are dividing by $\sqrt{2^{j-1}+1}$ each term.. but then the variable $u\in[-1/\sqrt{2},1/\sqrt{2}]$, not $u\in[-1/2,1/2]$.. Probably I'm making some obvious mistakes, but I would like if you could help me and explain this step. Thank you in advance! | |
| May 13 at 14:24 | comment | added | MathRevenge | Hi @Iosif Pinelis, is there a "straight forward" way to approximate the limit $q = \lim_{j\to\infty}q_j$? | |
| May 13 at 6:08 | vote | accept | MathRevenge | ||
| May 13 at 6:08 | |||||
| May 12 at 18:20 | comment | added | MathRevenge | I understand now! Ok.. thank you for your time and for your precious help. | |
| May 12 at 18:05 | comment | added | Iosif Pinelis | @MathRevenge : Well, again: it follows from the comment by mathworker21 that $p_{1,1}$ can be however small and hence $p_{1,n}$ can be however small. So, no nonzero lower bound on $p_{1,n}$ exists in general, and the distribution of the $X_i$'s is not specified in your post. | |
| May 12 at 17:43 | comment | added | MathRevenge | Of course, in this case it doesn't work as I said.. but maybe there exist other bounds I don't know which we can use to solve this! My question is all about the existence of such a function $g$ given the nature of the $X_i$ which satisfy the bound above. Anyway your answer teaches me a good technique. So thank you! Upvoted. | |
| May 12 at 17:40 | comment | added | Iosif Pinelis | @MathRevenge : This approach will not work, because what you denoted by $g(n)$ in your latter comment will likely be greater than $1$. It is highly unlikely that it will be like $1-n^{-c}$, as it would have to be according to my answer. | |
| May 12 at 17:36 | comment | added | MathRevenge | Well, $P(\max_{k=1,\dots,n}\frac{S_k}{c_k} \leq 1) = 1 - P(\max_{k=1,\dots,n}\frac{S_k}{c_k} \geq 1) \geq 1 - h(n) = g(n)$ where $h(n)$ is given by that answer. But as I said, it gaves a "trivial" bound since $g(n)$ becames eventually a negative number. | |
| May 12 at 17:36 | comment | added | Iosif Pinelis | @MathRevenge : Your post does not have any specifics as $X_i=\cos\theta_i$. Let's just stick to what your post says. | |
| May 12 at 17:33 | comment | added | Iosif Pinelis | @MathRevenge : The bound in that answer is an upper bound on a large deviation probability, whereas your question is about a lower bound on a small deviation probability. The two questions have nothing to do with each other. | |
| May 12 at 17:31 | comment | added | MathRevenge | I'm trying to understand a more general case.. where we only have these hypothesis. Although this problem was born when I was considering $X_i = \cos{\theta_i}$ with $\theta_i$ i.i.d. uniformly distributed on $[0,2\pi]$. For example, in this case, is there such a function $g$? And if the answer is positive, based on the nature of the $X_i$ could we find a general $g$ (which depends on $X$) which atisfy the inequality above? | |
| May 12 at 17:29 | comment | added | Iosif Pinelis | @MathRevenge : What do you mean by "my case"? | |
| May 12 at 17:19 | comment | added | MathRevenge | In my case $p_{1,1} \approx 0.5$.. I'm searching for a bound similar to mathoverflow.net/questions/146397/… (look at the first answer). By the way, we can use the bound in this question by moving on to complementaries, but it gaves a "trivial" bound which is not useful at all. I need something like $p_{1,n} \geq g(n)$ for all $n\geq n_0$. | |
| May 12 at 17:00 | comment | added | Iosif Pinelis | @MathRevenge : Note that $p_{1,n}\le p_{1,1}$ and, as noted in the mentioned comment by mathworker21, $p_{1,1}$ can be however small. So, $p_{1,n}$ can be however small. | |
| May 12 at 16:28 | comment | added | MathRevenge | I meant that the inequality works for all sufficiently large $n$, but I would like $N = 1$ (using your notations).. although mathworker21 said that there are no such $g(n)$.. but I don't understand why it is not the case. By simulating it seems that $1/n$ works (for example). But we have to prove it, or at least find such a function $g$, eventually different but such that it satisfy the property. | |
| May 12 at 15:52 | comment | added | Iosif Pinelis | @MathRevenge : I followed your linked comment, made in response to the earlier comment by mathworker21. | |
| May 12 at 14:45 | comment | added | MathRevenge | Thank you very much for pointing out this idea! But..there is no way to adjust this argument in order to fix $N = 1$? So that we can consider the maximum of the first $n$ instead of the maximum between the $N$-th and $n$-th observation. Please let me know and thank you again. | |
| May 12 at 14:38 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 2 characters in body
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| May 12 at 14:28 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |