Timeline for Maximum of a sequence of $n$ positive random variables where variance is an increasing function of $n$
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2013 at 2:21 | comment | added | Bullmoose | Unfortunately, the family of distributions in question is unknown, and I cannot constrain it to be unimodal. | |
| Oct 6, 2013 at 23:55 | comment | added | Brendan McKay | I'm guessing that a lot more can be said if the distributions are constrained to be unimodal (either continuous or discrete). | |
| Oct 6, 2013 at 14:17 | answer | added | Chassaing | timeline score: 1 | |
| Oct 6, 2013 at 10:19 | comment | added | Chassaing | Since the asymptotic behavior of the maximum usually depends much more on the tail behavior than on the variance, a bounded variable should achieve the lowest possible maximum, if there exists such a thing: a $X_{max}$ that is stochastically minimal. | |
| Oct 6, 2013 at 10:12 | comment | added | Chassaing | I assume here that the maximal variance for a random variable $X$ with given expectation $\mu$ and $a\le X\le b$ is achieved when $X\in\{a,b\}$ a.s.. | |
| Oct 6, 2013 at 10:04 | comment | added | Chassaing | Sorry, the parameter should be $(1+\tfrac{\sigma_n^2}{\mu^2})^{-1}$, and $\sigma_n^2=\sigma(n)$. Then $\mathbb{P}(X_{max}<\tfrac{\sigma_n^2+\mu^2}\mu)=(1+\tfrac{\mu^2}{\sigma_n^2})^{-n}$, if I did not do another mistake ... | |
| Oct 6, 2013 at 8:49 | comment | added | Chassaing | Heuristically, it seems that we need to know more about the variance $\sigma_n^2$. For instance, if the $X_i^n$ have the smaller essential supremum possible, which I would guess is when $X_i^n$ is $\tfrac{\mu^2+\sigma_n^2}\mu$ times a Bernoulli random variable with parameter $(1+\tfrac{\mu^2}{\sigma_n^2})^{-1}$, then if $n=o(\sigma_n^2)$ the max is $\tfrac{\mu^2+\sigma_n^2}\mu$ with a probability close to one, which is my guess for a lower bound, in this case. But this is heuristic at many stages. | |
| Oct 6, 2013 at 4:45 | comment | added | Nate Eldredge | @Bullmoose: Probabilists are used to expressing your "sequence of sequences" as a triangular array: a family of random variables $X_i^n$ indexed by $n \ge 1$ and $1 \le i \le n$, as Stephan Sturm suggested. | |
| Oct 6, 2013 at 4:28 | comment | added | Bullmoose | @StephanSturm Yes, that is correct (I don't know whether I should modify the question again, as "sequence of sequences" sounds rather unwieldy...) | |
| Oct 6, 2013 at 4:19 | comment | added | Stephan Sturm | So in fact you want a statement about the sequence of maxima $X_{\max}^n = \max\{ X_1^n, \ldots, X_n^n\}$ where $X_1^n, \ldots, X_n^n$ are iid with variance $\sigma(n)$? | |
| Oct 6, 2013 at 4:13 | comment | added | Kevin P. Costello | You need something more than just the variance to get any sort of lower bound, or else you could something like all of the $X_i$ being $n^{9}$ with probability $n^{-16}$ and $0$ otherwise (so that the variance increases quickly, but with very high probability all of the variables are $0$). | |
| Oct 6, 2013 at 4:04 | comment | added | Bullmoose | Oh -- I am really sorry for lack of clarity, I see how my comment can be misinterpreted. I meant to say that each random variable $X_i$ has variance $\sigma(n)$, which is the same for all $X_i$'s but is a function of the number $n$ of $X_i$'s in the sequence. | |
| Oct 6, 2013 at 3:53 | comment | added | Bullmoose | They are i.i.d., but the twist is that the variance of each r.v. is the function of their number in the sequence. Basically, imagine a machine that lets you pick a number $n$, and produces $n$ positive random variables with constant means and variances that are some increasing function of the number $n$ that you chose. I edited the question to clarify this (I also noticed a typo in the last paragraph, which I fixed.) | |
| Oct 6, 2013 at 3:48 | history | edited | Bullmoose | CC BY-SA 3.0 |
clarified question based on a comment and fixed a typo.
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| Oct 6, 2013 at 2:22 | comment | added | Nate Eldredge | I assume iid is a typo; they are just independent? And talking about "a positive random variable with increasing variance" seems confusing. A random variable only has one variance, so you really want to talk about your sequence. | |
| Oct 5, 2013 at 23:50 | history | asked | Bullmoose | CC BY-SA 3.0 |