Timeline for median and mean of the sample mean of i.i.d. log-normal
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 5, 2023 at 15:37 | answer | added | user225256 | timeline score: 2 | |
| May 16, 2020 at 2:55 | comment | added | user225256 | Is there a clean proof for the case $n=2, \mu=0$? | |
| May 31, 2017 at 8:46 | history | tweeted | twitter.com/StackStats/status/869837464109801472 | ||
| May 31, 2017 at 8:43 | comment | added | Glen_b | Oh, this may have been it; it's more a generalization of the above case to a broader class: ... stats.stackexchange.com/questions/25010/… | |
| May 31, 2017 at 8:39 | comment | added | Glen_b | Well, there's one here. On the above example ... whuber draws some member-densities of that family related to the lognormal here. I thought there was another explicit example but I may have misremembered -- or maybe I just didn't locate it quickly | |
| May 31, 2017 at 7:51 | comment | added | Hans | @Glen_b: Can you post the links to those "other examples... on our site"? | |
| May 31, 2017 at 6:25 | comment | added | Hans | @Glen_b: You are right about the non-uniqueness of the function with the same moments when the Taylor expansion of the function does not exist. Your example is ingenious, since $$\sqrt{2\pi}\int_0^\infty f(x)\sin(2\pi k \log x)x^m dx = e^{\frac{m^2}2}\int_{-\infty}^\infty e^{-\frac{(z-m)^2}2}\sin(2\pi kz)\,dz = e^{\frac{m^2}2}\int_{-\infty}^\infty e^{-\frac{y^2}2}\sin(2\pi k(y+m))\,dy \equiv 0,\ \forall m,k\in\mathbb N.$$ | |
| May 31, 2017 at 6:07 | comment | added | Hans | @Glen_b: Sorry I reposted my comment because I thought there was something to edit. I am saying if the Taylor series exists or converges the function must be analytic in some disk and is unique. OK, you are saying if the Taylor series does not exist, there may be two function not analytic but with all the same moments, right? I will have to think about that. | |
| May 31, 2017 at 6:04 | comment | added | Hans | @Glen_b: Is your "here" not the same as what I said only in term of the characteristic function which only multiplying the parameter by the pure imaginary number i? I must have misunderstood your claim about the non-uniqueness of distribution given all the moments. The claim goes contrary to the uniqueness of analytic expansion or Taylor series expansion. | |
| May 31, 2017 at 5:55 | comment | added | Glen_b | It might be the same as what you said, I'd have to look more closely to determine that; I merely wanted to give some link to what I would have said if I had given a more complete statement. ... On the second thing, given the MGF doesn't exist for the lognormal (in the sense I gave), why would you assume the convergence of Taylor series? | |
| May 31, 2017 at 2:53 | comment | added | Glen_b | I mean "no mgf" in the sense expressed here. By "other distributions with the same moment sequence exist" I mean there are distributions with the same sequence of moments as the lognormal / the lognormal is an example of a distribution not uniquely defined by its moments: if $f$ is a standard lognormal density ($\mu=0$, $\sigma=1$) and $g(x)=f(x)\cdot (1+\epsilon\sin[2\pi k\log(x)])$ then the contribution of $g-f$ to the $n$th moment is $0$ for each $n=1,2,...$. This extends to the general lognormal (indeed to the 3 parameter case) | |
| May 31, 2017 at 1:51 | comment | added | Hans | @Glen_b: By "no mgf" you mean the characteristic function is not analytic at the origin on the complex plane, and therefore there is no convergent Taylor series at the origin, right? What do you mean by "other distributions with the same moment sequence exist"? If all moment are equal, you should get the same series, therefore the same function if the series converges, except the complex variable may be rotated in the complex plane. | |
| May 31, 2017 at 1:08 | comment | added | Glen_b | The lognormal is an interesting case; it's a distribution with all moments finite but no mgf (and indeed, other distributions with the same moment sequence exist). You may be able to get somewhere with characteristic functions. | |
| May 31, 2017 at 1:06 | comment | added | Glen_b | Incidentally, somewhat related (at least in the limit): Peter Hall, (1980), "On the Limiting Behaviour of the Mode and Median of a Sum of Independent Random Variables", Ann. Probab. Volume 8, Number 3, 419-430. Open Access at Project Euclid | |
| May 31, 2017 at 1:03 | history | edited | Hans | CC BY-SA 3.0 |
Make the question more a conjecture.
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| May 31, 2017 at 0:56 | comment | added | Hans | @Glen_b: I am computing the sample mean of the lognormal random variables via Monte Carlo. The sample mean seems tend to concentrate below the mean for large $\sigma$. I am wondering whether this is true for all cases. | |
| May 31, 2017 at 0:42 | history | reopened | whuber♦ | ||
| May 31, 2017 at 0:41 | history | closed | whuber♦ | Needs details or clarity | |
| May 31, 2017 at 0:38 | comment | added | Glen_b | One could show that the third moment skewness is positive easily enough, but that doesn't automatically imply that the mean will exceed the median (I certainly believe the mean will exceed the median here but we'd need to prove that it does). What's this for? | |
| May 31, 2017 at 0:28 | history | edited | Hans | CC BY-SA 3.0 |
edited title
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| May 31, 2017 at 0:22 | history | edited | Hans | CC BY-SA 3.0 |
Pose the question in rigor and detail.
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| May 30, 2017 at 21:19 | history | asked | Hans | CC BY-SA 3.0 |