Timeline for Maximum of two normal random variables
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2017 at 5:31 | comment | added | usul | Since this post has been bumped and I don't think the idea here is that obvious, I believe it helpful to explain: for any $t$ at all, $E \max(X_j) = t + E\max(X_j - t) \leq t + E\max((X_j-t)_+) \leq t + \sum_j E(X_j-t)_+$. The first inequality is an equality if the largest $X_j$ always exceeds $t$, and the second is an equality if only the largest ever exceeds $t$ (so other terms are always zero). This is only possible for fedja's choice of $t$ and a worst-case correlation where with probability $1$, exactly one $X_j$ exceeds $t$. | |
| Apr 22, 2017 at 12:39 | comment | added | rgrig | My intuition for your solution is this: Pick $t$ such that (in expectation) exactly one of the random variables is above it. Then, in the sum $\sum_j \mathbb{E}(X_j-t)_+$, we have (in expectation) that all the terms are $0$, apart from one which is $\mathbb{E} X_j - t$, and it must be the maximum. But, the proof is not obvious to me. Also, the intuition explains only why $\mathbb{E}\max_j X_j$ is roughly $t+\sum_j\mathbb{E}(X_j-t)_+$, but not why the latter is an upper bound. I could, perhaps, figure it out with some work. | |
| Jul 19, 2014 at 21:14 | review | Suggested edits | |||
| Jul 19, 2014 at 21:20 | |||||
| Jun 20, 2014 at 21:47 | comment | added | fedja | @Proba Anyway, putting all other things aside, if you had asked about the actual setup you have instead of all this, there would be much less room for misinterpretation and the responses would be much more beneficial to you. So, I suggest we scrap this discussion and start the one that is more interesting for everybody involved :-) | |
| Jun 20, 2014 at 21:44 | comment | added | user0820 | Thanks, this last comment is exactly something I was hoping to hear, i.e. whether there is something clever about $e^{\lambda x}$ or this is just something that's computable and at the same time gives something nontrivial. | |
| Jun 20, 2014 at 21:37 | comment | added | fedja | @Proba Good point. However, the answer stays: there is no intuition, just whoever did it recalled (consciously or unconsciously) the Bernstein estimate for the sum of independent Gaussians (where it makes sense because the exponential function is pretty much the only non-trivial one that allows an easy computation of the expectation) and decided to do the same and see what comes out of it. There is absolutely no reason to prefer $e^{\lambda x}$ to $x^p$ or $e^{\lambda x^2}$ and "Laplace transform" is nothing more than a buzzword in this particular case. | |
| Jun 20, 2014 at 20:37 | history | answered | fedja | CC BY-SA 3.0 |