12:02 AM
When you say "the dimensionality of the samples" do you mean the length of the vector mu?

12:14 AM
@nullgeppetto Your question sounds like it might be the same as this one:
9

Gaussian data distributed in a single dimension requires two parameters to characterise it (mean, variance), and rumour has it that around 30 randomly-selected samples is usually sufficient to estimate these parameters with reasonably high confidence. But what happens as the number of dimensions ...

@whuber not sure if my ping will work there so I thought I'd raise it here: are those actually asking the same thing?
0

The question I have is: Define X,Y to be two independent uniform(0,1) random variables and $Z:=\frac{Y}{X}$ Compute $P(X<x|\sigma(Z))$ The answer given apparently by "straightforward elementary computations" is for $x\geq 0$ $P(X<x|\sigma(Z))$=min$\{{x^2,1}\}\mathbb{I}_{Z\leq 1}$ $+$ min$\{{x... 12:45 AM @Glen_b Hi Glen, your ping worked in both places. If you think there's good cause to re-open the new question, then by all means please do so. I made the close decision as soon as I saw that the duplicate carried out the calculations that this new question appears to be asking for, but if for some readers (who are capable of understanding the new question) you think the duplication would not be evident, then the threads perhaps ought both be open. 12:56 AM @whuber No doubt I am exposing one of the (many) gaps in my knowledge, but it's not immediately clear to me; I expect I'm not alone in that. [That may well indicate I am not included in the subset you suggest, so I'm left unsure that I should reopen it either] ... I expected that a brief explanation of why it's an answer to the question (say in a comment) might do. I'll delete my comment there. Nice to know that the ping works (I felt like it should but that doesn't always mean much). 1:11 AM @whuber That's odd, the notification of your reply took a long time to show up for me. I'll now undelete my comment again so your reply has some context. I'm still not clear how this tells us the conditional distribution of$X$rather than the marginal distribution of$Z\$ (it's here that my ignorance would be the issue... or perhaps it's that I need to reread the answers at the older question more closely)

8 hours later…
9:35 AM
@gung I voted on that one too. I could not suggest it yesterday because I did not have enough rep in the tag, so by now I put this tag on two threads with my answers which touch on the topic of unbalanced classes, and now I could vote :)

9:57 AM
@amoeba: Thanks! I'm thinking that rather than go through & approve all those I'll wait for a bit & see how many others vote.

10:18 AM
@Scortchi Can you vote yourself or is your vote then immediately deciding as when mods are voting to close/open/etc?

11:05 AM
Looking at what other high-rep non-mods are frequenting chat room I see @Sycorax and @NickCox. Consider voting for the recent tag synonym suggestions here stats.stackexchange.com/tags/synonyms. The relevant discussions are in meta.stats.stackexchange.com/questions/1200. I am trying to see if we can implement some of these synonyms without mod intervention (I would guess not but still).

2 hours later…
1:04 PM
Sorry; everyone has to have small rules for what they don't do (for example, many high rep users edit rarely) and mine is that I am not (yet) interested in tags. No good reason why not; it's just a personal rule to limit time spent.

1:45 PM
Fair enough @NickCox. There are many things here that I don't do either.

2:13 PM
@amoeba: I checked with Deming-regression -> total-least-squares & my vote's binding. (And I can't turn bindingness off.)
If no-one's interested I'll of course go through the rest, but I think it's a good idea to try using the system we've got.
And using the suggestion mechanism is already a good start.

2:37 PM
@Scortchi I agree it makes sense to try using the existing system more. Let's see if we can make the community approve synonyms at least in the more popular tags such as [mean] or [confidence-interval]. For the less popular tags such as TLS it cannot work in principle: there are less than 5 people with over 5 rep points there.