Mathematics

6:18 AM

Suppose that n random variables $X_1,\dots,X_n$ form a random sample from a discrete distribution for which the probability function is $f$. Determine the value of $Pr(X_1\leq\dots\leq X_n)$ and $Pr(X_1<\dots<X_n)$

Is that $Pr(X_1<\dots<X_n)=\dfrac 1{n^n}$

?

@skullpatrol , HI!

@skullpatrol,HI!

skullpatrol

Hi @sush how are you?

Yes, Fine!

skullpatrol

:-)

After 2 hours of peace, we are here:D@skullpatrol

skullpatrol

:D

6:36 AM

I think if the distribution was continuous there, then $Pr(X_1<\dots<X_n)=Pr(X_1\leq\dots\leq X_n)=\dfrac 1{n!}$ regardless of what the probability function is. Am I right?

@Mike?

8:29 AM

@robjohn how would you prove the convergence of $\sum \dfrac{1}{2^n-1}$?

@WillHunting: In depth?

8:47 AM

@DanielFischer how would you prove the convergence of $\sum\dfrac{1}{2^n-1}$?

@Hawk Comparing it to $\sum \frac{1}{2^{n-1}}$ for example, or the ratio test, or root test, whatever my fancy at the moment prefers.

@DanielFischer Is it possible using Cauchy Condensation? I am having troubles using that...

@Hawk That would mean to establish the convergence of $\sum \frac{2^k}{2^{2^k}-1}$. Which of course is obvious, but not as obvious as the convergence of $\sum \frac{1}{2^n-1}$.

Why would you want to throw condensation at it?

@DanielFischer I don't know...apparently...trying to apply what I most recently learnt

@DanielFischer Can you please show your way?

@Hawk One should apply what gives the result most easily. Of course one doesn't always know in advance which that would be, so does the first step of the most-likely seeming candidates to see whether that leads to an obvious continuation.

8:57 AM

@DanielFischer Yes, that is logical...but what seems the preferable candidate in this case?

@Hawk Here, for $n > 0$ we have $2^n - 1 \geqslant 2^{n-1}$, so an obvious majorisation, and we know that $\sum \frac{1}{2^{n-1}}$ converges.

@DanielFischer Suppose, we do not know that...and we want to explicitly prove that...then what would you do?

@Hawk Take the explicit formula for the partial sums of a geometric series. Note that $q^n \to 0$ for $\lvert q\rvert < 1$.

@DanielFischer Yes, I almost forgot the infinite GP sum...yes...it is done...thank you for that!

@Hawk Hi.

9:02 AM

@DanielFischer I learnt now...that even if I learn advanced things, I should never forget the simple ones...thank you a lot for reminding that in such a sweet way...

@Sawarnik Hi

@Sawarnik I could not recognise the picture!! New?

@Hawk The simple ones are the ones you need most.

@DanielFischer Yes, advanced things are ultimately made of the simple things...

Suppose that n random variables $X_1,\dots,X_n$ form a random sample from a discrete distribution for which the probability function is $f$. Determine the value of $Pr(X_1\leq\dots\leq X_n)$ and $Pr(X_1<\dots<X_n)$
Is that $Pr(X_1<\dots<X_n)=\dfrac 1{n^n}$
?@DanielFischer

@Hawk Hey, new developments here math.stackexchange.com/questions/750614/… . If Ivan is correct, then it suffices to prove for $x+y+z=5$, which I think you can do it!

@Hawk Yes, stolen :D

@Sawarnik Really...lets see..

@Sawarnik Stolen? from whom?

9:04 AM

@Hawk PK :D But he blocked me here :( I am thinking of some creative way to take revenge!

@Sawarnik Now he actually blocked you!

@Hawk Yes :( On the General Math room. On FB. Wherever he could.

@Hawk, if r and s are inverse functions, then r(X) ≤ y implies X ≤ s(y). Am i right?

@Sawarnik :D nice way of revenge...

@Hawk No, no. He did that. I am yet to take revenge! But I will :D

... just because I was asking him why did he block me on FB.

9:07 AM

@Sush Looks dubious. One would expect it to depend on the distribution (if the distribution is Dirac, the probability would be $0$ for $n > 1$).

@Sush please use $\LaTeX$

@Sawarnik Yes, I am appreciating your way of revenge...

@DanielFischer, oh,! First time heard about Dirak!

@Hawk That q was clear enough.

@Hawk, if $r$ and $s$ are inverse functions, then $r(X) ≤ y$ implies $X ≤ s(y)$. Am i right?

@Sawarnik Interestingly, nowadays I cannot clearly understand anything without $\LaTeX$

9:10 AM

@Sush Well, I think the probabilists call it something else. How do you call a distribtution with $P(\{x_0\}) = 1$ and $P(A) = 0$ for all $A$ not containing $x_0$?

@Sush What is $X$ and what is $y$?

@Hawk Are you in a cheerful mood again. Look at that inequality q once again!

@Hawk ratio test would be easiest, but we also have $\frac1{2^n-1}\le\frac1{2^{n-1}}$ so we can compare to a geometric series.

@Hawk but I see that @DanielFischer has already shown these :-)

@robjohn Yes, :)

@robjohn Yes, so he has, I asked him because I thought you were sleeping, it is quite late night there I think...

@Sawarnik That inequality question?

@DanielFischer, Sorry! I don't know.

9:12 AM

Hi @robjohn, you're early, aren't you? Or late ;)

@DanielFischer,I think if the distribution was continuous there, then $Pr(X_1<\dots<X_n)=Pr(X_1\leq\dots\leq X_n)=\dfrac 1{n!}$ regardless of what the probability function is. Am I right?

@Hawk I was afk for a while. I am now looking at some questions. Feel free to ping me.

@DanielFischer both, I guess...

@Hawk, variabls.

@robjohn Yes, sure...

@Sush $r=s^{-1}$?

@robjohn, if r and s are inverse functions , then r(X) ≤ y implies X ≤ s(y). Am i right?

@Hawk, yes.

9:14 AM

@Sush If $P(X_1 = X_j) = 0$, then it should be $\frac{1}{n!}$ by symmetry, if the $X_i$ are iid. But for a discrete distribution, not so.

@Sush if they are monotonically increasing.

@Sush then you are correct.

@Sush consider $r(x)=s(x)=1/x$

@DanielFischer, so am i right for continuous case?

@robjohn, thanks!

Learning early for an exam. I really don't understand some people.

9:19 AM

@Hawk I m finding introductory analysis interesting now! Although its very hard :|

@Sawarnik Let me test you a little..

@Sush If the $X_i$ are iid, and the probability that two (or more) are equal is $0$, then yes. Every permutation of the values then has the same distribution, and since you [almost] certainly get a sequence of distinct results, the probability is $\frac{1}{n!}$.

@Hawk I haven't studied at all to the depth which you are thinking!

@Sawarnik Try atleast?

@Hawk :) Ok :|

9:21 AM

@DanielFischer, OK! I think i will have to take some advance course in probability theory.

@Sush re you good at inequalities?

@Sawarnik What can you say about $f$ if $|f(x)-f(y)|\le 99(x-y)^2$

@Sawarnik, no.

:(

@Hawk Give me some time, if there is any chance I can solve this.

@Sawarnik sure...

Freeze

9:23 AM

@Sawarnik Hello

@Freeze Hello.

@Hawk Meanwhile you can try the inequality, or verify what Ivan has done.

Freeze

@Sawarnik You're Opulent?

@Freeze Means?

@Sawarnik yes...

@DanielFischer Have you done any work regarding Sobolev spaces?

9:29 AM

@LucioD I've done some work avoiding them ;)

But I know a little, maybe worth a try.

@DanielFischer Okay at the moment that sounds like good work.

@DanielFischer Actually I wanted to know your take on a post 'Extension and trace operators for Sobolev spaces'. It deals with some questions that I am stuck on as well.

@Hawk No idea till now :| Any hints?

@LucioD Ah, those fall into the region I most eagerly avoided.

@Sawarnik MVT

@DanielFischer What area of maths are you primarily involved in?

9:34 AM

@LucioD Whatever takes my fancy, that's mostly complex analysis, topological vector spaces and functional analysis, point set topology, and a bit of number theory these days.

@DanielFischer Any reason you avoid that region?

@LucioD Didn't like it. One needs too many inequalities on the dimensions, exponents and orders, never could remember any of them.

@Sawarnik any luck?

@Hawk Not really. What kind of answer should I get?

@Sawarnik what type of function is $f$?

9:44 AM

@Hawk Is it a constant? Just a wild guess. Coming in 10 mins...

@Sawarnik why? okay...

@DanielFischer Are you doing research at a university?

@LucioD No, it's a hobby.

@DanielFischer Do you play chess?

Heyall =)

What set is $\mathbb{T}$ ? It is in regard to some fourier analysis

9:58 AM

@LucioD Not really. In my youth I played half-seriously for a while, but I dropped it when it collided with other pastimes.

@N3buchadnezzar Five internet points say it's the unit circle (one-dimensional torus, whence $\mathbb{T}$).

@DanielFischer I have seen the notation $S_1$ for the unit circle, but never $T$. In topology T is reffered to as the torus, but it seems a tad out of context here. Thanks for the quick reply though !

Chris's sis

@robjohn the best thing I could get (these are my limits)