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Avra
Avra
00:57
@vitamind. The excerises in the book are not enough :0
 
1 hour later…
user10478
user10478
02:21
I am trying to understand whether this post (forumserver.twoplustwo.com/…) makes any sense. Suppose a gambler starts with $100$ points, and repeatedly engages in a positive expected value wager, with the goal of reaching $100000$ points in as few rounds as possible. Also suppose the gambler always sticks to wagering a constant percentage of current points. How can it be the case that both
1) If that constant percentage is greater than would be prescribed by Kelly Criterion, the exponential curve upward (assuming no bankruptcy) is less steep than if Kelly had been used, and 2) the expected number of rounds is minimized by wagering 100% of current points (granted with a much higher bankruptcy risk)?
leslie townes
leslie townes
02:47
i don't have a lot of intuition for this kind of thing. kelly betting is discussed in thomas koerner's "naive decision making." the latter bit might be in dubins and savage, "how to gamble if you must."
user10478
user10478
I did read over how to gamble if you must. It mainly focused on negative expected value games, with a few comments about the positive case. I'm actually comfortable with 2) from intuition and my own code simulations. I am having more trouble believing that betting higher than Kelly actually slows your pace down (as opposed to merely increasing risk of ruin).
SAJW
SAJW
is Kelly only applyable when EV>0?
user10478
user10478
That's the only time I would care to use it. Maybe if you had some goal in a negative EV game to minimize loss it could work, not sure.
leslie townes
leslie townes
03:02
i'm contractually obligated to promote how to gamble if you must because i learned linear algebra from dubins.
it's been ages since i read the book.
user10478
user10478
:)
Rithaniel
Rithaniel
Is $\mathbb{Q}(x_i)_{i\in\mathbb{R}}$ isomorphic to any kind of subfield of $\mathbb{R}$? I feel that this is equivalent to just having a subfield of $\mathbb{R}$ with an uncountable basis of algebraically independent elements, but I also feel like that's a simplification
Ted Shifrin
Ted Shifrin
03:57
This got discussed a few days ago. There are in fact uncountably many algebraically independent transcendentals in $\Bbb R$.
There was a link to a post on main or Overflow, going back to von Neuman.
Rithaniel
Rithaniel
04:11
Yeah, it was me asking it. My question now is "is this really an equivalent representation, or is there something else in the structure of the reals that would prevent this particular field of rational functions from being a subfield of it?"
Or, perhaps a more useful question: "The reals are not isomorphic to any proper subfield of the reals, so would this field of rational functions be isomorphic to a subfield or to the reals themselves?"
(But, either way, I'm asking at the wrong time. I need to go get sleep)
 
2 hours later…
Ted Shifrin
Ted Shifrin
06:11
@Rithaniel It seems to me that the maximal such extension has to be $\Bbb R$.
 
1 hour later…
Euler 2
Euler 2
07:13
why are the most useful questions on forums asked 10 years ago
leslie townes
leslie townes
07:26
math finished about 10 years ago, now people are just fiddling around
Euler 2
Euler 2
07:53
not just math
everything
leslie townes
leslie townes
yes
vitamin d
vitamin d
08:38
I read a quote once, but I forgot how it went exactly and from whom it was. I think it was something like "It is better to work with a small class with nice properties (or objects) than with a large class with no specific property." I can't remember.
leslie townes
leslie townes
sound advice.
i had a dream last night, but i forget what it was. :)
BannedUser
BannedUser
08:58
(removed)
 
2 hours later…
Mohcine
Mohcine
10:47
Let {X(t), t€ [0,00)} be poisson process , please, why do we have Prob{X(t+Δt)-X(t)=0}=Prob{X(Δt)=0}
Vrouvrou
Vrouvrou
11:42
hello, someone have an idea on how to start :math.stackexchange.com/questions/4235649/…
Peter John
Peter John
12:07
Let $G$ be a group and $A,B<G$ s.t. $G = A\oplus B$. I think if $H<A,H<B$ then $(A\oplus B)/(H\oplus \{e\})\simeq (A\oplus B)/(\{e\}\oplus H)$. But what is an isomorphism between them?
Thorgott
Thorgott
12:42
@Rithaniel @Ted the reals are not purely transcendental over the rationals, so fields like this will always be proper subfields
if you pick a maximal algebraically independent set $(x_i)_{i\in I}$, you obtain a tower of field extensions $\mathbb{R}/\mathbb{Q}(x_i;i\in I)/\mathbb{Q}$ with the latter extension purely transcendental and the first one algebraic and non-trivial
Rithaniel
Rithaniel
Okie dokie. How then might you express the reals in these sort of terms, then? Starting from a rational polynomial ring in uncountably many variables, that is. Is it even possible?
Thorgott
Thorgott
you can't even explicitly write down a maximal algebraically independent set, so I'm gonna press doubt on that one
Rithaniel
Rithaniel
Hmmmm, interesting. So you can express any subfield of the reals in these terms, but probably not the whole thing
Thorgott
Thorgott
@PeterJohn this is false btw
Rithaniel
Rithaniel
12:57
Or, actually, is it "any subfield?" Maybe there are subfields of the reals that also can't be expressed in these terms
Thorgott
Thorgott
I'm not sure what you mean by that
Rithaniel
Rithaniel
So, starting from a rational polynomial ring in uncountably many variables and constructing fields from localizations and/or quotients at maximal ideals
Would you be able to construct an arbitrary proper subfield of the real numbers in this way? Do there exist proper subfields of the reals that couldn't be constructed this way (gotta go driving. I will return later)
Thorgott
Thorgott
you will only get stuff that's purely transcendental over Q
 
1 hour later…
Peter John
Peter John
14:12
@Thorgott Yes yes thanks
Ted Shifrin
Ted Shifrin
14:41
@Thorgott oh duh, of course.
Avra
Avra
user image
leslie townes
leslie townes
if you say so
Avra
Avra
Hello, I am just looking to figure out how we transitioned from the first sum to the second please?
Thorgott
Thorgott
distributivity
Avra
Avra
we distribute the sum to 1, amd to the second sum. Thanks. I got that.
One second please
For the second sum, we have
user image
This is where I have trouble with
user image
user image
user image
@Thorgott. This is correct please?
This is why we got sum of $n-i$.
Thorgott
Thorgott
15:00
yes
Avra
Avra
Thank you very much
epsilon-emperor
epsilon-emperor
@Thorgott Thanks, your idea for the solution worked!
Avra
Avra
15:12
If we have indicator random variable $I[f(k_1) = f(k_2)]$, and we have that $Probability\{f(k_1) = f(k_2)\}=\frac{1}{m}$, where $k_1, k_2 \in \mathbb{Z}$, then why expectation $E[I[f(k_1) = f(k_2)]] = \frac{1}{m}$ please? I understand logically what that means, but how we got nested of $I$ inside $E$ equals to $1/m$?
This nesting confuses me a little bit
Mohcine
Mohcine
15:49
0
Q: $\lim_{\Delta t \to 0}p_{0}(t)\frac{o(\Delta t)}{\Delta t}=0$

MohcineI know that ; the little oh landau notation $$f(\Delta t)=o(\Delta t) \iff \lim_{\Delta t \to 0}\frac{f(\Delta t)}{\Delta t}=0$$ but I can't figure out why $$\lim_{\Delta t \to 0}p_{0}(t)\frac{o(\Delta t)}{\Delta t}=0\; \rm{and}\;\lim_{\Delta t \to 0}\frac{o(\Delta t)}{\Delta t}=0 $$ Do we have...

real-analysis calculus asymptotics markov-chains poisson-process
Rithaniel
Rithaniel
@Thorgott Are reals purely transcendental over the ring of algebraic integers, perhaps?
Thorgott
Thorgott
16:01
pure transcendence is a field-theoretic concept, not a ring-theoretic one. you could ask whether they are purely transcendental over the field of real algebraic numbers. i dont have an argument, but my bet is on "no".
Rithaniel
Rithaniel
Well drat. Getting a good grip on the subfields of the reals is proving tricky
Perhaps I should think about $\mathbb{Q}[[x]]$. I know this is a field that has the same cardinality as $\mathbb{R}$, but I don't know if they're isomorphic or if one is contained in the other
Wolgwang
Wolgwang
16:32
What is parallel normal?
> If $y=2x+3$ is a tangent to the parabola, $y^2=24x$, find its distance from the parallel normal.
NVM
Charlie
Charlie
In the case of a short exact sequence, what is the $0$ at the start and the end generally?
Is it just a trivial object in the same category as the rest of the objects in the sequence?
Oh wait
Thorgott
Thorgott
17:08
it's a zero object
i.e. one that is both initial and terminal
 
1 hour later…
leslie townes
leslie townes
18:15
the alpha and the omega of the category
Alessandro Codenotti
Alessandro Codenotti
19:06
@Rithaniel you should be able to tell whether they are isomorphic
Rithaniel
Rithaniel
Well, I don't usually work directly with the reals, but my instinct would be that they aren't. Probably the key thing to point to would be the completeness of the space?
(And yeah, I think it's fairly obvious that $\mathbb{Q}[[x]]$ is not complete)
But, does $\mathbb{Q}[[x]]$ contain any complete subfield? I think there are some good candidates
(Wait, no, the one I was thinking of isn't closed)
Alessandro Codenotti
Alessandro Codenotti
19:33
Another way to approach the isomorphic or not problem is thinking about whether $\Bbb Q[[X]]$ is real closed
Rithaniel
Rithaniel
19:56
Hmmm, "real closed?" Is that like "closed under multiplication as an $\mathbb{R}$-module?"
Ted Shifrin
Ted Shifrin
20:11
Hi, old man devilish @Alessandro.
copper.hat
copper.hat
my daughter arrives at sfo in a couple of hours :-)
Ted Shifrin
Ted Shifrin
Oh wow. She escaped the clutches of your generalized relatives?
Does she have to quarantine?
Oops, @Alessandro. It's been so long I forgot. Demonic, not devilish.
Avra
Avra
I am trying to prove the following: Show that if $\left| U \right|>nm$, there is a subset of U of size n consisting of keys that all hash to the same slot. By contradiction, assume that $\left| U \right| \le (n-1)m$, then we have a contradiction. By contradition, we want to proof that if the negative of our assumption is true, then the original assumption is false. Our assumption in the question is $\left| U \right|>nm$,
so I see that by contradiction we show the negative statement is false that $\left| U \right| \le (n-1)m$
What do you think please about my proof?
copper.hat
copper.hat
20:28
@ted i dropped her off to return to london while i was in ireland :-). she had to return for some work, clear her flat and pack a bigger bag:-). Fully vaxed arrivals to the US do not need to quarantine, but she will for 3 days and then take a test just to have some assurance that nothing arose from the flight. (To return to the US you need a negative test in the 3 days before departure.)
Avra
Avra
where $n,m \in \mathbb{Z}$ and $U$ is a set of numbers.
Ted Shifrin
Ted Shifrin
Do you know what proof by contradiction is? I have no idea what your statement is/means, but your logic is totally incorrect.
Avra
Avra
By contradiction, we need to show that if we have a statement $P$, then we want to show by contradiction that if $\lnot P$ is true then our original statement $P$ is false please?
Ted Shifrin
Ted Shifrin
@Avra: When you do a proof by contradiction of $P$ implies $Q$, you do not mess with $P$. You assume $P$ holds and $Q$ fails, and arrive at a contradiction.
No, you're totally incorrect.
Avra
Avra
@TedShifrin. Here we have only $P$ prof?
Ted Shifrin
Ted Shifrin
20:30
You have if and then, do you not?
copper.hat
copper.hat
@TedShifrin A (fully vaxed) close friend (who is even more careful about hygiene that myself, and somewhat at risk) just tested positive :-(. She thinks she got it on a flight from Oregon (to the Bay Area).
Avra
Avra
One second prof please
Ted Shifrin
Ted Shifrin
Yeah, one of my good friends here, fully vaxxed, ended up in the hospital with high fever and got antibody treatment. I think he did in fact do some flying, but I'm not sure it was the cause. ... My best friend in Michigan intends to visit here in a month, and I'm trying to dissuade him. It just scares me for him then to spend 5 days in my apartment.
leslie townes
leslie townes
i have put off my plans to meet with my best friend until probably 2022.
Ted Shifrin
Ted Shifrin
Ugh.
leslie townes
leslie townes
20:37
she gets sick very easily and was hospitalized once last year (not for covid). no reason to chance it.
Ted Shifrin
Ted Shifrin
Yup. I am about to get my flu shot (over a month earlier than usual), and will inquire at the pharmacy about my booster Covid vaccine whilst I'm at it.
copper.hat
copper.hat
Apparently the antibody treatment is the thing to get if you are sick.
Ted Shifrin
Ted Shifrin
Yeah, my friend here got it and felt better after a few days.
Avra
Avra
Prof, I have this please: Show that if |U| > nm, there is a subset of U of size n consisting of keys that
all hash to the same slot. Then by contradiction, |U| > nm and there is a subset of U of size n-1 that hash to the same slot, so we have |U| <= (n-1)m, but this is a contradiction since we have statement $P = |U| > nm$?
leslie townes
leslie townes
my friend got her booster already. i don't think it's available for someone of my age without conditions.
Ted Shifrin
Ted Shifrin
20:39
Meanwhile the idiocy of US citizenry continues. Not the least of which is that we're quite likely to end up with the ignoramus Elder as our governor. Lovely democracy.
copper.hat
copper.hat
well, thankfully the $\inf$ has left for now, it is astonishing that the replacement is so poor.
Ted Shifrin
Ted Shifrin
@Avra I do not understand the actual problem at all, so all I can do is complain about your logic. You have not correctly contracted the conclusion. What is the negation of "there is a subset $U$ of size n with property blah"?
copper.hat
copper.hat
I must admit, I was amazed that the flights were sardine packed. I thought some nod to the pandemic might be observed.
Ted Shifrin
Ted Shifrin
They were months ago. No longer.
copper.hat
copper.hat
I presume we will have a resurgence shortly :-(.
Alessandro Codenotti
Alessandro Codenotti
20:42
@TedShifrin I can be both
Ted Shifrin
Ted Shifrin
With pride, no doubt.
copper.hat
copper.hat
I'm off for a short smoke filled ride in the oppressive Bay Area heat.
Alessandro Codenotti
Alessandro Codenotti
clearly
Ted Shifrin
Ted Shifrin
@copper At first I thought you were referring to riding on BART.
copper.hat
copper.hat
@TedShifrin Ironicially when I took Bart back to the Plaza it was practically empty.
Avra
Avra
20:43
@TedShifrin. Thanks Prof. Lastly please, is the negation of there is a subset of U of size n consisting of keys that all hash to the same slot correct: there subset of U of size LESS THAN n $<=(n-1)$ consisting of keys that all hash to the same slot correct?
Alessandro Codenotti
Alessandro Codenotti
@Rithaniel No it has (many equivalent) definitions. en.wikipedia.org/wiki/Real_closed_field
copper.hat
copper.hat
I did not realise the direct train takes longer than switching at mcarthur
Rithaniel
Rithaniel
Well, this is something worth reading, then
Vrouvrou
Vrouvrou
hello
Ted Shifrin
Ted Shifrin
No. The correct negation is that there is NO subset of size $n$ with the property.
copper.hat
copper.hat
20:44
My life is fulfilled now that I have seen the quaternion inscription on broombridge.
Ted Shifrin
Ted Shifrin
You have no idea if there's a subset with size $n-1$ with the property, or with size $n-2$. Or, indeed, with size $2$.
Ah, Alessandro always fits in set theory whenever possible . :)
Avra
Avra
@TedShifrin. Thank you professor, then the following proof is wrong for the problem of, "Show that if |U| > nm, there is a subset of U of size n consisting of keys that all hash to the same slot":
user image
Ted Shifrin
Ted Shifrin
In everything you've said up until that, @Avra, there was no meaning to the letter $m$. At least they're using the letter $m$ in the argument.
Alessandro Codenotti
Alessandro Codenotti
There are also algebraic characterizations. The real closed fields are precisely the fields for which the algebraic closure is a degree two extension for example
Avra
Avra
@TedShifrin. $m $ is an integer value
Ted Shifrin
Ted Shifrin
20:47
No, it's the number of slots. That appears nowhere in your problem statement.
Alessandro Codenotti
Alessandro Codenotti
Fun fact of the day: by a result of Artin-Schreier (I think) the degree of the algebraic closure over the original field can only be one, two or infinite
Vrouvrou
Vrouvrou
i need help, please, if $(A_n)$ is a Cauchy sequence of bounded and closed sets of a metric space , then i have $\forall \varepsilon>0, \exists n_0\in\mathbb{N}, \forall p,q\in\mathbb{N}, p,q>n_0\Rightarrow h(A_p,A_q)<\varepsilon$ were $h(A,B)=\max(\sup_{x\in A} d(x,B), \sup_{x\in B}d(x,A))$
Avra
Avra
@TedShifrin. Prof, number of slots from the context of the book is an integer, so it could be any integer $m$
Ted Shifrin
Ted Shifrin
@Vrouvrou. You do not need help. You need to understand the words in what you typed.
But you have to say there are $m$ slots, @Avra. As it stands, the problem is meaningless.
I kept telling you it made no sense. That's why.
If you say there are $m$ slots, then the problem makes complete sense. You have to use that!!!
Avra
Avra
@TedShifrin. Sorry. I did not notice. So now my solution is correct please?
Ted Shifrin
Ted Shifrin
20:50
Your solution where you negated the hypothesis? NO.
Are you talking about the typed one you pasted in?
Avra
Avra
@TedShifrin. The one in the image is correct but the one I posted is wrong?
Ted Shifrin
Ted Shifrin
Yes, what you posted was wrong for the reason I told you several times.
Avra
Avra
Thank you very much. I got that.
Vrouvrou
Vrouvrou
i want to prove that $(A_n)$ converge. i obtain this:\begin{align*}
\forall \varepsilon>0,\ \exists n_0\in \mathbb{N},\ \forall p,q\in\mathbb{N},\ p,q\geq n_0&\Rightarrow h(A_p,A_q)<\varepsilon\\
&\Rightarrow \begin{cases} \max\limits_{x\in A_p} d(x,A_q)<\varepsilon\\ \max\limits_{x\in A_q} d(x,A_p)<\varepsilon\end{cases}\\
&\Rightarrow \begin{cases} d(x,A_q)<\varepsilon, &\forall x\in A_p\\ d(x,A_p)<\varepsilon,&\forall x\in A_q\end{cases}\\
&\Rightarrow \begin{cases} \inf\limits_{y\in A_q}d(x,y)<\varepsilon,& \forall x\in A_p\\ \inf\limits_{z\in A_p}d(x,z)<\varepsilon,&\forall x\in A_q\e…
Ted Shifrin
Ted Shifrin
@Avra: This is just a version of the pigeonhole principle. Are you familiar with using that?
Vrouvrou
Vrouvrou
20:52
@TedShifrin what is wrong in what i write ?
Avra
Avra
@TedShifrin. Yes prof. Thanks
Ted Shifrin
Ted Shifrin
@Vrouvrou: If you know the definition of a Cauchy sequence, then you are done.
Vrouvrou
Vrouvrou
i know it
Ted Shifrin
Ted Shifrin
This is the same idea. If every one of the $m$ hashes has at most $n-1$ keys in it, then how many keys do I have? @Avra
Avra
Avra
we have at most $m(n-1)$
Ted Shifrin
Ted Shifrin
20:54
Correct. And that contradicts the hypothesis that $|U|>mn$. That's the end of the proof.
Avra
Avra
@TedShifrin. Thank you
Vrouvrou
Vrouvrou
but i don't know how to get that $(A_n)$ converge @TedShifrin
Ted Shifrin
Ted Shifrin
Do you have a complete metric space?
Vrouvrou
Vrouvrou
yes $(E,d)$ is complete and the question is to prove that the familly of closed and bounded sets with Hausdorff distance is complete
but how to get a Cauchy sequence in $(E,d)$
Ted Shifrin
Ted Shifrin
Getting a Cauchy sequence in $E$ isn't good enough. You need to figure out what set $A$ will be the limit of the $A_m$.
I need to leave for a few hours now.
Vrouvrou
Vrouvrou
21:02
how to know the set A ?
is it $A=\{a\in E, lim a_n=a, (a_n)\subset A_n\}$ ?
schn
schn
@robjohn , I assume the whole shebang is $o\!\left(h^m\right)$, at least if the notation $o\!\left(h^m\right)$ includes all functions of $h$ and possibly other variables. This assumption follows from equation 3 in your answer, $$o\!\left((hu)^m\right)=\left\{f:\lim_{h\to0}\frac{f(u,h)}{(hu)^m}=0\right\}=o\!\left(h^m\right).$$ So $u$ is treated as a constant.
Correction; not all functions of $h$, but simply functions of $h$ (and possibly other variables).
Maybe the above only holds for finite $u$.
schn
schn
22:21
@robjohn , a minor detail, but in this message of yours, did you mean $g(h,u)$ instead of $g(hu)$?
 
1 hour later…
schn
schn
23:25
I assume $g(h,u)$, although I don't see the requirement for a global bound since $\int_{\mathbb{R}}k(u)\,o\!\left((hu)^m\right)\,\mathrm{d}u=\lim_{c\to\infty}\int_{-c}^c k(u)\,o\!\left((hu)^m\right)\,\mathrm{d}u=o\!\left(h^m\right)$, i.e. $u$ is finite prior to taking the limit and thus $o\!\left((hu)^m\right)=o\!\left(h^m\right)$.
Peter John
Peter John
23:47
For prime $p$, let $H<S_P$ with $|H| = p$. If $H<N<S_p$ and $N$ is nilpotent, then $H= N$.
Since $N$ is nilpotent and $H$ a Sylow $p$-subgroup of $N$, $H$ is normal in $N$.
Also, I know $H<Z(N)$. Is there something I can do more?

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