yep, that was what i meant when i said $\cos$ is even.

@copper.hat That one didn’t cost soul.

@TedShifrin Could have been worse, after all, the wages of $\sin$ is death. I'll take Desmos...

@copper.hat sorry, I didn't see that. It had scrolled off the page. It seems that the same question was asked twice.

good evening

How do I show that the VC dimension of axis aligned hypercubes in $\mathbb{R}^d$ is less or equal to $d+1$?

I have no idea

good night

Can the maximum likelihood principle be interpreted as saying that maximizing $\Pr(\theta\ |\ X)$ over $\theta$ always gives the same answer as maximizing $\Pr(X\ |\ \theta)$ over $\theta$?

@AkivaWeinberger finite covers are enough (in paracompact spaces). Now this still gives an infinite set of sentences (one for each $n$, where $n$ is the size of the cover), but there must be a minimal $k$ for which $\Bbb R^3$ has a size $k$ covering witnessing dimension $\geq 3$, so a single sentence for that $k$ is enough

hard to say, depends on the function. i would compute a few points first and plot them.

@copper.hat So you wouldn't right away put the dots in the coordinate system but make a table?

Mmm, table dots are much the same from my perspective

@copper.hat The problem is often that it gets hard without a calculator, take for example x^3 + x^4, the values get big quickly and you need to work with x values with decimals

my coordinate systems usually have 4 units in every direction

if you want to do curve sketching properly you'll probably have to treat the sum as a sum, or also plot values of derivatives and second derivatives

or just hope that the level of detail you're working at with individual points is enough to capture the flavor of it

you can only 'eyeball' this stuff in easy cases where minor differences in the shapes of the graphs won't affect the outcome, which might be true in an exam setting, but there's no general guarantee that would be the case

depending on where you're encountering this type of problem, you might not be expected to have the graph of the sum accurately reflect behavior of the function other than at a finite number of points

is there ANY mathematical difference between k maps between 2 objects drawn on paper and k maps between 2 objects with 1 extra degree of freedom?

although I sort of doubt there's a difference, I do recall that a matrix is an array of vectors, and that a 3-tensor is like a 3d matrix. And there's obviously a valid generalisation here as well as a noticeable mathematical difference...

$f_1,\cdot\cdot\cdot,f_k: X \to Y$ is how you denote what i wrote by the way..

arxiv.org/abs/2209.09968

What can we say about product of a smooth function and a continuous function? Will it be smooth?

By continuous, I mean continuous everywhere.

And assume that the smooth function is not constant function

have you tried some examples?

Ok I got it.

The road to enlightenment is paved with examples.

There exists a smooth function which is 1 on A and 0 outside an open set containing A.

So on A, we can get non- differentiability of the product.

But what about a smooth function which is nowhere locally constant?

Ok got it.

Sorry, I wasted your time with my silly questions. But, I understand something better when I talk to others.

what it means cos is even function?..

@Obliv read this: en.wikipedia.org/wiki/Even_and_odd_functions#Even_functions

Hi, this might sound like a stupid question (and most likely it is). What do we know about representing homologies between singular chains via cobordisms? let me be more specific, given a smooth closed n-dim manifold $M$, and two homologous chains $\alpha$ and $\alpha'$ of degree $n-1$ which can be both represented by two codim 1 sub manifolds $V,V'$.

Can I claim there is an n-dim mfld W with boundary, together with a map W to M that sends the two boundary components to V and V' and represents the homology between them?

So @AkivaWeinberger if $f(k)$ is the smallest size of a cover of $\Bbb R^k$ witnessing that $\dim\Bbb R^k>k-1$ (I think it might be as low as $f(k)=k+1$), then I believe you can distinguish $\Bbb R^k$ and $\Bbb R^{k+1}$ with around $2f(k)$ quantifiers (plus one or two)

@Riccardo It is very much not a stupid question. If $V$ and $V'$ are forced to intersect, then there can be no (embedded) cobordism $W\subset M$, but you're basically asking for an abstract cobordism. I think you can get a mapping from $W$ to $M$ by standard extension/transversality methods.

So it becomes a topological question of whether $V\cup \overline{V'}$ is nullcobordant. The topological obstructions are standard (see Milnor's Characteristic Classes or Hirzebruch's Topological Methods in Algebraic Geometry, for example). Here's a warm-up question (given the h-cobordism theorem): Can you give examples of homologous $V,V'\subset M$ that are not diffeomorphic? I'll ping our resident topology guru, @Balarka. Perhaps he has concrete thoughts.

Since our resident topology guru hasn't been in the room for quite some time he won't get that ping.

Yeah, I noticed after I typed.

Just want to make sure of something... a valid integral for a continuous region of a function equal to zero in that region is zero, right?

And then if I have a function defining arbitrary boxcar regions that spike to 1 for a continuous region and are zero for all other values, then what we should expect to see upon integrating is a function $f(t)$ such that $f(t + 1) - f(t) = f'(t)$, right (it's domain is defined over $\Bbb{Z}^+$ for $f$ and $f'$)?

I'm thinking this would be a good way to create an arbitrary set of mapping functions wherein each derivative of each function in the set sums to 1.

However, I'm not sure how best to compactly represent this, either functionally or otherwise as an explicit set of values.

That is, in terms of computer memory, not mathematical representation.

Hey wait a sec, I know...

Ok, so this is a different problem now, but I can represent it as a single real-valued number, but then I also need a way to get the value of the nth digit of its numerical representation in binary.

is here someone who knows about conditional expectation?

Actually, never mind. I just realized a possibly better solution.

@Wave Knows what about it?

@TedShifrin sorry I wanted to ask if there is someone who feels comfortable to work with conditional expectation.

Better proofread that.

Oh, I see.

Sorry what?

So what’s the expected value of $X$ if $Y=y$?

@TedShifrin I know that E(X|Y) is by definition the unique \sigma(Y)-measurable random variable such that for all \sigma(Y)-measurable and bounded random variables Z we have E(XZ)=E(\xi Z) where then \xi=E(X|Y)

I know that E(X|Y=y)=E(X\cdot 1_{Y=y})/P(Y=y)

So?

@TedShifrin You mean the "so" to my first comment or to the one with $E(X|Y=y)$?

The first makes absolutely no sense to me. What’s the expectation of $X$ given $Y=y$ and given your hypothesis?

Maybe I should say that X,Y are not discrete

Because we have defined E(X|Y) like this if they are not discrete

Understand the discrete case first?

Sorry I'm confused. Even if I need to understand the discrete case, my comment need to be correct because it is the definition?

but nevermind let us consider the discrete case

The English seems off in the definition.

hmm okey

So i know that E(X|Y=y)=E(X\cdot 1_{Y=y})/P(Y=y) in the discrete case right?

then using that X=Y a.s. would give me E(X\cdot 1_{Y=y})/P(Y=y)=E(Y\cdot 1_{X=y})/P(X=y)?

Why put those $X$s in there?

I don't know I think I messed up

I'm sorry but I don't see what this X=Y a.s. gives me

If $X=Y$ and $Y=y$, then …

Then X=y

So what’s the expected value of $X$ given $Y=y$?

what are some non-trivial methods to extend a geometric structure from 2d to 3d?

@TedShifrin it is equal to E(Y|X=x)?

That isn’t helpful, is it?

no it isn't why?

?

Sorry I think I do not understand what you mean by "That isn’t helpful, is it?"

Is my statement wrong or did I say something strange or what?

@geocalc33 What kind of geometric structure?

It’s no different from the original question.

So do you want the formula E(X\cdot 1_{Y=y})/P(Y=y)?

It’s like saying $3+2$ is the answer to $2+3$.

No. I want the final simple answer.

5

Thank you, anak. Finally.

what?!

@TedShifrin I do not get it, I mean all I can do is rewrite the conditional expectation I can't compute anything

Reread our conversation.

You want me to tell you what E(X|Y=y) is but I don't know more than the formula...

What is $E(X|X=x)$?

@anak wish to generalise an arbitrary 2d vector field on the open disk with two diametrically opposite accumulation points, whilst preserving the two accumulation points going up a dimension. Maybe there's not enough information to do this?

is it E(X)?

Is it a real number or a value of $X$?

I can see how it might work if you had some reflectional symmetry for the 2d vector field on the disk, but devoid of that i don't see a solution

Go back to the actual definition of expected value.

it is a real number

If I throw a die, what is the expected value of the outcome?

@geocalc33 so in particular, you want to take a pair of $(B,v)$ (where $B$ is the open disk, and $v$ is some vector field) and find a pair $(M,w)$ where there is an embedding $e\colon B\hookrightarrow M$, and $w$ is a vector field such that $w|_{e[B]} = v$? Or do you want something stronger for the vector field?

it is 1/6(1+2+3+4+5+6)

OK. Given that the outcome is 5, what’s the expected value?

is it (1/6(1+2+3+4+5+6))/(1/6)

No.

shouldn't it be E(X1_{Y=5})/P(Y=5)?

@anak just to make sure, i want to obtain a 3d vector field in $D^3$ with exactly 2 diametrically opposite accumulation points

You don’t even understand what the notation means,

If I roll a 5, what’s the probability that I rolled a 2?

I don't understand the notation

$w|_{e[B]} = v$ I don't follow this notation could you provide aid?

What is the probability that the outcome is 2 and the outcome is 5?

@geocalc33 So you want $M = D^3$ in particular. That notation there means w restricted to the direct image of B under e is just v

Do I then need the formula for P(A|B)?

I guess the notation doesn't really make sense because v not on e[B]

It's on b, right

You need to go back and read the basics and work out basic examples. Forget measure spaces.

you could use your formula, but you'd need to know what the functions X and 1_{Y=5} are

The problem is that we haven't seen anything else we started with this formulas

We’re assuming $X=Y$, leslie.

I have no basics

ted: i'm talking about your example, where we know what X and 1_{Y=5} are

But $Y$ should be $X$.

So I guess we want the pullback of w along e, $v = e^*w$.

Much nicer notation

@TedShifrin Hi! yes an abstract cobordism is what I had in mind. For what concerns your warm question, if you think about the 2-torus as M, then the separating circle in the middle is cobordant via the pair of pants to 2 non separating ones, but a circle cannot be diffeomorphic to two circles.

not meaning to butt in, just noting that your example involving dice could be put in that notation

Of course.

and pointing out that the fact that it isn't being put in that notation, or the relation between the two seems to be as yet unrealized

this is just your point that one might need to work with examples

But needing $Y=X$.

of course, but nothing wrong with X = outcome of die and Y = outcome of die

Same throw!

so I'm compleatly lost now where are we and what are we doing because I have no idea what is going on since we have never talked about examples and basic stuff without formulas

I'm sorry

yes, same throw

wave all of this stuff does get subtle when the events Y = y may all have measure zero because the discrete analogy is not at its strongest there

Get a standard book like Ross or Pitman with lots of examples and start at the beginning.

but it does help to understand the discrete case completely first

in order to understand what the general formalism is doing, and avoiding, and cleverly handling

Yes, but worrying about measure theory seems hopeless here.

@anak yes i think I see partially what you're constructing here

I can't get one of the books in the next days and I would like to understand this. can maybe someone explain this a bit?

it's tricky enough in the discrete case

so give me some intuition

ah whait so if you want to throw a 5 and before a 2 then the probability is 1/36?

Same throw.

so in the same throw a 5 and a 2?

Given that I threw a 5, what’s probability I threw a 2?

@geocalc33 I think it's worth mentioning that I think if possible, this can introduce zeros to the vector field on D^3. Is this okay with you?

0?

There is only one throw.

Right.

Okey now what should this tell me?

Now compute the expected value given that $X=5$.

I don't get this one because I know how to compute E(X) but not if I have X=5 and want something with 2

Use your formula correctly.

@anak Was hoping for a way to avoid introducing zeros, but I think your method is definitely something I can work with going forward.

So you mean E(X|X=5)=6E(X\cdot1_{X=5})?

@geocalc33 I have a feeling that in some case it might be unavoidable to introduce zeros.

Figure out if that is right.

But if you are okay with introducing one zero, there is a very easy solution. :DD

Just add a predicted $z$-coordinate and you can be sure there are no extra zeroes.

How? I mean this is the formula isn't it

@TedShifrin What do you mean by this?

Go back to the definition of expected value and calculate. You need to work this out for yourself.

Take a function like $1-y^2$ for the $z$-coordinate?

But how can I calculate something using the formula if you tell me to figure out if I applied the formula correct?

Wave, I”m done.

@TedShifrin okey sorry, I think I do not understand what you want, sadly

what does support in a graph mean?

I'm trying to understand imgur.com/a/ecKTcDD

@geocalc33 it's also worth mentioning you probably want it to be a vector field on $D^2$, to start, not the open disk...

At least, I feel like that simplifies it.

hmmm

@TedShifrin sorry to disturb again but I thought about it. So the probability that I threw a 5 and threw a 2 is 0 as I said. Furthermore the expectation given X=5 is E(X|X=5)=E(X 1_{X=5})/P(X=5)= (5/6)/(1/6)=5 right? But now I don‘t see what this shows me in order to understand my original question.