12:22 AM
@Obliv looks like the definition of an inner product

6 hours later…
6:49 AM
I'm studying order statistics. Let there be a sample of just $2$. Then apparently if $X_1$ and $X_2$ are $\text{Ge}(p)$, then $X_{(1)}$ and $X_{(2)}-X_{(1)}$ are independent. How can this be? Seems like $X_{(2)}-X_{(1)}$ depends on $X_{(1)}$...
Here, as always, $X_{(2)}=\max\{X_1,X_2\}$ and $X_{(1)}=\min\{X_{1},X_{2}\}$.
I should add too that $X_1,X_2$ are independent.
7:29 AM
0

Consider the equation $$y^2 u_{xx} - x^2 u_{yy} = 0.$$ The method used to solve this problem is given below(, and for the real solution given in the book please scroll down): The solution given in the book is as follows: Here $$A = y^2, \quad B = 0, \quad C = -x^2.$$ Thus, $$B^2 - 4AC = ... I need some help with this. It's very frustrating when authors state something in such a vague way and then does not bother to come back at it. 8:07 AM Author and book please. I've also got a new question on the main site: Add your attempt at a solution please. 1 hour later… 9:21 AM @SineoftheTime Hi ! @BinkyMcSquigglebottom hi Are you free ? yes you can ask and then ping me, when I'm free I see if I'm able to help Okay :⁠-⁠) What does it mean for a function to be equilimited? I'm not familiar with this definition context? 9:30 AM I edited the question you mean a family of functions? like a sequence? I'll go into more detail @SineoftheTime yes Like f_n(x)= 1/nx in I=(0,1) Why doesn't it converge uniformly? And it says because it is an equilimited function (f_n)_n is equilimited if |f_n(x)|\le M for all n\in \mathbb N and x\in [a,b] f_n:[a,b] \to \mathbb R @BinkyMcSquigglebottom f_n=\frac 1{nx} or f_n=\frac 1n x The First 9:47 AM (f_n) is not equilimited (?) .... I read it wrong, it said that they are not equilimited @SineoftheTime yes are you having trouble for the uniform convergence? what does (f_n) converge to (pointwise) ? at f(x)=0 10:02 AM correct what is \sup_{x\in ]0,1[} | \frac 1{nx}| ? Do you find that this point is made by trivially making substitutions? @SineoftheTime 1/n what substitutions? @BinkyMcSquigglebottom are you sure? No In theory I should put 0 but then it comes out 1/0 So I think it doesn't exist it's +\infty 10:06 AM because the functions are not bounded on ]0,1[ for n=1, you have f_1=\frac 1x @BinkyMcSquigglebottom This is true , since \infty is not a number. Nevertheless , usually , it is formulated as "sup=\infty" meaning that there is no largest value , so no (finite) supremum. but in this case doesn't the sup occur for x close to 0? @BinkyMcSquigglebottom No, because we can beat every finite value by choosing a suitable x close enough to 0. since f_n are decreasing, the sup is near x=0 but the functions blow up This would be like making a limit lim x->0+ 1/n•0+ And so sup=+∞ 10:17 AM so (f_n) can't converge uniformly Okay Can I ask one last question? @SineoftheTime No, because the sup should tend towards 0 @BinkyMcSquigglebottom yes But when I go to see the pointwise/uniform convergence, do I also have to consider the case in which x depends on n? what do you mean? do you mean when \sup_x is a function of n ? No Mmm like I have f_n(x)=x/n, if I consider x in real numbers, it always converges to 0, but if I consider like x=n^2 I find that it doesn't converge, should I do this last thing? 10:25 AM when you consider the pointwise convergence, you have to fix x_0 \in \Bbb R and then take the limit. If you choose x_0=n^2, this is not fixed Right But what about that uniform? I don't think there either because if I'm not mistaken it's fixed there too when you study the uniform convergence you first find the sup and then take the limit Can you go more in detail pls what's not clear? I didn't understand how to find the sup Maybe I'm using the wrong method 10:35 AM there's isn't a general rule you have to use different methods if the sup is the max, you can use derivatives for example No, because I have seen that sometimes the derivative is used to find the maximum even when the maximum is not guaranteed Like in I=(0,+∞) And from this I wonder why we didn't do it in the exercise above because you don't need it and it does no work since f_n are decreasing, the sup is as x\to 0^+ Correct So I never have to consider the case where x depends on n? 10:55 AM no :⁠-⁠) Thank you very much for helping 🎉🎉🎉 Sine, Binky see this 🤣 @SoumikMukherjee that's what happens when opponent tries to humiliate you there's the summer marathon today btw @SineoftheTime Yeah @SineoftheTime are you participating? 11:05 AM I played a couple of games last year I was in top 500 Nice but I played like 200-300 games :| That's a lot yes it's competitive top 500 because then you have a trophy in your profile Ooh I will play then 11:07 AM but you have to be top 500 :) last year I scored 545 points 11:45 AM @SoumikMukherjee ahaha 12:19 PM @SineoftheTime I started, now I am in top 1800 I am just flagging:) @SoumikMukherjee that's the spirit Nf3 Ng5 Nxf7 3 hours later… 3:32 PM glurbingston 4:03 PM by fixed, does that mean f doesn't change at all for any \phi? this theorem implies the existence of this f for any dual space and its vector space? @Obliv Yes? But I think that you are kind of thinking about it the wrong way. F lives in the dual space. What the theorem says is that there is a unique element of the original space, f, such that the action of F on any vector \varphi is the same as the inner product pairing of f with that \varphi. the inner product pairing is the isomorphism right? i.e., F(\phi+\psi) = (f,\phi +\psi) = F(\phi)+F(\psi)=(f,\phi)+(f,\psi) or something? No...? \langle f, \varphi\rangle is always a real (or complex) number, n'est-ce pas? It does not send bounded linear functionals to vectors (or vice versa). The theorem says that any bounded linear functional is determined by an element of V linear functionals are inner products 4:12 PM Obviously, y\mapsto \langle x, y\rangle is always a bounded linear functional for x\in V, the Riesz theorem gives the converse @SineoftheTime Not by definition. This requires a proof. @SineoftheTime not how I'd put it. That can cause misunderstandings and is somewhat confusing @Jakobian that every f \in F can be written in the form \langle x,y\rangle for some x \in V? what is F As Jakobian says (minus the "obviously"), the map f \mapsto \langle f, \varphi\rangle is a bounded linear functional. But the theorem says that these are the only such bounded linear functionals. 4:14 PM sry i wrote that wrong. should be F \in V' confusing because F can mean functional but also field @Obliv If your book is using F for both a linear functional and a field, find a better book. @Obliv if anything, I'd write it as, for every F\in V' there is x\in V such that F(y) =\langle x, y\rangle for all y\in V @XanderHenderson it was my mistake because I'm looking at different books at the same time. this one says V' is the dual space and F are functionals. Axler says \textbf{F} is the field @Obliv The standard notation would be something like \langle x, \cdot\rangle. The point is that feeding some arbitrary vector to F is the same as feeding that arbitrary vector to \langle x, \cdot \rangle (i.e. taking the inner product pairing of that arbitrary vector and x). @Jakobian yeah that's best. 4:16 PM or another, for every F\in V' there is x\in V such that the functions F and y\mapsto \langle x, y\rangle coincide @Obliv Yeah, I've previously advised folk not to try to learn a new topic out of multiple books at the same time. It causes confusion, not least of which is caused by differences in notation. I think Obliv might be already dealing with distinct notations if they're trying to apply it to something like physics :P @Jakobian Sure, but why make life harder than you have to? Also, bra-kets can burn in hell. ;) I actually agree with that I meant that you can express a linear functional on a Hilbert space as an inner product 4:24 PM Weirdly enough, I don't actually believe that bra-kets should burn in hell---they are clearly useful for people doing physics. And I kind of grok how physicists like to view all linear functionals as "nice" linear functionals which come from "nice" inner product pairings, but it always feels so very, very hand-wavy to me. I am not big brained enough to really get bra-kets. @SineoftheTime Sure, but, again, that is a theorem. I was under the assumption most of quantum theory was inspired by the understanding of linear algebra at the time of the development of the schrodinger equation since it was a linear eq. so besides being handwavy with proofs, most of the theory was built from math and deductive reasoning @SineoftheTime that's too imprecise to convey important information and I think it's not useful for me, neither for someone who learns the subject that's what I object to a linear functional is not an inner product, its an inner product with some fixed vector I am proposing there is two kinds of hand-waving in physics. The undergraduate hand-waving, and the graduate hand-waving. The latter, still too imprecise, probably makes sense and is educated. The former, not so much is this double arrow the same thing as iff? or is there some other meaning it denotes a bijection I'm pretty sure also this notation wasn't confusing when they were separating vectors from linear functionals but now for some reason they're writing linear functionals in both places. 4:35 PM they mean the map that sends x\in V to a linear functional y\mapsto \langle x, y\rangle I thought they'd at least distinguish them by upper and lower case characters. :\ or maybe a super/sub-script character they're writing \langle x| to mean the functional y\mapsto \langle x, y\rangle and |x\rangle simply to mean the vector x in similar way, if F is a linear functional, then \langle F| will just be F, and |F\rangle will be the element of V as per Riesz theorem @Jakobian Got it @Jakobian I think this is basically what u just said I don't see how this is the same they're just saying that you can treat \langle F|\phi\rangle  as \langle F, \phi\rangle but there are situations where it matters to not do that oh true I guess it fails to explicitly remind the reader that it's an inner product on a fixed vector on the space of ket vectors. 4:45 PM I wouldn't say it fails since even though it didn't mention details of Riesz theorem, it still says what it wants to say > Exercise Suppose n points are chosen uniformly and independently of each other on the unit disc. Compute the expected value of the area of the annulus obtained by drawing circles through the extremes. I just want to check if I understood the question correctly. We generate n points in the unit circle, and then we look at X_{(1)}=\min\{|X_1|,\ldots,|X_n|\} and X_{(n)}=\max\{|X_1|,\ldots,|X_n|\}. Then we draw circles, centered at the origin, and going through X_{(1)} and X_{(n)} respectively. Is that right? Wait a minute... @psie the question does seem to talk about X_{(1)} and X_{(n)} you just defined, but the drawing part, of course this is just about the area "...going through X_{(1)} and X_{(n)}..." doesn't make sense. the circles with radius X_{(1)} and X_{(n)} respectively will go through the points of least and most absolute value ah ok, thanks! Now this is clearer 4:49 PM maybe better notation would be to call it r and R because those are not order statistics of X_1, ..., X_n yeah is a linear operator and a vector space automorphism the same thing? (or module automorphism if vector space isn't an abstract algebra term) @Jakobian maybe I shouldn't go there (for this problem), but for X_1,\ldots,X_n being two dimensional random vectors, is there such a thing as order statistics? As there is no commonly agreed order in \mathbb R^2 I guess talking about \min and \max wouldn't make sense. Feel free to ignore this :) also why are they called operators and not functions? Does 'operator' suggest we're talking about linear vector spaces only? @Obliv yes a linear operator is sometimes referred to as a linear function T:V\to V where the domain and codomain are the same other times its just any linear function T:W\to V between two vector spaces a linear automorphism is a bijection T:V\to V where domain and codomain is the same this forces T^{-1} to also be a linear automorphism the difference between the first definition of linear operator and linear automorphism is that the latter is bijective @Jakobian my "yes" is supposed to be a "no" @Obliv operator sometimes means we mean maps V\to V they don't have to be linear in general, although the word is mostly for linear stuff In mathematics, the Hardy–Littlewood maximal operator M is a significant non-linear operator used in real analysis and harmonic analysis. == Definition == The operator takes a locally integrable function f : Rd → C and returns another function Mf. For any point x ∈ Rd, the function Mf returns the maximum of a set of reals, namely the set of average values of f for all the balls B(x, r) of any radius r at x. Formally, M f ( x ) = sup r > 0... for example the Hardy-Littlewood maximal operator isn't linear @psie given that the definition of order statistics depends on the order, I would say no I'm not an expert on different generalizations of what order statistics means, but I doubt they use them for anything more general than the case of \mathbb{R}-valued random variables (possible, that they generalize them in different ways than just depending on the order) 5:09 PM Ok. If r_1=\min\{|X_1|,\ldots,|X_n|\} and r_n=\max\{|X_1|,\ldots,|X_n|\}, I should compute E (\pi(r_n^2-r_1^2)), right? still having some doubts on what I actually need to do @psie yep ok, great and I think, if one generates a point X uniformly in the unit disc, |X| will not be uniformly distributed in [0,1]. Intuitively, this is clear, but it is tempting to think that way Yes, and you can see this pretty clearly, for a\leq |X|\leq b happens with probability equal to the area of the annulus a\leq x\leq b, which varies in a non-uniform way as a, b vary In fact its clear that P(|X|\leq t) = t^2 for t\in [0, 1], and with obvious values otherwise Indeed 5:33 PM how does operating with \langle u_i| transform the left side into the right M is a square matrix here? In between a basis vector and the linear functional associated with that basis vector? by the way as a mathematical aside, do all linear vector spaces have a set of orthonormal basis vectors? Or is there just the guarantee of a set of basis vectors to span the space I'd imagine only for discrete vector spaces, maybe only by relaxing the definition to 'discrete' modules I started reading The infinite dimensional topology of function spaces by J. van Mill. 6:05 PM @Obliv nvm, it's just an application of riesz theorem @SoumikMukherjee cool! well I understand how M|u_j\rangle a_j \vdash |u_k\rangle b_k but not how we get the form of a matrix equation in 1.11 how do we represent the inner product \langle u_i|M|u_k\rangle as a matrix? 6:45 PM @Jakobian You suggested this book long ago:), the exposition is very clear, thanks 7:06 PM what is the backward shift operator on F^{\infty}? Is that like mapping each index of the tuple to the one before it? Or I guess it would be the other way around since idk how i'd map the first index. i.e., (x_1,x_2,...) \mapsto (x_2,x_3,...) obliv: yes, something like that is often called the backward shift the 'forward shift' or just the 'shift' goes the other way and you add a 0 in the first entry (the only fixed scalar that you could add, to make a linear map) if S is the forward shift and T is the backward shift defined via your formula then TS is the identity but ST is not, odd little difference from finite dimensional linear algebra 7:22 PM oh okay, thank you. I'm guessing 0 is just the standard, but you could theoretically put any element of the field in there @leslietownes because 0 replaces the first element after ST ? kind of just deletes x_1 well, assuming you wanted a linear map, you couldn't put any other fixed element of F there. the zero vector has to go to the zero vector yes Oh true 8:17 PM @leslietownes are matrix representations of a linear map unique? if you specify enough ambient context (e.g. fix bases of the space that the map acts on), which people often do, and which some people do so frequently they don't realize they are doing it, they certainly can be so in the context of the image like you pasted above, the answer would be, that M is uniquely represented with respect to that chosen orthonormal basis by that matrix but choose different orthonormal bases and you will generally get different matrix entries M_ij what if the two spaces have differing size of bases, can that happen? @leslietownes but if you prove that your matrix commutes with every invertible matrix then it's also unique sure, there are interesting special cases like scalar multiples of the identity where the matrix does not change obliv: can what happen? @leslietownes yep like a \mathbb{I}, a = const @Obliv if you have a map from \mathbb{R}^n \to \mathbb{R}^n, and you determined a basis of the space how can that happen? If I understand what you're trying to say that would mean that the dimension of the space has changed anyways, I came here to ask a question about a sentence in this paper I'm reading. Please do not make fun of me :P I'm already ashamed to even ask this in the first place arxiv.org/pdf/2310.19665 It is said that: The prescribed recipe works in all cases except when the Z-axis is in the direction of the south pole, so we exclude this case for the time being 8:29 PM I don't see why this could be true. It's clear to me that, at the south pole, I can still make sure that the x-axis is tangent to the 0° meridian, unless the poles are somehow 'ill-defined points' on the sphere, thus creating problems @Obliv I had this anwer saved btw math.stackexchange.com/a/4774862/1096913 Nvm, I figured it out. I was thinking M(T) had to represent T inverse as well so I was confused if M(T) wasnt a square matrix, then it'd be unable to account for different size bases @Claudio thanks you're welcome does every compact metric space embeds isometrically into hilbert cube? 8:47 PM @monoidaltransform what do you mean by isometrically there is no "the Hilbert cube" as a metric space Hilbert cube is simply [0, 1]^{\aleph_0} and as a metric space it can be given a lot of different metrics it can also be treated as a certain subset of \ell^2 I meant, does there exists a metric on the hilbert cube compatible with its topology so that the emebedding is an isometric one? yes, there exist a lot of metrics on the Hilbert cube so you mean to extend that metric So that the embedding is isometric..? yup I need to think I believe the answer is yes Could you elaborate on why you think that is? 8:54 PM because you can extend continuous functions from a compact subspace, and there should be a theorem like that for when you can extend continuous functions, you can also extend pseudometrics and by obvious adjustment this becomes a metric on the Hilbert cube according to this article, if we got a metric subspace (X, d)\subseteq [0, 1]^{\aleph_0} of the Hilbert cube, then one can extend d to a continuous pseudometric \tilde{d} on the Hilbert cube okay my reasoning with the extension can be made a metric was somewhat flawed actually that there exists such pseudometric is something I even know how to prove actually no, I don't think I can prove something like that 9:13 PM ofcourse, there exists an infinite dimensional metric space Z such that X embeds isometrically into. But im interested in HC @monoidaltransform apparently, Hausdorff proved that you can extend a continuous metric from a closed subspace, to a continuous metric on the whole space. This doesn't answer your question however, since the metrics have to be admissible as well Admissible metric? Are these natural @Jakobian oh no it does, since for compact spaces a continuous metric is admissible @monoidaltransform it means that the topology it generates coincides with the topology of the topological space i.e. a metric d:X\times X\to \mathbb{R} on a topological space X is admissible if the open balls with respect to d, B_d(x, r), form a basis of X saying that such metric is continuous only gives you that B_d(x, r) must be open but since we are dealing with a compact space, thankfully we can stop at this and say it must be admissible So since (X,d) embeds into H topologically, as a closed subset, there exists a metric d' on H such that (X,d) isometrically embeds into (H,d') because d is admissible on X, since X is compact ? that d is admissible on X is without a question, its that d' is admissible on H that is important Hausdorff's theorem above only gives you continuity of d', while compactness gives that its admissible 9:19 PM So there exists an admissible metric d' on H such that (X,d) embeds iso into (H,d') and d' is a metric because H is compact right? yes but not d' is a metric because H is compact d' is admissible because H is compact. The fact that d' is a metric will always hold, even if H isn't compact I meant d' is a metric compatible with the topology of H then yes okay perfect thank you, if you want to write this as an answer in main, ill accept it without question (Ive asked it there) I am working the following problem; > Exercise: Independent repetitions of an experiment are performed. A is an event that occurs with probability p, 0<p<1. Let T_k be the number of the performance at which A occurs the kth time, k=1,2,\ldots. Compute (a) E(T_3\mid T_1=5) (b) E(T_1\mid T_3=5). It is clear T_1\in \text{Ge}(p), but how do I determine the distributions of T_2 and T_3? 9:27 PM @Jakobian also, do you know if the hilbert cube an ANR? Yes it is turns out it is true, but what about C(X,\mathbb{R}) @monoidaltransform I think a product of ANR's should be an ANR this holds at least partially but countable products of AR's are AR's which [0, 1] is Ah I see which is a stronger property, as AR's are contractible ANR's 9:37 PM what about C(X,\mathbb{R})? continuous functions X->R. Atleast when X is compact @Jakobian Also, could I ask what reference is this from? I think its an AR because its a Banach space @monoidaltransform Infinite-dimensional topology by van Mill Thank you @monoidaltransform every convex subspace of a locally convex topological vector space is an absolute retract Ah interesting class of ANR's is huge then here the subspace is C(X, \mathbb{R}) itself @monoidaltransform van Mill does everything for separable metrizable spaces by the way. So it might not exactly hold in this form in general 9:46 PM Two close enough maps into an ANR are homotopic, I was wondering if one could somehow improve this to homotopy equivalence, but it was a random thought maybe its highly false homotopy equivalence of what? of the domain space of the maps and the ANR I(s)=\zeta(s)+sf(s) If I want to analytical continue a function defined by an integral I(s) how would I do so? Assume \zeta(s) is the Riemann zetas function convergent for real s>1 and f(s) is some real analytic function convergent for real s>1 I(s) is clearly real analytic (as the sum of real analytic functions is real analytic). I thought I might analytically continue \zeta(s) and then f(s) seperately but I am not certain that would work 10:28 PM 3 By Tonelli's theorem we can move the summation outside of the integral$$ L=\lim_{k\to \infty} \frac{\sum_{n=1}^k \int_0^{e^{-\sqrt{\log n}}}e^{\frac{\log n}{\log r}}~dr}{\sum_{n=1}^k\int_0^1 e^{\frac{\log n}{\log r}}~dr} $$The numerator can be realized as$$ \sum_{n=1}^k \bigg(\frac{1}{2}e^{-...

a crisp clean proof