Mathematics - 2021-02-26 (page 1 of 2)

user2103480

00:17

@MikeMiller >calling scooter techno

Semiclassical

This is a nice little question:

1

Q: Wrapping a Rubber Band around a Cube

Imagine a cube wrapped with string so that each of the six sides has two linked "L's" on it, like this image (or some reflection or rotation thereof). You can get a lot of different knots and links depending on the particular choices made. Are any of them the unknot or the unlink? In other words...

Ted Shifrin

01:25

Looks like a good question for DogAteMy!! @Akiva

Karim Mansour

Hi @TedShifrin

Ted Shifrin

Hi, Karim.

Karim Mansour

02:05

@Astyx are you learning algebraic geometry ?

robjohn

02:19

@TedShifrin: I reposted the ruled lozenge in a CW post.

Ted Shifrin

@robjohn Yes, giving it all away.

Karim Mansour

so cool

Ted Shifrin

@robjohn: I no longer have patience with assholes like this.

robjohn

@TedShifrin Eh, it's buried at the bottom of 3 pages of prior posts

Karim Mansour

I have a feeling that some constructions in mathematics you don't actually understand them unless you see what they are built for.

@TedShifrin I would just ignore him.

Ted Shifrin

02:33

I am now, after voting to close.

robjohn

@TedShifrin I had one almost like that

Ted Shifrin

I need a proof ... even if I have no idea what I'm talking about ... and certainly have consulted nary a book.

robjohn

Yep. Spoon feed me, damn it!

Ted Shifrin

That might becan exam, too, now that I think about it. ... Before revision, it was ruder and much less explicit.

robjohn

When they want it fast, that is a dead give-away.

Ted Shifrin

02:45

Oh, I didn’t recall temporal exigency.

Jakobian

03:31

Is that diagram on wiki on the bands article a lattice of all subvarieties of regular bands, or just some?

Band (algebra)

In mathematics, a band (also called idempotent semigroup) is a semigroup in which every element is idempotent (in other words equal to its own square). Bands were first studied and named by A. H. Clifford (1954); the lattice of varieties of bands was described independently in the early 1970s by Biryukov, Fennemore and Gerhard. Semilattices, left-zero bands, right-zero bands, rectangular bands, normal bands, left-regular bands, right-regular bands and regular bands, specific subclasses of bands that lie near the bottom of this lattice, are of particular interest and are briefly described below...

Jakobian

03:55

It is

Nico A

04:13

Hey quick sanity-check group theory q: if we have f: X->X is injective but not surjective, can we show there exists a map g : X->X such that g(f(x)) is the identity?

Thorgott

what happens if f:Z->Z is the multiplication by 2 map

Nico A

i think the domains are just confusing me because neccesarily we have elements in the codomain of f that aren't mapped to, so how could g o f be the identity?

@Thorgott so the inverse would be g(x)=x/2 but thats not a map from Z->Z, right?

Thorgott

that's the idea, yeah

Nico A

okay so we can't, that's what i thought

for some reason our homework is to prove that we can so i was a bit confused lol

oh wait X being a set doesn’t change anything right?

rather than a group

Ted Shifrin

04:36

It changes everything.

Nico A

Yup I just got it thank you

We can literally take f^-1 for each element that’s mapped to right?

And even though the domain will be a strict subset of X we still consider that a function X ->X?

Ted Shifrin

05:46

@NicoA No, you have to define the function everywhere.

dc3rd

Hi @TedShifrin. I was having a little difficulty figuring out how to solve this. It is in a chapter about Arc measures. All that was done in the chapter was defining major/minor arcs, congruent arcs, arc addition, and congruent chords having congruent minor arcs.

So one idea that came to mind was perhaps using that part of the rope covers each semi circle, but I get jammed up from there.....

But since the diameter was given maybe I could use the circumference in some way, but since this is in a chapter about arc lengths then this wouldn't really apply

by the way...........I NEED an answer to this............🤣🤣

Ted Shifrin

06:20

You need to stop guessing and figure out what you actually know.

@dc3rd What's the configuration of the three circles?

The other major hint is: If a line is tangent to a circle at a point, what angle does the radius to that point make with the line?

dc3rd

What I know is that each circle has a diameter of 14in, the circles are tangent to one another as well

Ted Shifrin

Yes, so radius is 7".

What shape do the circles make?

dc3rd

A line tangent to a circle means the point of intersection will be perpendicular to the radius.

Ted Shifrin

OK, good. You'll need that.

Now what shape do you see when you look at the picture?

dc3rd

I'm trying to "see" the right triangle that the hints are getting at. But it hasn't sunk in yet

let me draw it out

Ted Shifrin

06:31

Not right triangle.

The general shape of the picture should jump out at you. If you make me give another hint, I'll say draw in the centers of the circles. Now what do you see?

Yes, you always have to draw pictures. You can't do math just looking at it.

dc3rd

So drwing in the center points if I were to attach the radii of each circle I could see an equilateral triangle

Ted Shifrin

So that should jump out at you from the beginning.

OK. So how do the line segments tangent to the circle relate to this triangle you've drawn?

No more hints.

dc3rd

Yes I did see the rounded edges of the rope forming an equilateral triangle, but I didn't want to jump to any conclusions

Ted Shifrin

No, the equilateral triangle is inside.

The outside shape is three line segments and three circular arcs.

The question is: How many degrees in those circular arcs? I.e., how much of the circle's circumference does each one give you?

dc3rd

well it's not perfectly drawn by me, but those line segments that are perpendicular to my radii formed some rectangles. So it appears that the "gaps" between the circles could be calculated by getting the correct side for the rectangles. And as you said I also have to find the sum of the arcs

Ted Shifrin

06:40

OK, so you know enough information to finish the problem now.

How long are those line segment pieces tangent to the circles?

dc3rd

I haven't calculated them yet, but by inspecting them it seems like they're going to be 14in each based on the radii

Ted Shifrin

Correct.

Now figure out the angle of the circular arc that's in each circle.

dc3rd

Doing that as we speak....just trying to recall what theorems I read to establish the calculation

Ted Shifrin

You don't need any theorems.

Use the information we've discussed.

(Look at the center of the top circle. Look at angles, of course.)

dc3rd

I shouldn't have used the word theorem, I am using the ideas discussed actually.

So...going by my drawing....thos radii perpendicular to the lines which I used to draw the rectangles would be right angles......so it appears that each of those arc lengths are going to be 90 degrees ($\frac{\pi}{2}$)

Ted Shifrin

06:54

No.

dc3rd

good...it didn't feel right

Ted Shifrin

I want the arc at the very top of the picture.

robjohn

dc3rd

Fancy picture that captures what my pen is struggling to do nicely

What program do you use?

robjohn

Mathematica

dc3rd

06:57

Outside my budget..........😝

Ted Shifrin

He's showing off.

robjohn

If you're a student, there are decent prices

Ted Shifrin

I would have done it in Illustrator in the olden days. But by hand is good.

dc3rd

I agree, but I don't think I would use it enough within a year to warrant it, at least currently.

robjohn

just thought that image gives the length pretty clearly.

Ted Shifrin

06:59

Doesn't replace thinking!

dc3rd

it does and takes away the impreciseness of my drawing....but back to this arc length

robjohn

what is the sum of the three marked angles?

dc3rd

ooooo..........what if I were to draw a line straight up and "bisect" the intended angle? So now I would have a right triangle and could use trig becuse I have radius lengths

right angle with the line tangent to the circle

no it woun't work....looked like it though...

Ted Shifrin

I told you to look at the angles at the center of the top circle. Tell me what the 3 angles we know are.

dc3rd

Oooooo....well, from the nice drawing it seems the sum of those angles is going to be 360

Ted Shifrin

07:06

Yes. But that includes the one we are looking for.

robjohn

Does this feel better?

Ted Shifrin

Robjohn, that truly is cheating.

He needs to do basics.

robjohn

Hey, you wanted something that looked hand drawn

dc3rd

So going back, the angles I know are the right angles formed from each radii....

Ted Shifrin

smack

And what else is in your picture?

dc3rd

07:09

the rectangles.

Ted Shifrin

You already did those angles.

robjohn

It's the same picture as before, just ran it through xkcdConvert

dc3rd

Oh jeez....I don't want to keep you up even if you're on the west coast.....

Ted Shifrin

You should be done in seconds.

dc3rd

well it seems I have to go back to the idea of the sum of the angles being the 360 and specifically for the arc of the top circle the angle would be 120.........

Ted Shifrin

07:23

Right. Now three of those ...

dc3rd

isn't three of those 360?........

so total length would be $3 \times 14 + 3 \times 120^{\circ}$

robjohn

How long is $120^{\large\circ}$?

dc3rd

Good question..............

robjohn

or you could ask how long is $360^{\large\circ}$...

dc3rd

this is where the diameter idea comes into play..............

I see

so with that being the case.....we would have total length being

$3 \times 14 + 14\pi = 42 + 14\pi$.....out of curiosity there isn't really a nice way to express that in the degree form is there?

Well that was fun. Piecing together all the pieces and using the previous concepts I learned....except for struggling more than I should on the final part, it was enjoyable. Thanks again for the help gentlemen. 😎

Felix Crazzolara

07:46

Hi everyone, I need to know what an "isometric embedding" is. What kind of textbook would contain a definition for this term and related ones?

To clarify, I have an engineering background and just had some basic Analysis and Linear Algebra. My problem is that I don't even know what topic "topology, functional analysis,...." the aforementioned term belongs to.

dc3rd

That sounds like it would be topological......look in Munkres Topology

acutally it is in the index of Munkre's Topology

but the more experienced folk could probably guide you better

Leaky Nun

@FelixCrazzolara https://en.wikipedia.org/wiki/Embedding#Riemannian_and_pseudo-Riemannian_geometry

love_sodam

any nonempty CW complex contains 0-cell

robjohn

@dc3rd No, but using $\pi\approx\frac{22}7$ you get close to $86$.

love_sodam

is this the correct statement?

by the way thank you @LeakyNun

Leaky Nun

07:53

what did i do

robjohn

Now you've done it

love_sodam

you gave me some counterexample yesterday

dc3rd

Cool...thanks @robjohn

Leaky Nun

oh

dc3rd

off to bed I go.

robjohn

07:55

$85.98229715$

Felix Crazzolara

Thank you!

love_sodam

09:22

0

Q: If $X=\bigcup_\alpha X_\alpha$ is a CW complex then $SX\simeq Y\vee_\alpha SX_\alpha$ where $Y$ is some graph

If $X$ is a CW complex with component $X_\alpha$, show that the suspension $SX$ is homotopy equivalent to $Y\vee_\alpha SX_\alpha$ for some graph $Y$. In this case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and 2-spheres. I think the second state...

algebraic-topology cw-complexes

Anyone can answer this problem?

user2103480

11:56

smh the one logic kid in my topology II class complained that our local degree computations are not rigorous and it wouldn't be possible to implement this in a theorem prover

Alessandro Codenotti

Hit himwith some presheaf categories and cubical types

user2103480

12:16

Graphical theorem prover goes brrr

Leonhard Euler

12:44

what are sets that are neither open nor closed called?

Edward Evans

abgeschloffen

Leonhard Euler

Noclopen probably

Lol

Edward Evans

ah I didn't see the neither

Jeff

Hi folks. Can anyone offer a hint for finding the value of $b$ for which
$$1+e^b+e^{2b}+e^{3b}+\cdots=9?$$

Leonhard Euler

Unclopen

Anticlopen

Antonymnclopen

Jeff

12:48

(Where we haven't gotten Taylor series yet)

Astyx

@Jeff $1+x+x^2+\dots = {1\over 1-x}$

Jeff

@Astyx We haven't been given that yet.

Astyx

Prove it

user2103480

@EdwardEvans omg i never thought of this term

Jeff

@Astyx Actually, we have (just not so explicitly)

@Astyx (by which I mean 'thanks')

Astyx

12:54

glad to help

Edward Evans

@user2103480 lmao

Astyx

I am @KarimMansour

Edward Evans

13:15

An artinian ring $R$ has finitely maximal ideals: let $(\mathfrak{m}_i)_{i \in I}$ be all the maximal ideals of $R$ and define $M = \prod_{i \in I} \mathfrak{m}_i = \bigcap_{i \in I} \mathfrak{m}_i$. Then $R/M$ is finitely cogenerated so there are finitely many $\mathfrak{m_1}, \dots , \mathfrak{m}_n$ such that $M = \bigcap_{j = 1}^n \mathfrak{m}_j$.

Since $\mathfrak{m_i} \supseteq \prod_{i \in I} \mathfrak{m}_i = \bigcap_{j=1}^n \mathfrak{m}_j$ we get that every $\mathfrak{m}_i$ is one of the $\mathfrak{m}_j$.

(this as a proof instead of $\mathfrak{m}_1 \supseteq \mathfrak{m}_1 \cap \mathfrak{m}_2 \supseteq \dots$)

Thorgott

13:55

what on earth is cogenerated

Edward Evans

If you've got a family of submodules $(M_i)_{i \in I}$ such that $\bigcap_{i \in I} M_i = 0$ then there's a finite subset $J \subset I$ such that $\bigcap_{j \in J} M_j = 0$

ergh

Turns out that being artinian (dually to noetherian) is equivalent to all quotient modules being finitely cogenerated

lol I just read that I wrote "has finitely maximal ideals"

Mike

14:12

Hello, I have a question about proving the following is true in Markov chain. I'm trying to prove by Chapman-Kolmogorov equations, but I failed. Any hints and suggestions are welcome!

0

Q: Why the recurrence relation of Markov chain can be written as the following?

I am reading E-Li-Vanden-Eijnden's Applied Stochastic Analysis, and the author claims: The recurrence relation of $\boldsymbol{\mu}_n=\boldsymbol{\mu}_{n-1}\boldsymbol{P}$, where $\boldsymbol{\mu}_n$ is the distribution of $X_n$, and $\boldsymbol{P}$ is the transition matrix, can be written as $$...

probability probability-theory markov-chains

user2103480

14:36

@Mike What is your state space? Countable? What is $\mu_{n,i}?$

Mike

Let's assume the state space is countable. $\mu_{n,i}$ is the ith component of the column vector $\mu_n$

Hope this clarifies

user2103480

So this is the probability that at time n the chain is in state i

Mike

Correct. Any hints or comments on how to show this rigorously?

user2103480

$\Bbb P(X_n = i) = \sum_j \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j) = \Bbb P(X_n = i \mid X_{n-1} = i) \cdot \Bbb P(X_{n-1} = i) + \sum_{j\neq i} \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j)$

$\sum_j \Bbb P(X_n = j \mid X_{n-1} = i) = 1$ so $\Bbb P(X_n = i \mid X_{n-1} = i) = 1 - \sum_{j \neq i} \Bbb P(X_n = j \mid X_{n-1} = i)$

I hate markov chain notation #

Mike Miller

@user2103480 "techno" seemed like a better descriptor than "whatever, man"

which is the more accurate description of what scooter is

user2103480

14:58

@MikeMiller I treat trash electronic music like metal fans treat metal

so I'd say this is eurobeat slash harddance

although, nah, eurobeat is too trashy need for speed type stuff. but something something

but the most accurate description is "100% diamond quality music"

Edward Evans

hYpErPoP

user2103480

calling scooter techno is like calling Die Antwoord rap

Mike Miller

Yeah that's what I'm saying

It's not accurate but it's still the best one word description

Edward Evans

Calling Vorga ancient black metal is like calling Maahes cosmic black metal

3

user2103480

@MikeMiller that would be trance I guess

Edward Evans

15:18

@user2103480 do you know Callush?

user2103480

no wossat

Edward Evans

She's a dj from Monnem

makes really really good techno

soundcloud.com/dvskrecords/dusk-pdcst-05-l-callush

user2103480

fuckin mannheimers stealing the term monnem from monheimers

Edward Evans

hahaha

user2103480

Monnem am Rhing

►

rhing = local dialect for rhine

Edward Evans

15:21

Yeah I assumed

user2103480

there can only be one

Edward Evans

It's not disgustingly industrialised enough to deserve to keep Monnem

user2103480

in fact it is

Edward Evans

ye but Mannheim and Ludwigshafen are fused by the industrial glue that is BASF

user2103480

the whole of dormagen's bayer is just across the river, and quite a few bayer things are in monheim

so bayer vs basf death match

Edward Evans

15:23

hahah fair

user2103480

@EdwardEvans lmao of course berlin based

Edward Evans

Apparently every fourth mannheimer is employed by BASF

user2103480

@EdwardEvans either I have no standards anymore or this really blasts

Edward Evans

it's so good init

user2103480

yup

yeh just skipped through this is really good

at least the selection, with regards to mixing I'm a pleb anyways

Edward Evans

15:26

Got really, really drunk the other night on the roof terrace and embarrassed myself dancing to this

user2103480

everybody else embarassed themselves by not dancing to this

Edward Evans

The correct perspective

user2103480

although mannheimers probably think sven väth sets at time warp are "super hard going"

Edward Evans

I don't know him but idk hahaha

user2103480

I wanted to say charlotte de witte first but then remembered that she sometimes actually plays pretty rough stuff, I think

Edward Evans

15:28

I've only been in one techno club in Mannheim cuz Covid hit straight away

user2103480

@EdwardEvans he's one of the DJ "legends", although that title is mostly given to people who came early enough and managed to stay long enough. Like carl cox, ricardo villalobos and such, but not on jeff mills or juan atkins level

Edward Evans

I don't know any of those people lmaoooo I only know the techno that my mates show me

user2103480

hahaha okay fair enough, doesn't matter anyways if your friends show you the right stuff and drag you to the right parties (which they seem to do)

Edward Evans

aye, I show them metal in return and they resent me

user2103480

rightfully so

Edward Evans

15:34

Couple of them haven't been to a metal concert before and I've roped them into seeing Meshuggah in Wiesbaden in November

which will be an experience

user2103480

"Did you hear that crazy 12.4/sqrt(2) rhythm switching into the chuck e. cheese drumroll?"

^how metal fans sound when they talk about meshuggah or any other nerd band

chad techno fan: "THASSA MAD KICK" - "YEH" sound of grinding teeth

Edward Evans

Ackchyually I think you'll find that Meshuggah perform in 4/4 with metric modulation of riffs

also true tho

user2103480

@EdwardEvans lmao

I don't wanna do arithmetic in $\Bbb Q$ when I listen to music

4/4

absolute unit

Edward Evans

all rational time sigs are units

user2103480

are they also absolute units

Edward Evans

15:41

idek what an absolute unit would be

the equivalence class of 1 under the relation a ~ b iff a and b are associates

user2103480

@EdwardEvans there won't ever be one, everytime you come up with one, anglo-saxons mess it up by using an absurd alternative metric

Edward Evans

wtf

user2103480

but maybe the size of imperial pints drove down the prices

either that or manchester is just cheap as sh*t

Edward Evans

it's the second one

user2103480

guess I wont ever visit scotland or southern england

is there any other cheap region in the UK

Edward Evans

15:47

Omg I got a crate of Bitburger for 99c on Tuesday

also anywhere in the north is cheap af

user2103480

@EdwardEvans does that include scotland

wouldve thought its more expensive there

@EdwardEvans whao was that just before expiration

Edward Evans

I don't think so, I know the North of England is cheap af for beer but I haven't been to Scotland

user2103480

also, how far did you have to carry that lmao

aren't you like 2km away from the next supermarket

Edward Evans

@user2103480 No, I put one bottle on the Band and told the lady I have a Kasten and I don't think she heard but my mate heard me, picked up the crate and carried it out to put in my rucksack, then I paid and left and then looked at the Beleg and I only paid 99c for it lmao

Leaky Nun

@EdwardEvans wenn alle Konjugierten einer algebraischen Nummer Norme eins hat, dann sie eine Einheitswurzel ist

Edward Evans

15:49

Yeah I put the whole crate in my Bundeswehronline Rucksack

@LeakyNun Ja

Leaky Nun

(hab ich meine Kasus wohl benutzt?)

Edward Evans

@Leaky Ja hast du, aber man sagt normal "algebraische Zahl" statt Nummer

Leaky Nun

ok danke

die Kasus sind sehr schwerig

Edward Evans

(Und ANT heißt darum AZT auf Deutsch)

Ja das stimmt, und man sagt "Fälle" für die Cases

Leaky Nun

aha, ich kenne die Zahlentheorie

Edward Evans

15:52

Genau

Leaky Nun

haben "Fall" und "Kasus" eine gleiche Bedeutung?

Edward Evans

Ich glaube schon aber ich würde nie Kasus sagen, vielleicht ist das ein Fachbegriff oder so

Leaky Nun

ok

Edward Evans

@user2103480 Hast du die 4 Fälle in der Schule je als Kasus bezeichnet?

user2103480

@EdwardEvans Probably? I honestly don't know the exact rules of german grammar lol

Edward Evans

15:54

Nice Deutsch

user2103480

I just do applied german

Edward Evans

lmao

Leaky Nun

what on earth is applied german

Edward Evans

German without knowing why what you're speaking works

Leaky Nun

lol

Leonhard Euler

15:55

Every peaceful german conversation sounds like a fight

Edward Evans

that's also fairly true

Leonhard Euler

Die zthdyfr fhdejgfdthcer!

user2103480

really depends on the dialect

Leaky Nun

lernt ihr Deutsch in Schule nicht?

Deutsche grammatische Regeln

user2103480

We do, but everyone forgets about it

Edward Evans

15:57

In German we don't say "that's fine thank you", we say "Dieser verfickte Hurensohn lässt mich einfach nicht in Ruhe, wieso versteht er das Wort 'nein' nicht!" and I think that's beautiful

4

user2103480

The sad thing about language is that you can know the rules in and out and still won't top a native speaker

@EdwardEvans our curse words are quite nice

Edward Evans

I think I could make a good case to say that I speak better German than some natives

e.g. anyone aus dem Saarland

Leonhard Euler

Ich kann mit Google Übersetzer Deutsch sprechen

user2103480

@EdwardEvans says the one with an austrian dialect

Edward Evans

I speak Hochdeitsch too

user2103480

16:07

@EdwardEvans is that austrian for hochdeutsch

Leaky Nun

@LeonhardEuler DeepL ist besser als Google Translate, aber DeepL hat wenige Sprache

Edward Evans

yeah I said that for comic effect

and it really, really worked, brilliant

In Vorarlberg people say Dütsch, while in the rest of Austria people say Deitsch

user2103480

you are behind 7 layers of comedy

ok gonna be off for a min

Edward Evans

Treachery, the 9th layer of comedy

alright bye

amWhy

@robjohn Get your "mean green Shamrock machine" ready for St. Patrick's day!!!

maths student

16:32

About 25% automobiles in Chennai are foreign made and nearly 40 % of automobiles in Mumbai are foreign made. A sample of 100 automobiles are drawn from Chennai and a sample of 200 are drawn from Mumbai. Let $X$ stand for the number of foreign made cars in the sample from Chennai and $Y$ for the number of foreign cars in the Mumbai sample and Z = X+Y. Then

I know that X and Y are binomial distribution

I know that sum of indepednen binomial distribution is binomial

So I want to know where X and Y are independent here ?

@LeakyNun

Any suggestions ?

Clarinetist

It's reasonable to assume they're independent. For one thing, the number of foreign-made cars you draw from the Mumbai sample does not influence how many foreign-made cars you draw from Chennai, or vice versa.

maths student

So Z would be linearly independent right ?

@Clarinetist

Clarinetist

Linear independence is a completely different concept

maths student

@Clarinetist sorry Z would be binomial right .

Clarinetist

What you can say is that $X$ and $Y$ are independent, so that $Z$ is a sum of independent random variables

maths student

16:43

@Clarinetist Linear Algebra hangover

Clarinetist

I don't know for sure, actually. The thing with summing binomials is that you usually assume the success probabilities are the same; they are not in this case.

maths student

@Clarinetist Right , I forgot that point ?

Clarinetist

What are you aiming to do with $Z$? Get its distribution? I don't know if it's binomial.

My guess is it's not.

porridgemathematics

hi, if I have a function that is measurable on $[a,b]$ and is s.t. it is bounded on $[a,b]$, specifically is less than $1$ (in absolute value) on that interval, can I always get a sequence of $C^{\infty}$ functions approximating this in $L^p$ that are also bounded below $1$ in absolute value on $[a,b]$? I know I can get compactly supported functions to approximate it in $L^p$, but am sturggling with the bounded part

maths student

Yaa I got it , but I want to know intutive explaination why it is not binomial ?

Clarinetist

16:48

Well, let's think of it this way

Let's say I give you two independent experiments with different success probabilities, and I tell you to add the number of successes together

Thorgott

measurable functions can always be approximate by a sequence of step functions increasing in absolute value

maths student

@Clarinetist I can always do summation manipulation like x takes value k then z takes value z-k. and then sum over it to show it is not binomial

Clarinetist

Would it be possible to model that as one single experiment with one constant success probability? That's the only reason that would make sense.

At the end of the day, it's primarily mathematical reasons for why this doesn't work.

porridgemathematics

@Thorgott oh yeah I see, I forgot the details of this construction, I only recall the way to prove that compactly supported smooth functions are dense is to show that they approximate $\mathbf{1}_{K}$ for any $K$-cell , but the detail involves finding a nondecreasing sequence

maths student

@Clarinetist so probablity of sucess changes depending on whether it is from which location mumbai or chennai .

Clarinetist

16:53

Right, 25% in Chennai, 40% in Mumbai

maths student

@Clarinetist so it might fluctuate right for z variable .

Clarinetist

Right, just try computing some probabilities

maths student

to get distribution for z we need to do proper binomial sum argument .

robjohn

@amWhy Perhaps you were not around for the fashion show. The only one that is still unseen is the Easter avatar.

Clarinetist

@mathsstudent Have fun:

12

Q: Addition of two Binomial Distribution

What is the distribution of the variable $X$ given $$ X = Y + Z, $$where $Y \sim $ Binomial($n$, $P_Y$) and $Z\sim$ Binomial($n$, $P_Z$)? For the special case, when $P_Y = P_Z = P$, I think that X~Binomial($2n$, $P$) is correct. If $P_A ≠ P_B$, the distribution might eventually just be Binomia...

probability probability-theory probability-distributions random-variables

Actually, it's more complex than this

because the number of trials differ

Here we go:

2

Q: Sum of Binomial distribution when the success rate is different.

Is there any easy way to calculate the probability of the sum of two binomial random variable if the success rates of them are different each other? I mean that $X \sim Bin(n,p_0)%$, $Y \sim Bin(m, p_1)$, $Z = X+Y$, $p_0 \neq p_1$ and hope to calculate the distribution function of $Z$. I know t...

probability probability-distributions binomial-distribution

Thorgott

16:58

this is simpler than that

the idea is simply to split the codomain into smaller and smaller intervals

maths student

@Clarinetist nice one thank you.

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