Mathematics - 2024-07-04

ephe

04:30

@Thorgott Sorry to ping you. But regarding your answer on the filtered module morphism factorization, how can I show that the regular kernel and cokernel modules with the induced filtrations are the kernel and cokernel of a map. I thought about using the universal property in the category of modules and then showing that all arrows must be morphisms but I couldn't figure out how to show that the arrow from the object that satisfies the universal property to the domain module is a morphism

By that I mean say $u$ is from $M$ to $N$ filtered modules. Then we have the usual commutative diagram for kernel. Say $K'$ is another filtered module that satisfies the universal property of $\mathrm{Ker}(u)$. Then there is a map $k'$ from $K'$ to $M$ such that $u\circ k'=0_{K'N}$. Then there exists a unique module homomorphism $z\colon K'\to \mathrm{Ker}(u)$ such that the whole thing commutes. I want to show that $z$ and $k'$ must also be morphisms of filtered modules. How can I do it?

On second thought I should be assuming that $k'$ is a morphism and showing that $z$ is a morphism

copper.hat

06:06

yellow

user 85795

06:19

mellow

TIL :-/

psie

08:47

I have a fundamental doubt. Suppose $M\in \text{Exp}(1)$. What does it mean when someone writes $X\mid M=m\in\text{Po}(m)$? In particular, why is $X\mid M=m$ a random variable? I'm familiar with the more abstract definition of conditional expectation, i.e. that $E[X\mid M]$ is a random variable, but $X\mid M=m$ on its own confuses me.

psie

09:40

I'd appreciate any viewpoint on this, as this is done repeatedly in the book I'm reading, but without introducing really what $X\mid M=m$ means.

Jakobian

10:18

@psie that $\mu(A) = P(X\in A | M = m)$ is $\text{Po}(m)$

Think of this not as an introduction of new object but rather of notation

as in, $\mu$ is exactly given as a measure by $\text{Po}(m)$ i.e. $\mu(\{x\}) = \frac{m^x e^{-m}}{x!}$ for $x = 0, 1, 2, ...$

Binky McSquigglebottom

10:33

i Need help with the domain in Polar coordinates

Sine of the Time

did you choose polar coordinates centered in $(0,0)$ or in $(0,2)$?

Binky McSquigglebottom

But aren't Polar coordinates always at the point (0,0)?

Unknown x

0

Q: Question on Continuum Mechanics: How does Eulerian derivatives of $f$ and Lagrangian derivatives of $f$ different?

Consider a particle $M$ occupying the position $\vec{a}$ at time $t = 0$ and the position $\vec{x}$ at time $t$. A function $f = f(M, t)$ associated with a particle $M$ (e.g., its velocity, acceleration, or other physical quantities to be defined subsequently) can be represented in two different ...

Sine of the Time

@BinkyMcSquigglebottom no why?

Binky McSquigglebottom

I have always centered them at (0,0)

To center them in (0,2) I had to do x=r cos(theta) , y=2+sin(theta)???

Sine of the Time

10:43

right

Binky McSquigglebottom

Could you help me search for the domain?

Sine of the Time

rewrite the domain using the transformation you wrote before

Binky McSquigglebottom

Okay

$x^2 +y^2 -4y$ ≤ 0 $\to$ $x^2 -4 +4 \cdot x \cdot sin(x) -4sin(x)$≤0

Sine of the Time

what?

Binky McSquigglebottom

The First equation

I replaced the polar coordinates

Sine of the Time

10:52

since $y=2+r\sin \theta$, $x^2+(y-2)^2=r^2 \cos^2 \theta +r^2 \sin^2 \theta=r^2$

So $0\le r \le 2$

Binky McSquigglebottom

Don't I have to see the other equations too?

Sine of the Time

yes

Binky McSquigglebottom

$y≤2-x \to r•(sin(x)+cos(x))$≤0

$x≥0 \to r\cdot cos(x) ≥0$

Instead of x it would be theta, when I transform to polar coordinates

Sine of the Time

yes

actually, there might be a problem using poar coordinates center at $(0,2)$

Binky McSquigglebottom

@SineoftheTime But here to see that it was between 0 and 2 you did: $r^2 ≤ 4 \to -2≤r≤2$ , However, since the radius had to be greater than 0, did you substitute 0 for -2?

Sine of the Time

11:02

yes

Binky McSquigglebottom

So I have the conditions: $0≤r≤2$ , $r≥0$ , $r \cdot (sin(x)+cos(x))≤0$???

x would be theta

Sine of the Time

11:20

If you use this change of variables, the integral is not straightforward, I don't know maybe polar coordinates centered in $(0,0)$ lead to an easier integral

sorry :)

psie

@Jakobian ok, thanks. So $\mu(A)$ here is actually a regular conditional probability of $P$ given $M$ for the set $X\in A$. Makes sense.

So technically I guess $X\mid M=m$ is not a random variable.

Binky McSquigglebottom

@SineoftheTime Don't worry, but solving the equations in the order of how I wrote them, I find that in the second one it is tan(x)≤-1, while in the last one cos(x)≥0, therefore -π/2≤theta≤-π/4

Sine of the Time

yes $3\pi/2 \le \theta \le 7\pi/4$

but that's not the problem

Binky McSquigglebottom

So $0≤r≤2$ and $-π/2≤theta≤-π/4$

Sine of the Time

usually $\theta \in [0,2\pi[$

Binky McSquigglebottom

11:38

I don't really understand what the problem is

Sine of the Time

never mind

you have to solve $\iint_D \frac{r^2 \cos \theta}{2+r\sin \theta}d\theta dr$

it's a bit lengthy, but you should be able to do it

Binky McSquigglebottom

but shouldn't theta go from 3π/4 to π?

psie

@psie on the other hand, if $\mu(A) = P(X\in A\mid M = m)$ is $\text{Po}(m)$, then I'd assume it is the law of some random variable. Do you know which one, @Jakobian?

Binky McSquigglebottom

11:56

@SineoftheTime I didn't quite understand which one is our D

Sine of the Time

@BinkyMcSquigglebottom $\int_0^2\int_{3\pi/2}^{7\pi/4}$

Binky McSquigglebottom

how did you find theta??? It's not very clear to me

Sine of the Time

it's the same you found

plus you can see it drawing $D$

Binky McSquigglebottom

but could I also get the radius from another equation?

Sine of the Time

why?

you already have the bounds for r

Binky McSquigglebottom

12:07

I mean

For example, could I also find the radius from the 2nd inequality, and theta from the 1st and the 3rd???

Sine of the Time

you have to use all the conditions to find the bounds

Binky McSquigglebottom

Yes but

So I have the conditions: $0≤r≤2$ , $r≥0$ , $r \cdot (sin(x)+cos(x))≤0$

Wouldn't the last one be r≤0???

Sine of the Time

$r\cos \theta \ge0$ so $\cos \theta \ge 0$

Binky McSquigglebottom

😓

Sine of the Time

@BinkyMcSquigglebottom the radius is always positive

Binky McSquigglebottom

12:16

Why did you solve for theta and not r?

Sine of the Time

because $r$ is positive so you can divide

if you draw the region, you don't have to compute anything

I've to go

Binky McSquigglebottom

While the cos(theta) could also be negative or zero, here's why?

@SineoftheTime okay , anyway thanks ^⁠_⁠^

Sahaj

12:55

Hi guys. Is Douglas West's book on graph theory a good resource for the subject ?

Jakobian

@psie what do you mean

any probability measure on $\mathbb{R}$ is a law of some random variable

and not just one either

psie

@Jakobian right, I meant that every random variable $X$ on a probability space gives rise to a probability measure (the law of $X$). Does every probability measure correspond to some random variable? I.e. is the converse true? Apparently yes...

Jakobian

it does, but maybe not on the original probability space

but in probability the thing is that we don't care if our original probability space changes

a mindset different than that of measure theory

by probability measure I suppose you mean a one on $\mathbb{R}$

Skorokhod's representation theorem

In mathematics and statistics, Skorokhod's representation theorem is a result that shows that a weakly convergent sequence of probability measures whose limit measure is sufficiently well-behaved can be represented as the distribution/law of a pointwise convergent sequence of random variables defined on a common probability space. It is named for the Ukrainian mathematician A. V. Skorokhod. == Statement == Let ( μ n ) n ∈ N...

Kolmogorov extension theorem

In mathematics, the Kolmogorov extension theorem (also known as Kolmogorov existence theorem, the Kolmogorov consistency theorem or the Daniell-Kolmogorov theorem) is a theorem that guarantees that a suitably "consistent" collection of finite-dimensional distributions will define a stochastic process. It is credited to the English mathematician Percy John Daniell and the Russian mathematician Andrey Nikolaevich Kolmogorov. == Statement of the theorem == Let T {\displaystyle T} denote some interval (thought of as "time"), and let...

psie

13:14

ok, I will take a look at these. The reason I was asking is because in my example we write $X\mid M=m\in\text{Po}(m)$ and more generally $Y\in F$, for some distribution $F$. But I don't think $X\mid M=m$ is a random variable, hence I wonder if one can replace $X\mid M=m$ with the actual random variable(s) that correspond to $\mu(A) = P(X\in A \mid M = m)$?

Jakobian

13:35

@psie maybe you've missed that, but I've said that $X|M = m$ is not a random variable

the notation $X|M = m \sim \mu$ merely means that $\mu(A) = P(X\in A | M = m)$

this means that the two measures are equal and nothing else

psie

ok, but people write $X \sim N(\mu,\sigma^2)$, where $X$ is a random variable. By $X \sim N(\mu,\sigma^2)$ I suppose they mean that the density admitted by the law of $X$ is given by a certain formula that we call the density of the normal distribution. I understand that $X\mid M=m$ is not a random variable, and so I was wondering if it is possible to write $X\mid M = m \sim \mu$ as $\star\sim\mu$, where $\star$ should be replaced by some notation that represents a random variable

Jakobian

14:00

that for specific $m$ we can find a random variable $Y$ with law $\mu$ is not a problem. We can probably even find a whole family $Y_m$ of them, and a copy of $X$ and $M$

but what is the purpose of this

Is it so that $Y_M = X$? Then we are probably entering some kind of regular probability territory

utility is one thing. This can probably be answered if we are experts on probability theory

If by doing so we are going over some complex mathematics, then why just not stick to the simple thing

psie

ok, yeah I don't know. I just wanted to better understand the notation. I do understand it better now, but I find it a bit inconsistent when someone writes $X\mid M=m\sim \text{Po}(m)$ and then $M\sim \text{Exp}(1)$. $M$ is a random variable, whereas $X\mid M=m$ is not.

Jakobian

So what

in essence it means the same thing

you put $\in A$ into the notation, embrace it into "$P$" and claim, the two measures are the same

$P(X\in A | M = m)$ or $P(M\in B)$

this is quite intuitive

$M\sim Exp(1)$ merely means that the measure $B\mapsto P(M\in B)$ is the same as that of exponential measure with parameter $1$ on the real line

the information that $M$ is a random variable, perhaps isn't as significant for this notation as you would like to think

probability theory has lots of such hidden unexplained notational issues

psie

it does, regarding this one you have me convinced :) I give up

Sine of the Time

14:27

@BinkyMcSquigglebottom it's the same reason why when you solve $2x\ge 0$, you divide by $2$

ZaWarudo

15:15

Problem: construct a sequence $x_n \in \mathbb{R}^2$ having a subsequence that converges to $x \in \mathbb{R}^2$ for every $x$. Isn't enough to consider for each $n\in\mathbb{N}$ the sequence $x_n=x+(1/(n+1),0)$ and its subsequence $x_{2n}=x+(1/(2n+1),0)$, so that $|x_{2n}-x|=1/(2n+1) \to 0$ as $n\to +\infty$?

leslie townes

there is some quantifier confusion there. i am guessing that the desired condition on the single sequence x_n is that for all x in R^2 there be a subsequence of x_n convergent to x. i.e. one sequence has to get the job done for every x in R^2.

the alternative interpretation is maybe suggested by the sloppiness of putting the "for every x" quantifier at the end of the statement instead of the beginning, but as you point out, it is not an interesting problem, given x, to construct a sequence that converges to x (or indeed, has every subsequence convergent to x)

you might try the example in R^1 first. find a single sequence whose set of subsequential limits is R

ZaWarudo

@leslietownes Indeed, I was confused because it seemed to easy and I was unsure about the quantifiers as well. Thanks for the answer, I will think a little more about that.

Jakobian

In other words, the set of limit points of the sequence $x_n$ has to be whole $\mathbb{R}^2$

Binky McSquigglebottom

@SineoftheTime but is there a way to understand from the domain if it is better to integrate for example to (0,0) or (0,2)?

ZaWarudo

@Jakobian: Thanks, I will think about it. So I solved the problem: "Show that for each $x \in \mathbb{R}^2$ there exists a sequence $(x_n) \subseteq \mathbb{R}^2$ such that $x_n \to x$"? In other words, this latter problem fixes an $x$ and then construct a sequence that tends to $x$; the other interpretation is completely different. Right?

Jakobian