Mathematics - 2019-12-06 (page 1 of 3)

12:35 AM

Ugh, this problem appears daunting: Let $(X,Y)^\top$ have a two-dimensional normal distribution with means 0, variances 1, and correlation coefficient $\rho$, $|\rho|<1$. Compute the moment generating function of $(X^2-2\rho XY+Y^2)/(1-\rho^2)$.

Rithaniel

1:13 AM

Honestly, I'm having difficulty figuring this one out

Leaky Nun

@Rithaniel express (X, Y) in terms of a two-dimensional independent normal distribution

Ultradark

How many functions can be represented as projections?

Leaky Nun

about 7

Ultradark

hyperbola, parabola, ellipse, what are the other 4?

Rithaniel

So, in terms of $(X+Y,X-Y)^T$?

I feel as though another two-dimensional distribution might have been intended, actually

Baker013273213

1:39 AM

hi

Leaky Nun

@Rithaniel that still isn't independent

Rithaniel

Yeah they are, when Var(X)=Var(Y) (and here they both equal 1) where X and Y are normal random variables, X+Y and X-Y are independent.

Leaky Nun

that requires X and Y to be independent to start with

Rithaniel

Nah, they can be dependent. Just calculate the covariance matrix

If $\sigma=Var(X)=Var(Y)$ and $\hat{\sigma}=Cov(X,Y)$ you have that $\left(\begin{array}{c c}
1 & 1 \cr
1 & -1 \cr
\end{array}\right)\left(\begin{array}{c c}
\sigma & \hat{\sigma} \cr
\hat{\sigma} & \sigma \cr
\end{array}\right)\left(\begin{array}{c c}
1 & 1 \cr
1 & -1 \cr
\end{array}\right)=\left(\begin{array}{c c}
2(\sigma+\hat{\sigma}) & 0 \cr
0 & 2(\sigma-\hat{\sigma}) \cr
\end{array}\right)$

Non-diagonal entries are zero and uncorrelated normal random variables are independent

Although, clearly this isn't what you had in mind, so perhaps I should keep looking for other potential independent representations

Adam

what do the numbers that appear below our icons in the chat window represent?\

Rithaniel

1:59 AM

Alright, I've managed to get it into the form $y^2+\frac{(x-\rho y)^2}{1-\rho^2}$

Rithaniel

2:10 AM

Now I've gotten it into the form $(\frac{y(\sqrt{\rho^2-1}+p)-x}{\sqrt{\rho^2-1}})(\frac{y(\sqrt{\rho^2-1}-p)+x}{\sqrt{\rho^2-1}})$

Math geek

1

Q: How do I eliminate the repeating cases?

Let $K$ be the field with exactly $7$ elements. Let $\mathscr M$ be the set of all $2×2$ matrices with entries in $K$. How many elements of $\mathscr M$ are similar to the following matrix? $ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ My attempt: Answer is given is $56$. We need to find the...

Can you please help me with ancientmathematician's comment?

How did he use that formula?

Is that formula same as math.stackexchange.com/questions/1600432/…

Rithaniel

Unfortunately I feel like I'm just chasing my tail

Hawk

In a field, does the element $-1_F$ necessary have to exist?

or that is yes because that is the additive inverse?

actually yeah it has to nvm

Rithaniel

Technically yes. Every element needs to have an additive inverse. However, it could equal $1_F$

Hawk

yeah in F_2

Rithaniel

2:22 AM

Or $F_4$ or $F_8$. Any field with characteristic 2

Hawk

This is another strange question. If F is a finite field, it contains a copy F_p. But is F_p the only finite subfield?

can I have another subfield F_q where q is not p?

Rithaniel

Not necessarily. I don't remember off the top of my head what the sizes of subfields can be, but a finite field can have multiple distinct, nontrivial subfields

Hawk

so lets say it does, F_q, would that also imply F would be the finite extension of F_q?

Rithaniel

I believe so

Hawk

But does q have to be prime?

characteristics are unique right?

Rithaniel

2:28 AM

So, a field of order $p^n$ has exactly one subfield of order $p^m$ for each divisor $m$ of $n$

Characteristics are unique, yes

A finite field must have prime power order

Hawk

so q has to be prime still?

but that would imply q = p?

Rithaniel

So, there exist fields of orders 2,4,8,16,32,64, etc, but no fields of order 6

q has to divide the order of F

Hawk

so all my questions are because of the proof |F| = p^n. It just appears to me if we can find a subfield F_q where q isn't prime, we can show |F| isn't a prime power.

Rithaniel

So, for $F_{64}$ (The field of 64 elements), there is a subfield of order $2$, order $4$ and order $8$

Well, there $F_8$ in $F_{64}$ and $8$ isn't prime.

Heya Semi

Hawk

well its a prime order

Rithaniel

2:34 AM

Well yes, the order of a subfield has to divide the order of the parent field

Hawk

that's only because we know ahead of time the finite filed must have prime power

Rithaniel

Well technically it falls from the fact that fields are, first and foremost, groups, and a theorem by Lagrange that the order of a subgroup must divide the order of the parent group

Hawk

well i understood that

the issue is the proofs that show |F| = p^n assmes F contains only the subfield F_p.

Rithaniel

Well, are you comfortable with the thought that the characteristic must be prime?

Hawk

yeah

n*1_F = 0 has to be prime because otherwise n would not be the smallest positive integer

Rithaniel

2:48 AM

Alright, let $p$ be the characteristic of $F$. Since $1$ has order $p$, we know that $p$ divides $\vert F\vert$. Now assume that $q$ divides the order of $F$ where $q$ is a prime distinct from $p$. Then, by Cauchy, there exists an element $x\in F$ of order $q$. So $qx=0$ and $px=0$. Also, by bezout, there exist integers $a$ and $b$ such that $aq+bp=1$. So $x=(aq+bp)x=a(qx)+b(px)=0$, but this is a contradiction, because $x$ is assumed to have positive order.

So, the only fields that exist require that they only have one prime divisor

anakhro

@TedE waves frantically.

Rithaniel

@Semiclassical If you have a moment, I have this probability problem which is giving me a real headache: Let $(X,Y)^\top$ have a two-dimensional normal distribution with means 0, variances 1, and correlation coefficient $\rho$, $|\rho|<1$. Compute the moment generating function of $(X^2-2\rho XY+Y^2)/(1-\rho^2)$.

Jeff

Hi folks. The answer to this question uses that $(A \cup B) \cap C=\emptyset$. But how do they know that?

Let A, B, and C be events in a sample space S such that S = A ∪ B ∪ C. Suppose that
P(A) = 0.3, P(B) = 0.6, and P(A ∩ B) = 0.2.

Hawk

@Rithaniel in the last step, x = 0_F, you say contradiction is because it is assume positive order. Is the order not 1?

Leaky Nun

@Hawk $0_F \ne 1_F$ is one of the field axioms

Rithaniel

2:56 AM

Yeah, I misspoke. The order is 1.

Then the contradiction would be that 1 is not prime

Hawk

i think th contradiction is that the characteristic is not p

lol

Rithaniel

That was something we assumed off the bat

anakhro

@Jeff what is the question?

Hawk

the contradiction is that 1*x = 0_F, instead of the characteristic p. So q does no exist. So what if we don't assume q is prime? Then q is composite, and we just let another prime, say b, divide q?

Rithaniel

Yeah, that's the stuff. Now you just repeat the steps from there

By Cauchy there is a y of order b in F

Hawk

3:02 AM

if jus leave q as composite, can bezout's equation force out the prime b? or do i have to make that argument ahead of time? i am trying to see if i can reach a contradiction by allowing q to be composite

Rithaniel

You don't need to. If you have a composite q, you just pick out a prime divisor of it: b, and work exclusively with it.

You can then find an element of order b, and show that it has to equal 0

Just pick b such that $b\neq p$. If you can't, that means $q=p^n$ for some $n$

Jeff

Let A, B, and C be events in a sample space S such that S = A ∪ B ∪ C. Suppose that
P(A) = 0.3, P(B) = 0.6, and P(A ∩ B) = 0.2.

anakhro

@Jeff that is a bunch of information. It doesn't have a question.

Hawk

hang on if b does not exist. this means q is not composite since we assumed it has a prime divisor, so q is prime. Therefore q also does not exist by repeated argument. So the remainder factor m in |F| = m*p^n is also prime but it cannot be distinct from p, so it must be p.

@Rithaniel i think i forced it out!

Jeff

@anakhro The question has multiple parts. The part I'm answering is "Find $P(C)$". But my question is "Why is $(A \cup B) \cap C$ the emptyset?

Hawk

3:09 AM

because if it is distinc from p, it has only m and 1 as divisor and if proceed with Cauchy and Bezout, we would arrive at the same contradiction that 1*x = 0_F

Rithaniel

Alright, so you agree that all fields have order p^n?

Hawk

if what I wrote is correct

anakhro

@Jeff can you show the full solution as written?

Jeff

20. a. By the formula for the probability of a general union,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.3 + 0.6 − 0.2 = 0.7
b. Because (A ∪ B) ∩ C = ∅, then
P(C) = P(S) − P(A ∪ B) = 1 − .07.
c. By DeMorgan’s law for sets, Ac ∪ Bc = (A ∩ B)c = S − (A ∩ B). Hence, by the formula for
the probability of the complement of an event,
P(Ac ∪ Bc) = P(S − (A ∩ B)) = P(S) − P(A ∩ B) = 1 − 0.2 = 0.8.

anakhro

I don't see any reason to conclude that that's an empty intersection.

Rithaniel

3:21 AM

I believe it is, @Hawk, just needs to be more thoroughly stated if it's for an assignment or anything like that

Hawk

No not an assignment lol

I am literally self studying galois theory

i think this would be a sick assignment question though

Rithaniel

Actually, I'm taking Galois theory next semester.

I'm familiar enough with a few results

Jeff

@anakhro Oh? That's the textbook's answer.

anakhro

Weird.

Ultradark

hahah

Hawk

3:32 AM

lol i think u know more then me

Ultradark

I found some fun points

$(a,b)$ and $(1/b,1/a)$

Semiclassical

3:55 AM

@Rithaniel neat. So you've got $E[XY]=\rho$

(the joy of having zero mean and unit variances)

Note that this does uniquely specify the normal distribution, i.e. you can write down the pdf

Ultradark

Say for all $m<1,$ you take all points $(a,b)$ and $(1/b,1/a)$ such that line $y=mx$ contains those points.

Does that make sense so far?

Rithaniel

4:10 AM

I think I figured it out.

Semiclassical

Nice.

mathematica gives the answer as just 1/(1-2t)

Rithaniel

$\frac{X^2-2pXY+Y^2}{1-\rho^2}=\left(\begin{array}{c c}
X & Y \cr
\end{array}\right)\left(\begin{array}{c c}
1 & -\rho \cr
-\rho & 1 \cr
\end{array}\right)^{-1}\left(\begin{array}{c}
X \cr
Y \cr
\end{array}\right)$

Semiclassical

Right.

Wait...

there you go

well, probably shouldn't have the minuses when taking the inverse

Shine On You Crazy Diamond

@Ultradark still up to study tonight?

Sorry pinging you back late, today got some work from my dad

Rithaniel

By Lotus, we just get that this is the integral of the pdf of $(X,Y)^T$ with an extra factor of $(1-2t)$ which pops out when you integrate

Semiclassical

4:15 AM

ya

A cute observation to be had in this vein: Let $A=\begin{pmatrix} 1 & \rho \\ 0 & \sqrt{1-\rho^2}\end{pmatrix}$

Ultradark

@shi probably too late for tonight but can you help me with a question

Shine On You Crazy Diamond

@Ultradark which one?

Semiclassical

Then $A^\top A=\begin{pmatrix} 1 & \rho \\ \rho & 1\end{pmatrix}$

So you can factor the above as $(X,Y)A^\top A(X,Y)^\top=|A(X,Y)^\top|^2$

Shine On You Crazy Diamond

@Ultradark I can only help you if you communicate fast enough

It's frustrating sometimes waiting for responses

Rithaniel

Oh snap

Ultradark

4:17 AM

okay

@shi

Shine On You Crazy Diamond

Let's start a room

Ultradark

ok

Jacksoja

4:28 AM

@LeakyNun Hello

@LeakyNun chess games?

Leaky Nun

sorry not now

Jacksoja

okay no problem

i was trying to solve this

gf = hf then h=g

i could not find a couter example

Semiclassical

Context?

Jacksoja

f: A--> B

g , h : B--> C

the compositon is same

If I argue by contradiction

i need to find b in B

Semiclassical

well, that means that g=h on the image of f

Jacksoja

4:31 AM

such that g (b) is not equal to h (b)

Semiclassical

so you want g,h which do have the same image on some subset of B but not on all of it

Jacksoja

I did not find a way to prove or disprove it

Semiclassical

The simplest counterexample I see is this: f(x)=x^2, g(x)=x, h(x)=|x|

g and h agree on positive x, so they'll agree on the entire image of f

Balarka Sen

@LeakyNun I looked at Stewart. That's a cool, elementary proof. I should have noticed that.

Leaky Nun

@BalarkaSen nice

Jacksoja

4:35 AM

@Semiclassical I see thank you !

but how does one prove this

Leaky Nun

@BalarkaSen it's "essentially the same idea"

Jacksoja

without giving counter example

I keep thinking thiws

which is wrong

for gf : A-->C

Rithaniel

Alright, I have this expression $ne^{-x}(1-e^{-x})^{n-1}-\text{log}(n)$ and I'm trying to find a value as $n$ goes to $\infty$. How should I try and manipulate this?

Jacksoja

in my mind i think of , all of B must have an image

but that is clearly not the case for the composition , we need only to work with image of f

@Semiclassical You see what confuses me?

Balarka Sen

@LeakyNun Yeah I see the point.

Leaky Nun

4:41 AM

I like your proof better.

Balarka Sen

@Thorgott Ah, you wrote down Stewart's proof (essentially). Right, the key point is any embedding of $K$ in a bigger field restricts to an automorphism of $K$ again, since $K$ is a splitting field.

Thanks @Leaky :)

Leaky Nun

@BalarkaSen is K/F normal equivalent to "any two F-embeddings K -> L have the same image"?

Balarka Sen

They both have image K, right?

Oh, equivalence.

Let's see.

(Don't tell me if you know the answer, I should be able to figure this one out)

Leaky Nun

I know the answer.

Another (irrelevant) question: are all purely inseparable extensions normal? @BalarkaSen

Balarka Sen

So in particular any F-embedding K -> bar{K} has image K. Suppose f was an irreducible polynomial in F[x] which had one root, say a, in K. Then for any other root b, there is an isomorphism K = F(a) -> F(b) \subset \bar{K}. This is an F-embedding in the closure, so must have image in K, so F(b) = K = F(a), so b belongs to F(a).

So all roots of f are in K, and f splits completely in K

I know algebraic closure can be avoided here but it's slightly easier for me to think in that term

Leaky Nun

4:56 AM

nice

I don't object to the use of algebraic closure

it is an important philosophy

Balarka Sen

@LeakyNun Hmm. That's an interesting question.

Leaky Nun

btw K is not F(a)

Balarka Sen

Oh my bad.

Just ignore that "= F(a)" everywhere, it doesn't affect the solution.

$F(a) \to F(b) \subset \bar{K}$ extends to $K \to \bar{K}$, which has image $K$, so in particular $F(b) \subset K$.

Leaky Nun

how do you extend the map?

Balarka Sen

First extend to an automorphism $\bar{K} \to \bar{K}$ by what I wrote earlier, then restrict to $K$.

Leaky Nun

5:00 AM

nice

what extension can you replace $\overline{K}$ with?

I call this philosophy "reflection"

the proposition containing "for all L" is equivalent to the proposition where you substituted $L$ with $\overline{K}$

a universal proposition is reflected by a particular instance

in a ring, $\forall x, x = 0$ is reflected by $1$

Balarka Sen

That makes sense

@LeakyNun I'm thinking the splitting field of $f$.

Leaky Nun

$\forall a, \forall b, (a+b)^2 = a^2 + 2ab + b^2$ is reflected by $\Bbb Z[a,b]$

@BalarkaSen I don't think you can map K to the splitting field of $f$

K can be very big

Balarka Sen

Splitting field of $f$ over $K$, I mean. If that's $K$ I am done, otherwise it's some bigger extension $L/K$. $F(a) \to F(b)$ extends to $L \to L$ because automorphism group of splitting field acts transitively on roots. Restrict to $K$ to get $K \to L$.

Leaky Nun

cool

what if I want the field not to depend on $f$

(so it would have to be bigger)

Balarka Sen

Oof. Let me think.

Leaky Nun

5:12 AM

4 mins ago, by Balarka Sen

Splitting field of $f$ over $K$, I mean. If that's $K$ I am done, otherwise it's some bigger extension $L/K$. $F(a) \to F(b)$ extends to $L \to L$ because automorphism group of splitting field acts transitively on roots. Restrict to $K$ to get $K \to L$.

wait I don't think this works

Balarka Sen

Why so

Leaky Nun

take $F = \Bbb Q$, $K = \Bbb Q(2^{1/8})$, $f = x^4-2$, $L = \Bbb Q(2^{1/8}, i)$, $a = 2^{1/4}$, $b = i2^{1/4}$

it's because $L/F$ isn't the splitting field of $f$

"automorphism group of splitting field acts transitively on roots" requires the polynomial to be irreducible

$f$ may not be irreducible when transferred to $K$

wait does it work for my example

Balarka Sen

Splitting field of $x^4 - 2$ over $\Bbb Q$ is $\Bbb Q(2^{1/4}, i)$, a strict subfield of $L$. I see the point.

Leaky Nun

maybe replace $2^{1/8}$ with $2^{1/1024}$ just to be sure

Balarka Sen

Ok but this shouldn't be a problem I think.

Leaky Nun

5:22 AM

I just need $\zeta_{16} \notin \Bbb Q(2^{1/16}, i)$

(this fails for $8$ because $\zeta_8 = (1+i)/\sqrt2$)

Balarka Sen

We have some irreducible $f \in F[x]$ with a root $a \in K$ and we want to show all roots of $f$ are in $K$. Suppose not. Then $f$ has a nontrivial splitting field $L$ over $K$. The splitting field of $f$ over $F$ is contained in $L$, let's call that $E$.

For any other root $b$ of $f$, there is an element $\sigma \in \text{Aut}(E/F)$ such that $\sigma(a) = b$. But how to get from this $\sigma : E \to E$ to an embedding of $K$ in a larger field, I guess.

Leaky Nun

why not take $F = \Bbb Q$, $K = \Bbb Q(2^{1/4})$, $f = x^2-2$, $L = K$, $E = \Bbb Q(\sqrt2)$, $a = 2^{1/2}$, $b = -2^{1/2}$.

then $E \to E$ cannot be extended to $K \to L$

Balarka Sen

Because $E \to E$ need not leave $K$ fixed pointwise

Or rather $K \cap E$.

Leaky Nun

right

Balarka Sen

That's annoying.

Leaky Nun

5:30 AM

c'est la vie

Balarka Sen

Oh. But every $F$-automorphism $L \to L$ (where $L$ is the splitting field of $f$ over $K$) must preserve $K$, because you can restrict that to $K$, and the image of the restriction will be $K$.

Hm, does that really accomplish anything.

I want to somehow say $\text{Aut}(L/F)$ still acts transitively on the roots of $f$ over $F$.

$\text{Aut}(E/F)$ is a subgroup, which acts transitively on the roots of $f$ over $F$. Does that not imply so?

It's not a subgroup, it's a quotient

Very annoying.

Leaky Nun

it's a subquotient

Balarka Sen

yeah lol

I am getting confused by the picture. Need to smoke and think

It's like you can go to a higher extension but with lesser symmetries, I guess

@LeakyNun Which was the point of this example

Leaky Nun

I guess

Balarka Sen

Thanks for your patience!

Leaky Nun

5:42 AM

what does this mean geometrically

np

Balarka Sen

Non-normal covers can cover normal covers, I suppose.

Leaky Nun

Galois theory <-> covering space?

Hawk

I got a definition question. If a polynomial is separable, meaning it has no multiple roots. Does that immediately imply it is irreducible since we know it is separable it must mean it is completely broken down to its linear factors right?

Balarka Sen

I have never actually deeply thought about normal coverings.

Leaky Nun

can we construct an explicit example?

@Hawk more context needed

Hawk

5:44 AM

@LeakyNun what is the missing info i need to supply?

Leaky Nun

under my interpretation, separable means "no multiple roots in an extension containing all the oots"

well I don't know how your source defines separable

Hawk

oh

Leaky Nun

and whether your polynomial is over any arbitrary field

Hawk

let me just grab dummit and foote

Leaky Nun

I suspect it is, since you were asking about fields previously

but in any case (x-1)(x-2) is separable but reducible

Hawk

5:45 AM

okay let p(x) be in F[x] (F is a field)

wait (x-1)(x-2) is reducible?

Leaky Nun

yeah

Hawk

what does it reduce to?

Leaky Nun

x-1 times x-2

@BalarkaSen do you have an example of a normal cover and a non-normal cover?

(abnormal cover?)

Balarka Sen

@LeakyNun Yeah this shouldn't be hard to cook up. Consider the subgroup $\Bbb Z * 0$ of $\Bbb Z * \Bbb Z$ and the corresponding covering space of the wedge of two circles.

Leaky Nun

cool

so ...-o-o-o-o-o-...

or rather, ..._O_O_O_O_O_...

Balarka Sen

5:48 AM

I don't think it corresponds to that

That would be the subgroup generated by a^n b a^{-n}

Leaky Nun

oh you're right

Balarka Sen

Z * 0 corresponds to just "opening up" one of the petals of the figure 8

It's like a loop attached to a tree

That has fundamental group Z indeed so it checks out

Leaky Nun

or rather consider the Cayley graph of F2 and then modulo translation one unit to the "right"

Balarka Sen

Yeah

It's two copies of Cayley(F_2) attached to a loop

That loop projecting to the a-circle of 8

Leaky Nun

what does normality mean?

Balarka Sen

5:54 AM

@LeakyNun The way to immediately tell if a cover of 8 is irregular is if it's not even a regular graph, yeah?

because the action of the subgroup on Cayley(F_2) has to be transitive

Leaky Nun

googles regular graph

hmm

Balarka Sen

Oh I mean if it's regular every vertex has to have equal valence

I think

The converse need not be true

Leaky Nun

4

A: Normal Covering of a Space

A covering is normal or regular or Galois if the group of deck transformations act transitively on the fibers. It's called "normal" because this is the case if and only if the fundamental group of the covering space is a normal subgroup of that if the base. See wikipedia for more details.