help-0

[warped kindle]

Not sure what is that?😂

[onyx swallow]

ok, then in this case the problem is a bad problem

[warped kindle]

Oh..no

Anyway thanks for helping me, appreciate it a lot.

[onyx swallow]

if you want

ill sent you a pic with all the solutions

[warped kindle]

Ooo sure. I think by then i would need to learn it

Ahh

There is another question i can't seem to do

[onyx swallow]

here you go

[warped kindle]

Wow. Omg thank you!! Ya i have not learnt it

[onyx swallow]

(p/q)^(r-1)=p^(r-1)/q^(r-1)

this is yet another bad problem as it gives lots of solution lol

best to switch book

[warped kindle]

Haha, all the teachers would say that as long as your final answer is correct. Anything is acceptable

So what is r?

[onyx swallow]

for example, if p=q, any r works

[onyx swallow]

try and simplify it first

[warped kindle]

I seee

[onyx swallow]

put p on one side and q on the other

[warped kindle]

Like this?

[onyx swallow]

yes, this is the left hand side of the equation, can you do the same for the right hand side?

[warped kindle]

Sorry, not so sure what you meant by right hand side of the equation.

[onyx swallow]

you have (p/q)^(r-1)=(q/p)^(r-4)

[warped kindle]

Ya

[onyx swallow]

you can simplify this equation to p^(r-1)/q^(r-1)=q^(r-4)/p^(r-4)

[warped kindle]

Yaaa

But then what do i do?

[onyx swallow]

you can use cross product!

if a/b=c/d then ad=bc

[warped kindle]

Can't seem to get the answer

Well never mind thrn, i'll just leave thid question out

Thanks for helping though

Highly appreciated

[onyx swallow]

p^(2r-5)=q^(2r-5)

2r-5=0

[safe tartan]

Can’t you just make one of the sides negative exponent

So 1-r=r-4 r= 5/2

[onyx swallow]

yeah

[lone heart]

@warped kindle Has your question been resolved?

[warped kindle]

Oof, i already got the answer but i thought was wrong.

So it was correct all this while, haha.

[lone heart]

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[warped kindle]

Anyway thanks

[lone heart]

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[bright parrot]

Can someone explain me what A should look like?

[vale crag]

darn that's a mouthful lol

anyway

wdym by "look like" ?

what are you looking for ?

@bright parrot

[bright parrot]

I can not really imagine what is in this shit

Normally the languages are so simple that you can understand what you could build from them but with this task I don't even know what kind of subwords I can make from A

{001} is maybe in there bec it is |w| mod 2 = 1 is true and

(w)n = 1 -> (w)n-1 = 0 is also true

but what dose the ∨ |w| = 0 is doing there

[vale crag]

it's |w|_0 = 1 mod 2

it's talking about the number of zeros in the word

[bright parrot]

oh

[vale crag]

here you got 2 zeros

[bright parrot]

ahhh so {10} and {0100} ... is in there

[vale crag]

and |w| = 0 is prolly just a weird edge case they need idk

[vale crag]

yes

[bright parrot]

okay thank you alot i think i can now work an the Task!

./close

[vale crag]

.close

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[radiant dragon]

can someone give me step by step guide on how to find the inverse of a 3x3 matrix

[wet wigeon]

ok

[wet wigeon]

i can

show me it

do you mean inverse function or just function

[mortal trellis]

do you know how to row reduce/gaussian elimination

[sharp panther]

There are two ways of doing that

1. Using row and column transformation and this step is very tedious so I prefer never to do it
2. Inverse of A = 1/|A| * adj(A)

[mortal trellis]

not like calculating 9 2x2 dets is that much nicer

and for 4x4 or bigger the adj(A) is even worse

[lone heart]

@radiant dragon Has your question been resolved?

[vale crag]

more like 65487587412682615 2x2 dets

[lone heart]

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[static burrow]

Hello i need help with trigonometry i don't know how and what stuff means and that trigonometric syrcle is wierd to me cold someone help me with it?

[alpine sable]

yes

Be more specific

[static burrow]

this

look i get the red and the top pi/180 and things

i dont understand the formulas on the botom

this sin^2a...

are those formulas

[tardy stag]

Most people (including me) don't really know how to prove that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

[ocean seal]

Hayley

[tardy stag]

that's one of those formulas that's easiest to memorize, or you're often given it on tests and things

[static burrow]

what is theta?

[tardy stag]

any angle

you could use whatever variable you wanted but theta is a standard "angle" variable name

[static burrow]

look these are some of the questions on my exams that i could have

ill send them in a sec

and for the question 13. i have next answers to check in1, 1, 1/3, 1/9, 2/9, 7/9

[tardy stag]

these are somewhat challenging problems! First of all, do you understand what sin/cos/tg actually mean? Does the fact that they're restricting the first problem to an interval make sense to you?

[static burrow]

i think they do i do understand somewhat what they are and for what you use them for

[tardy stag]

and would you be able to fill this out? https://media.discordapp.net/attachments/1064643750493691967/1119874751561744404/Screen_Shot_2023-06-18_at_2.22.14_AM.png
you don't have to do it for me right now but it's a good idea to make something like this

oh and do you understand that when we say $\sin^2 x$ it's just a simpler way to write $(\sin x)^2$

[ocean seal]

Hayley

[static burrow]

this i do yes

this math stuff is exosting ive been studding some stuff from the scratch and thankfully the exam is in monday

been looking at math for way to long for my likeing (45days) straight

[tardy stag]

okay great (there is one error, tan(pi) is +infinity but I usually just say undefined)

let's look at that second one

can you rewrite $\sin(2a) = \frac23$ using those double angle formulas?

[ocean seal]

Hayley

[tardy stag]

you're replying to something from like four hours ago

[static burrow]

ye

sorry

[static burrow]

but i dont get this

[static burrow]

my bad

[tardy stag]

hmm. let's remind ourselves of the double angle formulas

,tex .double angle

[ocean seal]

Hayley

[static burrow]

so this 0 with the - over it is alpha?

[tardy stag]

this symbol is called theta $\theta$

[ocean seal]

Hayley

[tardy stag]

this is alpha $\alpha$

[ocean seal]

Hayley

[static burrow]

so its cos^2 (example. 90) -1

[tardy stag]

uh, sure, if you had $cos(180)$ that would be equal to $2cos^2(90) - 1$

[ocean seal]

Hayley

[tardy stag]

and you can check that with your chart you just made

[static burrow]

this is my full question

[plain wave]

is that sin (2 (alpha))

?

[tardy stag]

okay, I see a sin(2a) on the left, how about we rewrite that using our formulas

[static burrow]

its sin 2*alpha

[tardy stag]

right

(I don't have alpha on my keyboard so I'm using a in its place)

[static burrow]

okay

i got it

so i would say the sin2*a is when rewrote 2 sinx * cosx

[tardy stag]

well, you'd use a not x

but yes

[static burrow]

ye

🎊

[tardy stag]

if you solve that for cos(a) you might be able to evaluate the thing on the right

[static burrow]

so i need to find cosx

[unreal phoenix]

Actually, you don't

[static burrow]

damn nice

could you explain to me how did you get rid of sin2a=2/3

and what did you do with it

[unreal phoenix]

Get rid of it from where?

[static burrow]

is the sin^2a+cos^2a in the () rewritten sin2a=2/3

i mean how did you rewrite sin2a=2/3

[unreal phoenix]

No. Sin^2 a + cos^2 a =1, right?

[tardy stag]

yeah that's the solution I had in mind but you need some practice to see it I think

[unreal phoenix]

sin2a=2sin a cos a

[static burrow]

you got it right

i was doing it the wrong way the whole time

[unreal phoenix]

Yeah ig. Initial practice for students is just calculating sin and cos explicitly, but looking at the degree of exponent, ig it wasn't supposed to be solved by finding the trig ratios explicitly.

[static burrow]

oooo

i see

im not the best at reading mathematic problems with text our teachers always just gave us the problem with nothing else and we had to solve it

[tardy stag]

these problems are pretty challenging especially if you've just started learning about double angle formulas

[static burrow]

yeah somewhat i just started

but

yall cleared it up for me a lot

thanks for this @tardy stag and @unreal phoenix meant a lot

its hard on me from all the pressure from the university exam

i had to learn like 60 different matematic problems, still there is some left but this one is the worst,

,close

how do i do it

.close

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[willow brook]

I thought that $p(n) = (1-p)^2 p^n$

[ocean seal]

jobo6

[willow brook]

Where does the (n+1) come from

[lone heart]

@willow brook Has your question been resolved?

[vale wigeon]

@willow brook think about what X=n means

X=n means that, up until you had your second failure, you had exactly n successes

you also had 1 failure to make the 2nd failure the 2nd

1 failure and n successes -- there are (n+1) ways for these outcomes to be arranged in a row

[willow brook]

Wouldnt it be $(n+1)!$ than for the permutations

[ocean seal]

jobo6

[vale wigeon]

no

$\frac{(n+1)!}{n! \cdot 1!}$ perhaps

[ocean seal]

Ann

[willow brook]

ohh I got it

thanks

.close

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[patent swallow]

https://cdn.discordapp.com/attachments/326138772477575180/1124786428350574654/rn_image_picker_lib_temp_8ac09193-8a98-406c-8982-13c87c14fe18.jpg

[patent swallow]

didnt work

can someone help me with the R part? idk what I'm doing rly

basically I've this function

I need to develop Taylor of order 2 in x = 8

and prove that with this taylor I can estimate 2ln(1,3) with an error smaller than 0.02

[lone heart]

@patent swallow Has your question been resolved?

[lone heart]

@patent swallow Has your question been resolved?

[lone heart]

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[dawn quail]

why are brackets required on third line?

[dawn quail]

I thought addition was commutative

[unreal phoenix]

It still is

[tacit arch]

-(2+3) = ?

[dawn quail]

oh, it should really be -ln2 -lny^2 if we remove the brackets?

distribute the -

got it

that's why brackets are not needed for first part

only second

interesting..

thank you!

.close

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[rotund crater]

someone can explain me this

[mortal trellis]

explain what about it

graphically, your function increases or decreases with a few saddle points. so it passes the horizontal line test and is injective

and if you then restrict the codomain to the image you are automatically surjective, so you have an inverse

[lone heart]

@rotund crater Has your question been resolved?

[rotund crater]

i dont understand

[trim heron]

Start from this: if a function is increasing (or decrasing) then it is a one-to-one function.

Then the inverse of it is also defined.

like f(x) = 4x

[rotund crater]

yes i know but why is f"(x) >= 0 why has equal or f'(c)=0

[trim heron]

this was not included in your screenshot

what do you mean?

[rotund crater]

this part

[trim heron]

There is no f''(x), that was what I mean

[rotund crater]

i accedently put double comma

[trim heron]

consider $f(x)=x^3$ so f'(x)>0 when $x \neq 0$

[ocean seal]

ertansinansahin

[rotund crater]

Sooo if stricly positive must be f'(x)>0 no f'(x) >=0

[trim heron]

then it is still increasing function but it will just stop for a while at x=0 just for one point

[trim heron]

so the fuction is increasing

which means it is one-to-one

[rotund crater]

and has inverse

[trim heron]

consequently it will be invertible function, yes

[rotund crater]

soo thex for the help

[trim heron]

bu if it stays here for more than one point, then it is not invertible

[rotund crater]

wdym?

[trim heron]

I just wanted to explain why "infinitely many point" is stressed.

[trim heron]

here

this is also an increasing fuction's graph

but in the yellow region there are infinitely many points with zero derivatives.

[rotund crater]

thats points are derivates?\

and critical points?

[trim heron]

these points will yield zero for the derivatives since the slope is zero there

[lone heart]

@rotund crater Has your question been resolved?

[rotund crater]

thx for the help

.close

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[tacit jewel]

$$ {R} \ {0, 1} $$
Does this mean all reals excluding 0 and 1

[ocean seal]

sweet.pea

ZanarQ

[tacit jewel]

Yes sorry dont know how to properly type stuff

[blissful temple]

Yes

[tacit jewel]

thank you 😄

.close

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[chrome sorrel]

For #1.

I graphed the regulra function 4/(x^2+1). because i dont think its possible to graph integrals on graphing calcs right?

[blissful temple]

You only need to graph the integrand - that's what's inside the integral

[chrome sorrel]

Yes I graphed that

But like how does that tell me if the integral is positive/negative, or zero

[blissful temple]

Good

Well, what does the integral represent in the graph?

[chrome sorrel]

area under curve

[blissful temple]

Right

[chrome sorrel]

ah and they're saying from 0 to pi

so its positive

there

[blissful temple]

[chrome sorrel]

Ty

.close

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[median oar]

Is there an intuitive understanding for the covariance metric of a vector parameter estimate?

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[sand pawn]

dude u must be insane. Asking a advanced math question in such help channels is just ridiculus.

You should better ask in advanced section

[lone heart]

@median oar Has your question been resolved?

[median oar]

Perhaps a different perspective on the question is what is the use for the variance of a vector parameter estimate?

I think the covariance here is the MSE of an MVU estimate

If that is so, it answers the question what the covariance means, but then it raises the question of what the variance means

[lone heart]

@median oar Has your question been resolved?

[warm raven]

helo

[mortal trellis]

random message

[lone heart]

@median oar Has your question been resolved?

[subtle birch]

Meth

[raven haven]

poor frosst

wait why don't you just ask in #probability-statistics

[lime kindle]

isn't there is anyother method to solve this without L'hopital rule?

[jolly widget]

use a separate channel

[lime kindle]

which one ma'am?

[echo socket]

Any available help channel below #❓how-to-get-help

(Currently the ones with numbers 6, 7, 11, 13 and 14 are open)

[lime kindle]

pretty compilacted nvm

[echo socket]

Just read #❓how-to-get-help if you haven't yet

[lone heart]

@median oar Has your question been resolved?

[sinful star]

yo

[median oar]

How do I differentiate with respect to a vector

Is that even a thing

Look at the regularity condition it says that the expectation of derivative with respect to θ is 0

But θ here is a vector

[solemn juniper]

given that it appends "for all theta" I wonder if it's just sketchy notation and saying for every element in theta

[keen plinth]

grad = 0

[median oar]

Hmm I don’t really get the regularity condition

That line is saying that the expectation of the gradient of my log likelihood function is 0

So does that mean it’s a stationary point?

[keen plinth]

yes

the expectation log-likelihood function as a function of theta should have a stationary point at the true value of the parameter

[median oar]

Ohhh

Right

[keen plinth]

this is just to ensure that the MLE comes from a stationary point of the log-likelihood function i guess

rather than some boundary point

that notation C_theta - I^-1 is really cursed

since they're both matrices

[median oar]

Is this better

It’s using >= to represent that last line on the LA recap on the bottom right

[keen plinth]

it's meant to mean pos semi definite

[median oar]

,rotate

,rotate

[median oar]

That right side looks kinda weird

Perhaps the pdf part isn’t correct?

My friend said that the goal of the condition is to be able to interchange the integral from the expectation operator and the derivative from the gradient operator

[keen plinth]

yes that is the point of the condition

otherwise you don't really have a good way of calculating the derivatives

[median oar]

Could you elaborate on what exactly we want to find the derivative of, maybe I can see why we wouldn’t have a good way of doing it

[keen plinth]

uh

let's see

[ \nabla_\theta \E_\vartheta l(X, \theta) = \E_\vartheta \nabla_\theta l(X, \theta) ]

[keen plinth]

this i suppose

we want to be able to interchange the derivative and the integral

[median oar]

The subscript on expectation is for the likelihood function?

[keen plinth]

the subscript is for the X

i'm saying i'm taking the expectation with X distributed wrt the distribution using the true parameter value vartheta

[median oar]

So the expectation of X, when our pdf of X has θ = true parameter value

[keen plinth]

yes

[median oar]

Wait

So the expectation gives us a value of the sample space

Not parameter space?

[keen plinth]

sample space?

[median oar]

Like possible realisations of the distribution

[keen plinth]

the expectation is of a real function of X

you take the expectation of X plugged into the log likelihood for the density associated with parameter theta

and treat that whole thing as a function of theta

but the expectation is for X

X is the RV

[median oar]

Right there’s 2 pdfs here

1 for X and 1 for θ

[keen plinth]

no theta is a constant

[median oar]

Isn’t the density associated with parameter θ a pdf

[keen plinth]

theta parametrises a family a distributions

each distribution has a density so theta parametrises those as well

but theta is not a random variable

[analog gull]

In theory the theta could be random by law

[median oar]

I thought it would be in the sense that for some given realisation of the data, what is θ

[keen plinth]

this is not how it is modelled mathematically

[median oar]

Wait no that means θ is deterministic

[analog gull]

Correct

[keen plinth]

the true parameter value is a constant

[median oar]

Since it would otherwise make no sense to estimate it

[analog gull]

Perhaps a different perspective on the question is what your looking for

[median oar]

Ok let’s back up then, we want to do this

[analog gull]

Let's do this!!!

[median oar]

Ok let me write on paper

[analog gull]

Same with me

[median oar]

Ok this is this goal right?

[keen plinth]

uh no

[median oar]

I’m still not exactly sure why we want that condition to be met, the swapping of expectation and gradient but let’s run with it for now

[analog gull]

Okay I think the we should initialize θ, and gather demonstrations x then solve for optimal policy for pi(a|s) w.r.t cθ with value iteration then solve for state visitasion frequencies p(s | θ,T)

[median oar]

What’s pi(a|s)

[analog gull]

and then compute gradient of the equation on what we have so far and update θ with one gradient step of using nabla θ vartheta

[median oar]

And what’s cθ

[analog gull]

okay well i see that you are not as smart as me

[median oar]

Nope

I’m a first year uni student

[analog gull]

I have multiple degrees in Advanced Calculus and Advanced Physics

[median oar]

Ok if this isn’t the goal what is the goal

Isn’t this the point of checking the regularity condition?

[analog gull]

I mean it could be just takes some time

[median oar]

Because if it satisfies the condition then something cool happens (the ability swap the operators)

[keen plinth]

the first equation is still kinda sus, i'm pretty sure the usual assumption is that you can interchange derivatives and integrals

[analog gull]

Okay um mods

[median oar]

Well both these videos seem to suggest it as the condition

[analog gull]

okay well who is this guy first off

[median oar]

Daniel Romero

[keen plinth]

the notation is also sus

[median oar]

Ok what’s a better notation then

[keen plinth]

i can't tell what they're saying with the for all theta

[opal relic]

guys I'm a little bit lost here

[keen plinth]

because if it holds that the gradient is 0 for all theta your function should be constant

[analog gull]

True

So true

[median oar]

Let me see if he says anything specific about it

[rose sigil]

that sounds like an amazing idea to me

[analog gull]

Thank youu

[opal relic]

wait seriously?

can you explain limp bizkit

[rose sigil]

yea i wanna hear it explained too

[median oar]

Nothing in the scalar case either

[analog gull]

another day I would love to

[rose sigil]

i wasn't smart enough to understand it all but it sounds so interesting

[median oar]

But I see your point snow

If the likelihood had 0 gradient everywhere then it’s just a flat line

[keen plinth]

my notes on cramer rao are not very comprehensive unfortunately

[median oar]

Ok let’s back up even further then

Back to Wikipedia

Some condition allows for the top and bottom to be the same

Surely that statement is correct

[keen plinth]

yeah im pretty sure thats a consequence of being able to interchange derivatives and integrals

[median oar]

No I don’t really see it

Maybe I’m just a bit weak in this level of calculus

[keen plinth]

uh no its a slightly non trivial calculation

[median oar]

I don’t see how the derivatives combine and where the - comes from

[keen plinth]

let me see if i can find it in my notes

[median oar]

[keen plinth]

the univariate case is relatively painless to show

multivariate will probably get a bit awful

[median oar]

Well it seems I don’t get the univariate case either so that’s probably a good place to start

[keen plinth]

okay so

[keen plinth]

this notation is slightly terrible

lets use a slightly different notation

[median oar]

Sure

[keen plinth]

so we'll have a random vector $X$ which will represent our sample, and the underlying distribution of $X$ will be $X \sim \P_\theta$, where i'm using $\P_\theta$ to denote the probability measure, and $\theta \in \Theta$ is the parameter which varies over the parameter space $\Theta$

[keen plinth]

then we can call the true parameter value $\vartheta$

[keen plinth]

so when i write something like $\P_\theta(X \in B)$ i mean $X \sim \P_\theta$, the distribution picked out by $\theta \in \Theta$, and $\E_\theta g(X)$ will be the expectation taken wrt to this distribution

[median oar]

Expectation of g(X) on the bottom here?

[keen plinth]

yes

just some random function g

so here we can assume $X$ has a density $f_\theta$, so the log-likelihood function will be
[
l(x; \theta) = \log f_\theta(x)
]

[median oar]

And P_θ here means a distribution with parameter equal to θ, which is some element of all possible parameters

[keen plinth]

yes

so what we can do is

we just calculate

[median oar]

What’s B

[keen plinth]

a set

[median oar]

Any set?

[keen plinth]

a measureable set lol

[median oar]

The sample space?

[keen plinth]

yes

like

[median oar]

Should we use omega

[keen plinth]

[
\P_\theta(a \le X < b) = \P_\theta(X \in \coi ab)
]

[keen plinth]

for example

its just the probability function

[median oar]

So a subset of omega

[keen plinth]

yes

so we want to show that
[
I(\vartheta) = -\dnv 2 \theta\biggr\rvert_\vartheta \E_\vartheta l(X; \theta) = \E_\vartheta \biggl(\dv\theta\biggr\rvert_\vartheta , l(X; \theta)\biggr)^2
]

[keen plinth]

notation gets kinda confusing

but the equality only holds when $\theta = \vartheta$, so when the parameter is set to the true value

[median oar]

How do we know it only holds?

Also what does the \middle|_\vartheta mean

[keen plinth]

oh ig not only but

its a => implication

otherwise we dont really know in general

[median oar]

We want to find some condition that implies the equality?

[keen plinth]

well not really some condition

that is the condition

$\theta = \vartheta$

[keen plinth]

i've just given it to you

i'm using the notation
[
\dv x\biggr\vert_a f(x) = f'(a)
]
so evaluating the derivative of $f$ at $a$

theres kinda

no good way to write this lol

[median oar]

What’s the p got to do with a

[median oar]

Oh lmao

[keen plinth]

me being a dum dum switching notation halfway through typing

okay so

if we just assume these conditions

we should be able to evaluate these things

[median oar]

So I’ve got this

[keen plinth]

yeah sure ig

the top part was mostly just establishing notation

[median oar]

Right good

[keen plinth]

so basically we just expand out the integrals

assume you can interchange derivatives and integrals

and then it should work

[keen plinth]

so theses are the regularity assumptions you'll find on wikipedia i think

maybe

if it isnt well at least itll give you an idea of how it works lol

[median oar]

Ok I’ll have a go at it

[keen plinth]

\begin{align*}
  \MoveEqLeft
  \dnv2\theta \E_\vartheta l(X; \theta) \\
  & = \dnv2\theta \int (\log f_\theta(x)) \, f_\vartheta(x) \, dx \\
  & \overset{!!}= \int \pdnv2\theta \Bigl(\log f_\theta(x)\Bigr) \, f_\vartheta(x) \, dx \\
  & = \int \biggl(\f {f_\theta''(x)} {f_\theta(x)} - \Bigl(\f {f_\theta'(x)} {f_\theta(x)}\Bigr)^{\!2}\biggr) \, f_\vartheta(x) \, dx
\end{align*}

[keen plinth]

so the !! is where you need to assume you can bring the derivative inside

[median oar]

,rcw

,rcw

[median oar]

What happens to the limits

[keen plinth]

which limits?

[median oar]

Cos now that function isn’t a function of θ anymore it’s a function of some constant \vartheta

[keen plinth]

yeah

we're integrating wrt x

the expectation is of a function of X

[median oar]

The 2nd time I do the derivative

Wrt θ

[keen plinth]

you have to take the derivative wrt theta

so the notation gets really really messy

but like

[median oar]

And then evaluate it at vartheta no?

[keen plinth]

yeah

[median oar]

From this

[keen plinth]

yes

[median oar]

So if you look at the last line

Now it’s f_\vartheta

[keen plinth]

but here we're gonna use [ f_\theta'(x) = \dv\theta , f_\theta(x) ] rather than [ f_\theta'(x) = \dv x , f_\theta(x) ]

[median oar]

And I need to differentiate it again with respect to θ

Yeah but where does vartheta go

[median oar]

Does the evaluate only happens after 2 derivatives

[keen plinth]

yes

[median oar]

Oh

That’s where I went wrong

I shouldn’t evaluate it yet

It’s the evaluation of the 2nd derivative

[keen plinth]

what i meant by the second derivative was you do the derivatives first

yeah

notation is bad

but i dont know of another way to write it that isnt worse

[median oar]

,rcw

Ok yes I got there

[median oar]

Missing - signs at the front

[keen plinth]

yeah alright

so we can just like

take that

\begin{align*}
  \dnv2\theta\biggr|_\vartheta \E_\vartheta l(X; \theta)
  & = \int \biggl(\f {f_\vartheta ''(x)} {f_\vartheta (x)} - \Bigl(\f {f_\vartheta '(x)} {f_\vartheta (x)}\Bigr)^{\!2}\biggr) \, f_\vartheta(x) \, dx \\
  & = \int f_\vartheta''(x) \, dx - \int \Bigl(\f {f_\vartheta '(x)} {f_\vartheta (x)}\Bigr)^{\!2} \, f_\vartheta(x) \, dx \\
  & = \int f_\vartheta''(x) \, dx - \int \biggl(\dv\theta\biggr\rvert_\vartheta \log f_\theta(x)\biggr)^{\!2} \, f_\vartheta(x) \, dx \\
  & = \int f_\vartheta''(x) \, dx - \E_\vartheta \biggl(\dv\theta\biggr\rvert_\vartheta l(X; \theta)\biggr)^{\!2}
\end{align*}

[median oar]

My steps were the right ones

Ok I am not crazy

[keen plinth]

yep

[
\int f_\vartheta''(x) , dx \overset?= 0
]

[keen plinth]

so this is the question we need to answer

[median oar]

So that’s the regularity condition?

If it’s true then the equality holds

[keen plinth]

what we're assuming is the ability to interchange derivatives and integrals

[median oar]

It’ll mean we can swap the order of operators?

[keen plinth]

so this is actually just a result of f being a density

[
\int \pdnv2\theta , f_\theta(x) , dx = \dnv2\theta \int f_\theta(x) , dx = \dnv2\theta , 1 = 0
]

[median oar]

,rcw

[median oar]

So we have this then

[keen plinth]

ye

[median oar]

So now the problem has moved to why can we swap the operators

Or how do we know we can swap it

[keen plinth]

the answer to that is we cant in general

we have to assume that the density is nice enough

[median oar]

What do you mean by nice enough

[keen plinth]

the derivative is defined through a limit

[median oar]

So we have that x and θ are unrelated?

[keen plinth]

so you basically need to know that you can interchange limits and integrals

and there are a whole bunch of theorems which tell you when you can

[median oar]

The integral is also a limit

And sum

[keen plinth]

well

[median oar]

Since it’s definite no?

[keen plinth]

kinda

in probability all the integrals are lebesgue integrals

[median oar]

Great, that sounds like a detour for another time

[keen plinth]

lol

[median oar]

Ok I will let it be for now I guess

[keen plinth]

yeah

but this is like

part of interpreting what the fisher information actually means

[ \dnv2\theta \E_\vartheta l(X; \theta) ]

[keen plinth]

this quantity is the second derivative of like

your MLE essentially

when you evaluate at the true parameter vartheta

you get the curvature of the curve

[median oar]

And we want that to be 0

No

We want it to be negative

[keen plinth]

second derivative is curvature

[median oar]

The more negative the better

[keen plinth]

yes

[median oar]

Since it’ll mean it’s more “correct”

[keen plinth]

that corresponds to having a sharp point

[median oar]

Hence we want with the - at the front

And the inverse of the fisher information to be big

[keen plinth]

when you graph the function $\E_\vartheta l(X; \theta)$ as a function of $\theta$ and the curvature at $\theta = \vartheta$ is very large

[keen plinth]

then the graph will be a sharp point

[median oar]

And that’s a good estimate

[keen plinth]

what that means is when you're estimating

like

[median oar]

It’ll use a lot of the information

[keen plinth]

you won't hover around a flat bit of the graph essentially

[median oar]

The more the merrier

[keen plinth]

because like say you're doing an optimisation problem

[median oar]

As in I won’t get a big interval of same ish possible values?

[keen plinth]

yeah

like

if there was a large flat region

near your max

then there would be a whole bunch of values of theta that are all "pretty" good

but they might be quite far from vartheta just because you have a large flat region

[median oar]

Right right

[keen plinth]

but if you have a sharp pointy region

theres gonna only be a small patch of theta that are good

[median oar]

Ok so I just need to understand this in higher dimensions

It sounds like I need multivariate calculus

[keen plinth]

in higher dimensions you have to look at the hessian

[median oar]

Those funny upsidedown triangles

With a 2 on top

[keen plinth]

uh

maybe

[median oar]

Isn’t \nabla^2 the hessian

[keen plinth]

no thats laplacian

[median oar]

Oh

[keen plinth]

its the trace of the hessian

which does tell you stuff

[median oar]

I saw something about the trace

Something about the variance is the trace of the covariance matrix

So maybe the covariance of the estimate vector is the hessian

(Now I am just saying words that I don’t understand)

[keen plinth]

its the one with the mixed 2nd derivatives

hessian should be neg semidefinite or something at the true parameter value

for the log likelihood function

so then if you take its trace you get sum of eigenvalues, which will all be negative

and that sum can tell you about how sharp that max is

[median oar]

Something eigenvalues something I haven’t really delved into yet

Which I’m thinking should be the next topic I look at

[keen plinth]

[median oar]

Look there’s only so much I can delve into with 24 hours in a day and I only started really going at it a few months ago

[keen plinth]

this is true

[rose sigil]

snow has like 60 hours every day to devote to math

[keen plinth]

this is false

[median oar]

ok LA time

looking at "similarity of matrices" here

so for context, e1, e2, e3 are the standard bases of R^3, and f1, f2, f3 are on top, them being (1, 1, 0), (1, 0, 1) and (0, 1, 1)

Does it make sense to write this?

The e’s here are basis vectors themselves…

[acoustic wharf]

what's f tho

is it a matrix whose columns are f1, f2 and f3?

[median oar]

f is a vector

[acoustic wharf]

just any vector in R^3?

[median oar]

hmm

yes

f is a vector of R^3 written as a linear combination of f1, f2, f3 which are linearly independent and form a basis of R^3

in this specific case it is (1,1,0), (1,0,1) and (0,1,1)

[acoustic wharf]

is I supposed to be the identity matrix?

[median oar]

yes

[acoustic wharf]

bruh then that last equality doesn't make any sense

because to the right you've got a constant vector, whilst to the left you've got f

[median oar]

hmm

[acoustic wharf]

can you send the video?

[median oar]

well the goal here is to "define" P

https://youtu.be/MJic8o5ph5M?list=PLW3u28VuDAHJNrf3JCgT0GG_rjFVz0-j9&t=2467

[vale crag]

yeah why are you multiplying by (e_1, e_2, e_3) on your pic ?

your input vector is in basis F

[median oar]

ok

[vale crag]

(looking at the I(f) = ... part)

[median oar]

then where does P come from

surely it must be because i have a system of linear equations

that i can form a matrix out of

and here I(f_1), I(f_2), I(f_3) are equations

and it would give me the right numbers for P

but im not too sure what's happening with the f and the e

[vale crag]

[I]_F^E is a change of basis matrix

i.e. it's the identity transformation I, but it takes a vector in basis F, and outputs its representation in basis E

if you remember, last time we talked about the [v]_B notation, for representation of vector v in basis B

[vale crag]

if we write that in equation we get, [v]_E = [I]_F^E [v]_F

just trying to explain the [I]_F^E notation a bit

you alright ? @median oar

[median oar]

ngl i nearly fell asleep

[vale crag]

o well

[median oar]

does "same vector" mean if i translated both to the same basis i get the same vector

[keen plinth]

its 4 in the morning

[median oar]

that's the point right?

[vale crag]

well it's supposed to be the identity transformation yeah

so you put them back in the same basis

you get the same vector yes

[median oar]

ok so [v]_E = [I]_F^E [v]_F

and we want to know what [I]_F^E is

well i know $[v]_E = a_1e_1 + a_2e_2 + a_3e_3$

[ocean seal]

frosst

[median oar]

then $[v]_F=b_1f_1+b_2f_2+b_3f_3$

[ocean seal]

frosst

[median oar]

hmm, that's not very helpful

[vale crag]

if you're able to write the e_i's in terms of the f_i you're pretty much done (1st pic)

[median oar]

let's instead have $[I]_F^E(f_1)=(x_1, y_1, z_1)^T$

[ocean seal]

frosst

[vale crag]

since you'll have something of the form [v]_E = (...)f_1 + (...)f_2 + (...)f_3

and you can construct the [I]_F^E matrix from there

[median oar]

then we get
$[I]_F^E(f_1)=(x_1, y_1, z_1)^T\$
$\left[I\right]_F^E(f_2)=(x_2, y_2, z_2)^T\$
$\left[I\right]_F^E(f_3)=(x_3, y_3, z_3)^T$

[ocean seal]

frosst

[median oar]

ok anyway

[vale crag]

well actually no your notation is a bit improper

[v]_E = (a_1, a_2, a_3)^T means that v itself = a_1e_1 + a_2e_2 + a_3e_3

[v]_E is just a coordinate representation of v

I agree it's easy to confuse them if you're working with R^n

[median oar]

ah

in the video it says we are only working in R^n

but i suppose you can extend this to any finite dimensional vector space

[vale crag]

yeah

[median oar]

hmm how is that wrong

[median oar]

i wrote this to represent that some vector v, in the basis of E, is a linear combination of the basis vectors

[vale crag]

yeah but [v]_E is not the lin combo itself

it's just the vector of coefficients

[median oar]

ahhh

[vale crag]

if you think V = polynomials

[median oar]

that's a bit comfusing

[vale crag]

those are two different types of objects

v would be a polynomial

[median oar]

so [v]_E and v are not the same

[vale crag]

[v]_B would live in some R^n

yes

[median oar]

[v]_E is just the coefficients

[vale crag]

yup exactly

if you don't differentiate those, you're bound to confuse shit

and not be able to go where you want to go

[median oar]

right ok

so if $v = a_1e_1 + a_2e_2 + a_3e_3$ then $[v]_E=(a_1, a_2, a_3)^T$

[ocean seal]

frosst

[vale crag]

yup

[median oar]

ok

[vale crag]

so now our goal is to write v = (...)f_1 + (...)f_2 + (...)f_3

like that we can get [v]_F

and we'll be able to link the two coordinate vectors [v]_E and [v]_F

and find [I]_F^E

[median oar]

so then consider a vector $v$, where $v = b_1f_1 + b_2f_2 + b_3f_3$, then $[v]_F = (b_1, b_2, b_3)^T$

[ocean seal]

frosst

[median oar]

which is the same as with E, but with F now

now we need to find out what the b's are

[vale crag]

yup

[median oar]

so the idea is to consider a linear transformation from v to v

ie. the identity map

[vale crag]

and the b's will depend on the a's we have (for basis E)

[median oar]

so the identity is going to send b_1f_1 to some LC of e_1, e_2, e_3's

[vale crag]

yeah

[median oar]

and same for b_2f_2 and b_3f_3

[vale crag]

exactly

[median oar]

hmm

so what would we denote the identity with

would it be [I]_F^E

noting that it sends a vector from F to the same vector but in E

[vale crag]

the identity map itself is just I

[I]_F^E is the matrix representation of I

for input basis F and output basis E

[median oar]

yeah im kinda lost there lol

[vale crag]

it's a bit like the v vs [v]_B

given a choice of basis, you can map a vector (which might not even be from R^n) to R^n

now for I vs [I]_F^E

you have a linear map (let's say V->V)

given a choice of 2 bases (one for input, one for output), you can represent this linear map as a matrix

(when you interpret the input vectors and output vectors correctly, i.e. with the correct bases)

and this representation is not unique

still depends on the choice of bases (like for [v]_B)

@median oar

[median oar]

wdym it's not unique

given 2 basis

how would the matrix representation not be unique

[vale crag]

^

ofc if you give 2 bases it's unique

[median oar]

oh

yeah right

for every 2 basis vector set, there's a unique matrix that maps vectors between the 2 basis

[vale crag]

yep

so yeah to come back to our problem

let's actually go about it the other way

i.e. start from [v]_F

(figured out it's a pain in the butt if you start from [v]_E, you're computing the inverse transformation instead)

[median oar]

ok this is what i got

so the b's are known because that's what [v]_F is

we just need to find out what x y z's are

and those corresponds to where the identitiy is sending the vector to, in terms of the other basis, E

so then this would make sense

[vale crag]

yep exactly

[median oar]

i've used P here but i guess it should be [I]_F^E?

[vale crag]

if you want to keep the notation yes

[median oar]

that's wack

instead of having the values of our vector in the argument of the matrix

we have instead the coefficients

and this is about what happens to the coefficients

[vale crag]

you've done this all the time

it's just that if you work in the standard basis for R^n, the two notions coincide

[median oar]

i suppose it actually never mattered

all the LA i've done could've been for a vector space of polynomials and i wouldn't have known

[vale crag]

it's relativity

but for vector spaces

[median oar]

all the x y z's could've been the coefficients of the variables

[vale crag]

everything depends on the basis you're looking at

[median oar]

right

so if i wanted to find I(f_1) = x_1e_1 + x_2e_2 + x_3e_3

do i just bang my head against the wall to get that

[vale crag]

well I(f_1) = f_1

so you want to write f_1 in the E basis

[median oar]

then i match each component and solve the system?

so i'd be solving 3 3x3 augmented matrices for the identity from F to E

like suppose f_1 was (1, 3, -2)

and i had e_1 = (-2, 0 3), e_2 = (1, 1, 0), e_3 = (4, 2, -1)

[vale crag]

yeah the coefficients of your system would be the same each time tho

those are your e_i

[median oar]

but i'd be solving for different f's

[vale crag]

yup

[median oar]

$\begin{pmatrix}
-2 & 1 & 4 & | & 1 \
0 & 1 & 2 & | & 3 \
3 & 0 & -1 & | & -2
\end{pmatrix}$

man i have no idea how to do augmented matrices

[vale crag]

yeah you'd have to do a table for the ccc|c I guess

,tex \left(\begin{tabular}{ccc|c}
-2 & 1 & 4 & 1 \
0 & 1 & 2 & 3 \
3 & 0 & -1 & -2
\end{tabular}\right)

fu

[median oar]

ok it's scuffed but we know what i mean

[vale crag]

wtf

[ocean seal]

frosst

[median oar]

dont let snow see it

[vale crag]

ah alright

[median oar]

ok that works too

[ocean seal]

_aplatypus
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vale crag]

nvm compile error ig

it works lol

[median oar]

ok

[median oar]

so let's say we have this set up

[median oar]

and this

then i need to pretty much solve for

,tex \left(\begin{tabular}{ccc|c}
-2 & 1 & 4 & 1 \
0 & 1 & 2 & 3 \
3 & 0 & -1 & -2
\end{tabular}\right)

[ocean seal]

frosst
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vale crag]

yeah

[median oar]

then if f_2 = (-2, 0, 3)

then solve

,tex \left(\begin{tabular}{ccc|c}
-2 & 1 & 4 & -2 \
0 & 1 & 2 & 0 \
3 & 0 & -1 & 3
\end{tabular}\right)

[ocean seal]

frosst
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[median oar]

and whatever the 3rd one is

[vale crag]

ya

[median oar]

and that's whatever's in my matrix [I]_F^E

[vale crag]

yeah the augmented col you get after RREF'ing (for example), goes in your [I]_F^E

[median oar]

right right

it's then obvious that [I]_F^E must be full rank

[vale crag]

since I has full rank yea

[median oar]

since if it weren't something must've been a linear combination of what we started

but we started with basis vectors

which are by definition linearly independent

so it's not possible, after a transformation to suddenly have "lost" dimensions

[median oar]

since this left side here also is full rank

[vale crag]

yea if your linear transformation has full rank, its matrix rep will also have full rank

[median oar]

well we're going from V to W where dim(V) = dim(W)

so i'd hope i dont lose vectors lol

i hope they dont just disappear into thin air

[vale crag]

for identity transformation sure

if your transformation is, project on some subspace

[median oar]

yeah for anything else it can

[vale crag]

I sure hope you lose vectors

[median oar]

right

it should be the same amount of loss too

[vale crag]

yea

[median oar]

like if i lost 2 dimensions using basis E i should as well in basis F

ok this makes more sense

although it's still a bit rough to wrap my head around tbh

[vale crag]

now for more general transformations it's almost the same thing

[median oar]

it being 5:30am might have something to do about that

[vale crag]

ah yeah

[vale crag]

anyway the technion guy will prolly talk about it soon after

[median oar]

well, for any other transformation that's not the identity

[vale crag]

go to sleep now

[median oar]

i can probably apply the transformation to the identity right?

like suppose i have A[v]_F

and [I]_F^E[v]_F = [v]_E

[vale crag]

the basic equation would be [Tv]_E = [T]_F^E [v]_F

i.e. the matrix of T in bases F and E takes the coordinate rep of v in basis F

[median oar]

then A[I]_F^E[v]_F = A[v]_E

[vale crag]

and sends it to Tv in basis E

[median oar]

this sounds right to me

anyway i'll probably go for now

[vale crag]

yea

g'night

[median oar]

thanks!

or morning 💀

[stark tiger]

hello can i get some help?

[tight locust]

no.

go away.

[upbeat hornet]

!help

[lone heart]

Please read #❓how-to-get-help

[lone heart]

@median oar Has your question been resolved?

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[wide trout]

.close

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[alpine sable]

Hi so the volume of a cylinder was 43.6577cm^3, and question was to round it off to 3 significant digits

[alpine sable]

my answer was 43.6 cm^3

because of the odd-even rule

but my physics teacher said taht its 43.7 cm^3

is he right in saying so?

if yes, then why?

[vale wigeon]

odd-even rule?

[alpine sable]

like where we have
numbers like 3.55 or 3.65, where you are to round off to say 2 significant digits. 3.55 rounds to 3.6 and 3.65 to 3.6

[cinder thicket]

You are right

Maybe they meant 4 significant rules and 43.66

[alpine sable]

nah it was 3

thank you

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[warped topaz]

https://i.imgur.com/fLGsc1p.png

[warped topaz]

I remember how to do these but I just forget the last bit
Like how do I decide whether its positive or negative infinity
I know its (-x + 3 - (1/x) - (3/x^2)) / ((7/x) + (5/x^2))

https://i.imgur.com/T5x6eK6.png

Which becomes (-x+3)/0 so its DNE
But I forget what to do next to figure out if its positive infinity or negative infinity

[tardy stag]

essentially is f(99999) positive or negative?

[warped topaz]

It will be DNE no?

(-x+3)/0 makes everything DNE

[tardy stag]

that's the expression "at infinity" but the original function very much has a value at 99999

[warped topaz]

Wym sry I dont understand

[tardy stag]

this thing here? if you put x = (whatever number) into that then a number will come out

[warped topaz]

Ok makes sense

[warped topaz]

But from what I remember doing these questions I need to do something like this no?

[tardy stag]

sure, yeah, that's one way to do it as well

[warped topaz]

https://i.imgur.com/jaMupY0.png

Or is it this?

[tardy stag]

everything that is over x can just be canceled to 0

[warped topaz]

(-x+3)/0 Thats how I got this

[tardy stag]

is the denominator positive 0 or negative 0?

as x -> +inf

[warped topaz]

Positive right?

[tardy stag]

correct since 7/x is positive and 5/x^2 is positive

[warped topaz]

Wait

I think I misunderstood the question and just got lucky haha

where do you see 7/x and 5/x^2

[tardy stag]

in the denominator

[warped topaz]

Ohhh yeyeye

[tardy stag]

so you can think of it as $\frac{\text{a really big negative number}}{\text{a really tiny positive number}}$

[ocean seal]

Hayley

[warped topaz]

How did you get that the numerator is a really big negative number

[tardy stag]

oh because it's (-x + 3) and as x -> +inf that becomes a pretty big negative number!

[warped topaz]

But its negative no?

[tardy stag]

yes, that's why I said "negative number"

like -139203238238

[warped topaz]

Oh hahaha oops

Im blind today

Bro now im confused

What would $\frac{\text{a really big negative number}}{\text{a really tiny positive number}}$ be

[ocean seal]

odietje

[warped topaz]

Positive or negative infinity

It would be negative right?

Yep I got the answer!!

Thank you

❤️

.close

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[warped topaz]

.reopen

[lone heart]

✅

[warped topaz]

https://i.imgur.com/ArONYuV.png

Apparently I dont get it hahaha