advanced-probability

[young hawk]

@remote rune pretty much any one lmao

[remote rune]

Ah the two examples i had in mind were unbiased lol, thanks

[young hawk]

@remote rune what about the uniform distribution

[remote rune]

Well if it is on [0, theta] the mom for theta is unbiased right since its like 2 times the mean

[young hawk]

@remote rune no I mean I’m (a,b)

On*

[remote rune]

Ah yeah why didnt i think of that lol thanks

[grave oasis]

If I have 2 covariance matrices A and B of n jointly gaussian distributions each. Does there exist some result where if the diagonal entries of A are greater or equal to the ones of B and the off diagonal entries of A are smaller in absolute value than the ones of B.
Then the max of the n gaussians represented by A is larger in expectation than the max of the other gaussians.

My kind of intuition for that is, the more independent the gaussians are, the larger is the max in expectation. I'm looking for such a result.

[cedar iris]

https://www.math.ucla.edu/~biskup/PIMS/PDFs/lecture5.pdf

try Sudakov--Fernique

@grave oasis

maybe some care with the signs of the covarian.ces

[grave oasis]

thanks a lot. This seems to be exactly what I was looking for. However, sth feels off with that result. I'm not sure where I'm going wrong with this, but if you look at the covariance matrix of 2 independent standard gaussians (identity matrix) and the covariance matrix of a standard gaussian and its negative (so ones on diagonal, -1 off diagonal), then the independent gaussians fulfil the role of X and the coupled ones as Y. The inequality should hold, however that would mean that the expected max of the independent gaussians is smaller, which shouldn't be the case should it?

[grave oasis]

ok maybe for the case I just stated it still works if I didn't mess up my calculations

so I guess my intuition is wrong for some cases, which are better captured by this inequality

[lament hawk]

Any good texts on semimartingale theory? Ive been using Eberlein and Kallsen, but im not sure if its the best text

[lament hawk]

Thank you

[tulip tree]

is there like a "brownian motion" for continuous functions? like, suppose $f_t \colon [0,1] \to \bR$ is a family of functions such that

• $f_0 \equiv 0$

• $\int (f_t - f_s) \sim N(0,t-s)$

• the $f_t$ are continuous functions, and increment independently

[livid sand]

Average J∘du=du∘j enjoyer

[tulip tree]

is this a thing?

[lament hawk]

Need more information on the integral (set you are integrating over, is it a stochastic integral, is it a Lebesgue integral, etc)

[white sundial]

Given that this question is a "is this a thing" question, I'm guessing it doesn't particularly matter.

But if I were to formalize this, I'd say this seems like you're asking about a random walk in an L1-space.

This is related: https://mathoverflow.net/questions/135654/brownian-motion-on-metric-spaces

[tulip tree]

my bad, I meant a lebesgue integral

probably f_{t-s} ~ f_t - f_s ?

that it's a continuous map? idk

maybe I should have prefaced this by saying I don't know what I'm talking about (which is probably obvious by now)

[tulip tree]

ty this is helpful

[lament hawk]

kind of related: the structure of martingales on a banach space is used to study the banach space's geometry

[white sundial]

oh, that sounds neat, do you have an example?

[lament hawk]

https://webusers.imj-prg.fr/~gilles.pisier/ihp-pisier.pdf

[white sundial]

thanks!

[pseudo schooner]

Question related to lemma 1.42 in Koralov & Sinai

[radiant cloud]

Hi this might be a naive question and I hope it is relevant to this channel as it falls in the domain of financial mathematics.
Why in most time series models like ARCH,GARCH etc we assume the stationarity (in weak sense) of the series? Like does it have any physical intuition?

For example in the Black Scholes Model we assume that the log returns of stock prices to be normally distributed where the stock price follows the Geometric Brownian Motion process. The log normality of returns in this case as I have read is due to the efficient market hypothesis and an application of central limit theorem to the returns.

I was wondering is there any similar line of reasoning for the assumption of stationarity of time series in ARCH/GARCH models.

[lament hawk]

stationarity has the implication that the parameters (mean, variance, etc) don't change over time

in essence it's statistical properties are constant

[radiant cloud]

thanks

[keen loom]

Hi, does anyone have a clue how to solve this?

[broken umbra]

Define the variation distance between two probability measures on say, a LCTVS by
$$\abs{\abs{\nu-\mu}} = \sup_{A}(\abs{\nu(A)-\mu(A)} + \abs{\nu(X-A) - \mu(X-A)})$$

[livid sand]

ShiN

[broken umbra]

Then two measures are mutually singular iff their variation distance is 2

can someone help me with this? I'm not sure how to approach this (haven't done any probability in a bit)

intuitively the statement is very clear. I'm specifically interested in distance 2 implying mutually singular

i'm not sure how to procure the required set from the supremum

[rugged tartan]

I have question regarding couses in my university

one is called 'stochastic processes'

the other is called 'advanced probability 1'

durrett's essentials of stochastic processes

https://www.amazon.com/Essentials-Stochastic-Processes-Springer-Statistics/dp/331945613X

the latter uses billingsley

but by looking at table of contents

I cant really seem to sea differences

[pliant seal]

But if I were to formalize this, I'd say this seems like you're asking about a random walk in an L1-space.

Beyond what the first answer on that MO post says (written by a guy who specializes in this stuff). BM on a Riemannian manifold is characterized by an extension of Levy's characterization to tangent spaces: $\forall f\in C^\infty:\mathrm{d}\langle f(W)\rangle_t=\lVert\nabla f(X)\rVert^2,\mathrm{d}t$.

[livid sand]

teafortwo

[lament hawk]

for $|\vert\nu-\mu\vert|=2\implies\nu\perp\mu$: under what circumstances does $|\vert\nu-\mu\vert|=2$? What does $|\nu(A)-\mu(A)|$ have to be? What does $|\nu(A^c)-\mu(A^c)|$ have to be? (given the measure of the whole space is one)

[livid sand]

VMM

[broken umbra]

It's a supremum though, You don't know that there's a single set that realises the supremum

That's my point

Obviously if you could find such a set that would ahow they're mutually singular

[lament hawk]

mb

hmm

[broken umbra]

My problem is the weight can bounce around as you approach the supremum

From measure to measure

[brittle plank]

Hi.

Is Feller's book on probability a good one? Has anyone here read it?

[keen loom]

It's definitely old

And doesn't cover the standard approach to probability iirc (with Lebesgue integral)

the first part is mostly about discrete probability iirc, and the second one is about continuous random variables, although like I said, the approach is a bit outdated, and in second part it's especially visible

nonetheless, because of how this book is written, it's still considered a worth experience to read it

at least the first volume

I didn't read a lot of it, but my opinion is based on other people opinions about it

[sharp patio]

For the setup that you've already done I think, take sets A_n approaching the sup of 2. Since both quantities |nu(A) - mu(A)| and |nu(X-A) - mu(X-A)| are bounded by 1, |nu(A_n) - mu(A_n)| and |nu(X-A_n) - mu(X - A_n)| both must approach 1. Now, max(nu(A_n), mu(A_n)) approaches 1 and min(nu(A_n), mu(A_n)) approaches 0 (I think this is what you're worried about in terms of the measure bouncing around).
I'm assuming you don't want the answer immediately so I'm going to give you 3 vague steps to fix this and put in spoilers the specific thing that you should be doing

1. Pass to a subsequence of A_ns ||so that the max is always attained by nu and the min is attained by mu, wlog||
2. Pass to a further subsequence of A_ns ||so that the sum of the mu(A_n)'s is finite||
3. Pick your set to be ||the limsup of the A_ns||. this works because ||the nu(A_ns) go to 1|| and ||borel cantelli says mu(limsup A_n) = 0||.

Sorry, I should have seen this 2 days ago

[broken umbra]

Thanks!

I already finished the talk

That this was a part o

Of

But I'll think about this

[sharp patio]

I see

[broken umbra]

Does anyone know if the cameron-martin space of a gaussian measure defined on a LCTVS on the sigma algebra making all continuous functionals measurable is always measurable?

The cameron-martin space being
The space of all elements $h$ where the supremum
$$sup{f(h): f\in X^{\ast}, , \operatorname{Cov}(f,f)\leq 1}$$
Is finite.
Where we are assuming functionals are $L^2$. Equivalently it is the space of all elements h such that the translation by h measure is equivalent to the original measure

[livid sand]

ShiN

[broken umbra]

Maybe someone has seen this in the context of cameron martin theorem

Real vector space

I think you might need to take completion

In general

[autumn seal]

hi guys

I have a question related to probability problem with Markov's chains.
I have a state matrix of first and second degree. If i know the state in which the object is the most probably in after 2 steps, how can I find the "starting state" the object started from?

[broken umbra]

What

[broken umbra]

The sup is the Cameron martin norm of a single element h

It's not the def of the space

The space is all those elements eith finite norm

So I don't understand what you mean

[young hawk]

Can anyone tell me if I did this right

[sweet raptor]

are densities always measurable maps?

[lament hawk]

the RN process is measurable

[sweet raptor]

what is the RN process?

[lament hawk]

radon nikodym

the density process

[sweet raptor]

ah ok

but the actual density need not be measurable?

wrt B(R) and whatever sigma algebra’s in the domain

[lament hawk]

The density is measurable

after all how do you define the lebesgue integral without measurability

[sweet raptor]

well ig that makes sense

thanks

[brittle plank]

@keen loom thank you

[keen loom]

You're welcome

[torpid owl]

Hello

I got this question

is my answer right?

Did you find any mistakes there? thank you

[lament hawk]

Is it true that $e^{X_{t}}$ is markovian if $X_{t}$ is levy?

[livid sand]

VMM

[shrewd raptor]

question

for part A is solving it that different that solving one with no drift

ignore part B i think ik how to do that

[shrewd raptor]

i mean do u have to go about it differently solving it with no drift then solving it with drift and is so, how

because solving it with drift is very easy

[lament hawk]

then set the drift to zero

[hard tundra]

hi, I understand why the conjugate prior of a gamma(a,b) [unknown b] is a gamma(a0 + na, b0 + sum(x)). but if I

*I'm told that b = 1/theta, how does that fit in?

[hard tundra]

nevermind, I think inverse gamma for the prior then

[grave oasis]

anyone got a good source on how strong the spectral norm of a symmetric matrix with i.i.d standard gaussian entries concentrates around its expectation.

Well I'm really just looking for user friendly upper bounds

I know there is this, if you give the diagonal entries a variance of 2 instead of 1. So I assume for my question this should be pretty similar apart from some constant factors somewhere

ah wait I think I can use this

sigma = 1 should work, edit: no it doesn't

[cedar iris]

Yes exactly since the norm is 1 lipschitz

[cedar iris]

although actually there is probably a sqrt(2) in there because of the symmetric requirement

[grave oasis]

edit: insert random rambling

but maybe I'm overlooking something

no, you're right, the norm irritated me nevermind

the sqrt(2) is needed

I somehow think of spectral norms every time I see two lines now, even though it's clearly the euclidean one here

[midnight spear]

Hi, this may or may not be an advanced question, but I am in an advanced math class just am struggling with a basic concept, if I have a piecewise cdf, and I want to try to generate a pdf from that cdf using the fundamental theorem of calculus, if y=x^2 do I need to consider any piecewise which is defined on x < 0?

[grave oasis]

generally you need to take the derivative everywhere yes, (well, everywhere where it is possible, it jumps at points where it is not differentiable) but y=x^2 is a bit of a weird choice since you have to cut it off at y=1 for it to become a cdf

"we immediately see", well I certainly don't
does anyone know why this is considered obvious

[sterile sail]

Maybe this is a dumb question, but what does it mean for Z = N(0,\sigma^2) here? I assume that is not a typo based on the pseudocode.

[keen loom]

Hello. If X is a stochastic process and T a stopping time, what does the notation $X_T$ mean? Is it maybe the random variable defined by $X_T(\omega) := X_{T(\omega)}(\omega)$?

[livid sand]

Gewisser Fler

[sterile sail]

That is how I would interpret that notation without any other context. Can't say I love the choice of T as a stopping time though

[cedar iris]

do you have more specific info on J?

[lament hawk]

The way I like to think of it is the limit as $t\to\infty$ of $X_{t\wedge{T}}$

[livid sand]

VMM

[grave oasis]

I actually found a more thorough explanation

[grave oasis]

No, this means that the w0 is a vector that was sampled from a gaussian distribution with independent entries and variance sigma

[sterile sail]

What are the probabilities associated with J?

[grave oasis]

this shouldn't be relevant according to lemma 2.3, I would like to post it, but that's like 2 pages of derivations

[sterile sail]

You said it was 1, -1 or 0? Does it require 2 pages of derivations to say J takes the value 1 with probability blah?

[grave oasis]

This is a more thorough explanation of how this algorithm is supposed to work

[grave oasis]

it's not gonna help you much, but I'm just gonna send it anyways

[sterile sail]

I can mostly see all the pieces with the additional context, but some bit of linear algebra is rusty enough that I can't tie it all together

By substitution you see that essentially $J_{x^{\prime}}^{\sigma^{\prime}} = J_{w_{i-1}}^{\sigma}$. It's not quite an equality, which is why you correct for it by moving the J units in the direction $v_i$. The choice of the $\sigma^{\prime}$ is exactly what you need so that in the exponential, there is a change of variables $j \mapsto j \lVert v_i \rVert$. I feel like there is some trivial linear algebra I have just forgotten to put it all together.

[livid sand]

Hexicle

[grave oasis]

@sterile sail this might make things more clear for you, just replace the sigma_t there by the J (I know inconvenient notation, but they are 2 different papers

[sterile sail]

I already gathered that much

I still don't quite see the finish line

[grave oasis]

I probably need a source on how gaussian vectors behave under projection, addition with dependence/independence etc. that should clear up a lot of the confusion

[jagged lynx]

does anyone know of a book or lecture notes on pde aimed at probabilists? i have no pde background and i want to learn spde, so im looking for a crash course on the minimal pde to get started with that

more specifically i want to read hairer’s rough paths book and his various lecture notes so that eventually i can study some of the more analysis-heavy papers on solving kpz equation and other important spdes.

i truly know no pde beyond a couple weeks in a diffeq class for engineers though haha

[lament hawk]

If you have a good background in real analysis use evans

[jagged lynx]

im looking for something a bit more succinct than evans, if it exists

i assume the entire book is not needed for what i’m aiming to learn?

but yes i have plenty of background in analysis, i just never learned pde

ah i never googled “pde for probabilists” until just now, but apparently there is this book https://www.cambridge.org/core/books/partial-differential-equations-for-probabilists/30BB576097CF5ECF7914FDBCAF69E1F2

has anyone here read it

[lament hawk]

I don't know anything about SPDEs, but iirc it is very technical and pde heavy, so you probably want to know PDEs very well (maybe everything in Evans is enough). Idk exactly though

[jagged lynx]

ok, but analysis-heavy does not imply pde-heavy

my impression that reading evans cover to cover is way overkill for getting started with the spde resources i mentioned, i just can’t really tell what can be blackboxed and what is too important to skip

does anyone here know about spde & can answer?

[keen loom]

Is there some way to obtain a probability distribution function from partial derivatives of the cumulative distribution function like the 1-dimensional case?

[south oxide]

@keen loom $f(x, y) = F_{xy}(x, y)$

[livid sand]

IlIIllIIIlllIIIIllll

[keen loom]

really? That's unexpected for me

[south oxide]

$F(x, y) = \int_{-\infty}^{x}\int_{-\infty}^{y}f(a, b),db,da$

[livid sand]

IlIIllIIIlllIIIIllll

[south oxide]

$\implies F_x(x, y) = \int_{-\infty}^{y}f(x, b),db$.

[livid sand]

IlIIllIIIlllIIIIllll

[south oxide]

this means that if $(X, Y)$ has a continuous density $f$, then $F_{xy} = f$.

[livid sand]

IlIIllIIIlllIIIIllll

[south oxide]

Conversely, if $F$ is $C^2$, then we can work backward and obtain that $F_{xy}$ is the density

[livid sand]

IlIIllIIIlllIIIIllll

[keen loom]

makes sense

[grave oasis]

The longer I look at this, the more I feel like there is a mistake in there, the x' should be divided once more by $||v_i||_2$, otherwise there is some variance problem

[livid sand]

Dr. J. Stockfish

[livid sand]

DvaNapasa

[keen loom]

Is there a theory of non-homogenous Markov chains?

People assume that they are homogenous, but it feels like what's really the interesting part is the non-homogenous ones

[stark chasm]

Check out reinforcement learning

As your policy improves, the transition probabilities of the environment will change over time

And Transition probabilities changing means non-homogenous markov chain

[midnight spear]

Can the binomial distribution approximate the guassian?

[midnight spear]

Can a uniform distribution have any bounds that we like?

[lament hawk]

the convolution of binomials is asymptotically gaussian

[lament hawk]

any bounds less than infinity/more than negative infinity

[lament hawk]

in fact one can prove that there is no 'uniform' finite countably additive measure on all of the reals

[midnight spear]

Easiest way to find the marginal cdf given a joint cdf?

[lament hawk]

I think #probability-statistics is a better place to ask

[stark chasm]

let the random-variable you don't care about be less than or equal to infinity and keep the one you care about less than whatever value you are focusing on

E.g. For joint CDF f(x, y) over random variables X and Y, marginal CDF of X is f(x, ∞)

[keen loom]

Im currently learning for my exam and i want to solve this exercise:

This exercise is from an older exam so it should be relatively easy to solve, however i dont know how

First off, we dont really know what the distribution of Z is (i looked it up, its called gamma distribution, which we dotn know yet), second of, how would the joint density of X and Z then look like? It sounds like a lot of work to compute it, however like i said, this exercise shouldnt take too long and esp no long calculations

I was thinking that maybe X has to be independent of Z, then things would be easier

However that doesnt seem to be true as well

What can i try here?

E[X|Z] = E[Y|Z] so E[X|Z] = (X+Y)/2

no need for any calculations

Ah yes that makes sense

If you compute E[X^2|Z] then you can similarly get E[XY|Z] = (X+Y)^2 /2 - (E[X^2 | Z] + E[Y^2 | Z])/2

[lament hawk]

Adding to Blitz's answer you can also easily obtain the conditional density of X given Z by setting it to (E(X|Z)f_X)/E(X)

[keen loom]

Ahh okay i now got E[XY|Z] = E[X^2|Z] = 1/4 * (X+Y)^2

Let me think about the density for a bit

Ahh yes now i understand

Thanks a lot guys

[keen loom]

Hey, i was thinking about the formula you gave, which is really useful since we dont need the joint density function at all, however i have never seen this formula before, intuitively it looks correct however im unable to prove that its correct, do you know a book where it is explained more?

I have a problem trying to prove that $$P(X_{k_n} = m_n | X_{k_{n-1}} = m_{n-1}, ..., X_{k_0} = m_0) =$$ $$ P(X_{k_n} = m_n | X_{k_{n-1}} = m_{n-1})$$ for a Markov chain $X_k$

[livid sand]

Blitz

[heady hornet]

this is the assumption of markov chain

[keen loom]

no

[stark chasm]

Intuitively it’s true

It’s saying that probability of next state depends only on previous state and not the history of states

Proof would be more challenging though I see what Kong is saying about it being the assumption

[stark chasm]

This is how markov chains are introduced in books

[heady hornet]

This thing is called as "Markov property", and one way to define Markov chain is "a sequence of
random variables that has the “Markov property”", see http://www.stat.yale.edu/~pollard/Courses/251.spring2013/Handouts/Chang-MarkovChains.pdf

[keen loom]

No. This is pretty much a corollary of exercise 1.5b of what you linked

You can call it Markov property, sure, but not even what you cite calls this Markov property

[stark chasm]

It does

[keen loom]

No, this is a different condition

anyway, exercise 1.5a helped a lot, thank you

[heady hornet]

emm, if you need help to prove this, then you need to give your definition of markov chain first

[keen loom]

not anymore

[olive moth]

we have function f in L^1 definied on domain [0,1). Is it true that we can find continuous function g defined on [0,1] such that f -g arbitrarily small ?

[ionic tree]

I think so. try choosing a simple function $s:[0,1]\to \mathbb R$ such that $|f-s|{L^1} < \varepsilon/2$, and then using appropriate bump functions, make a continuous function $g$ such that $|g-s|{L^1} < \varepsilon/2$.

[livid sand]

Joseph

[ionic tree]

you can choose the simple function by splitting $f=f^+ - f^-$, and choosing simple functions $s^+, s^-$ such that $0\leq s^+ \leq f^+$ and $0\leq s^- \leq f^-$ that satisfy
$$\int_{[0,1]} f^+ - s^+ \mathrm{d} \mu < \varepsilon/4, \qquad \int_{[0,1]} f^- - s^- \mathrm{d} \mu < \varepsilon/4,$$
using the definition of integrability

[livid sand]

Joseph

[ionic tree]

then, for $s=s^+ - s^-$, you get
$$|f-s|{L^1} = \int{[0,1]} |f-s| \mathrm{d} \mu$$
$$= \int_{[0,1]} |f^+ - s^+ - (f^- - s^-)| \mathrm{d} \mu$$
$$\leq \int_{[0,1]} f^+ - s^+ \mathrm{d} \mu + \int_{[0,1]} f^- - s^- \mathrm{d} \mu$$
$$< \varepsilon/2$$

[livid sand]

Joseph

[ionic tree]

now, we want to choose the continuous approximation of s

write $s = \sum_{i=1}^n a_i \chi_{E_i}$. The crux is approximating the indicator functions to arbitrary precision. hmmm how do we do this

[livid sand]

Joseph

[ionic tree]

ahh nice we can use regularity of Lebesgue measure and Urysohn's lemma

so the last thing that we need to accomplish is approximate an indicator function $\chi_E$ by a continuous function to arbitrary precision. Once we figure out how to do this, we can define our $g$ as a sum of indicator approximations with error less than $\varepsilon/2n|a_i|$ or something like that. So let's just abstract away the details and try to approximate $\chi_E$ by a continuous function with error less than $\varepsilon$

[livid sand]

Joseph

[ionic tree]

Since $E$ is Lebesgue measurable, we can pick a closed set $F\subseteq E$ and an open set $G\supseteq E$ such that $\mu(G\setminus F) < \varepsilon$

[livid sand]

Joseph

[strange marten]

Read rules and don’t post same question in 3 channels

[ionic tree]

oh oops did they?

[strange marten]

He was already told it was true in another channel

Ye

[ionic tree]

oh lol

[olive moth]

Thank you for help

[torn wyvern]

I'm looking for a little statistics related assistance. For brevity, I already have help-8 open where I've been working on providing any necessary background for those who may need to understand the notations or see an MWE or source file. Would be very much appreciated if someone could take a look at that room and see if it'd be possible to assist me in deciphering the conditional formulaic switch

[lament hawk]

I'm very confused about what exactly you need help with

[torn wyvern]

Hi, thank you, could you jump over to help channel 8, I'll clarify there

[lament hawk]

ok

[young hawk]

Anyone know how I’m supposed to do this

I forgot sums in the exponent in the last line

But idk where to go from here. This was on a qualifying exam and I would never have time to do all that algebra

And I don’t see anything simplifying

[lament hawk]

Work with the log likelihood

[grave oasis]

Does anyone know about families of rotationally invariant distributions on R^n, s.t. projections onto orthogonal subspaces of random vectors drawn from those distributions create independent random vectors.

The gaussian multinomal distribution for example belongs to that family as orthogonal components of gaussian vectors are independent.

However I'm looking for distributions with that property that possibly concentrate stronger around 0 than the gaussian distribution does

but I'm not sure if there are any

[grave oasis]

ok after looking a bit further I do in fact think the answer is probably no

Yeah, the result is called "Maxwell characterisation of Gaussian distributions"

[keen loom]

We don't define recurrent states in a Markov chain in general, right. Just for homogeneous ones

[stark chasm]

yeah recurrent states don't apply to nonhomogeneous markov chains

same goes for transient states

[keen loom]

thank you