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21:00
@leslietownes Oh, alright
Wonder how it'll go if all active members here one day met up for Pizza or something
21:15
$0^0\cdot -{1 \over 12}$
Looking for some help here. Any references would be useful: mathoverflow.net/questions/434889/…
I suggest asking on the site itself
A far greater number of people would see it as opposed to less than 20 here
I'm linking the original post. I haven't obtained any answers. Perhaps it's just not a tractable problem
it is not clear what you are asking, are the transition probabilities time deppendent?
Not sure it belongs on Overflow.
21:24
definitely not mo
Move it over to SE.
they are not time dependent; perhaps the question isn't clear
I'm trying to optimize Prob(state 4), and in each time period I can choose to apply the top markov chain or the bottom one
Ted: I should move this optimization problem question to stack exchange?
you need to figure out what you are asking first.
do you suggest I edit the question or post a new one? thanks
Fix it to clarify so copper approves, then move it. It is not MO level.
21:27
ok, thanks
runges & rouches theorems always seems a little like magic to me
more runges
Rouché is simple winding numbers.
a lot of complex analysis has a certain magic to it
the 'geometry' behind rouche is very clear, but it just surprises me for whatever simple minded reason
ok; could you please check out the post again? hopefully it's clear now
thank you
21:34
it is still a bit odd, so the transition probabilities are time varying and there is no feedback, correct? is $\alpha$ known?
it really is not mo material.
math.stackexchange.com/q/4580572/1092912 This question seems a little unclear.
alpha is known, yes
hmmm
not sure I'm able to clarify it further without feedback; to me it is well specified
of course, you're not inside my head :)
@Goku Raise your questions in a comment.
the control variable at each point in time is the choice of one of the transition probability matrices
@TedShifrin I planned to but the comment might be too long, at the time I don't want to post an "answer" that basically doesn't answer anything
21:39
Yeah, I have no idea what the question is. Is it math or is it programming?
it's a stochastic control problem
let me try again, this time removing all the text and just writing down the optimization problem
thanks for the help
good luck. i would try mse rather than mo.
thanks. I didn't know mse existed
@TedShifrin it is definitely programming, I copy pasted some of the code but the issue is he doesn't provide enough context about what the problem is that he's trying to solve. I can understand it better largely because I can understand the code and even then it isn't enough for me to answer the problem
ah, wait, you're not talking about my post
21:43
Anyone else tired of folks over at Stack Overflow using Math.SE to do all their math homework? Come on guys, you can do better
$Q(\beta) = YY^t - 2\beta^t X^t Y + \beta^t X^t X \beta$

Where $X$ is a fixed matrix and $Y$ a fixed vector.

So out of curiosity I wanted to derive the derivative expression which has been given:

$$DQ(\beta) = -2X^tY + 2X^t X \beta$$

So I began with the strategy of finding $Q(a+h) - Q(a)$.

Doing all the algebra I arrive at:

$$-(Xh)^t - y^tXh + (Xa)^t Xh + (Xh)^t Xa + (Xh)^t Xh$$

Recall that $x^ty = x \cdot y$. Using that I get:

$$-(Xh) \cdot t - y \cdot Xh + (Xa) \cdot Xh + (Xh) \cdot Xa + (Xh) \cdot Xh$$
Sorry, cat, I was talking to Goku.
@Goku If the OP can’t clarify an actual math question, then a programming issue isn’t for MSE.
Also to any admins, the norm command in Latex ($\|a\|$) is not rendering correctly on the main site
21:46
@TedShifrin yep, might have to flag it but I'll wait and see what judgment other qualified users make
@D.C.theIII I totally cannot follow.
@D.C.theIII i am a little unclear about your notation. is $Y$ meant to be a row or column vector?
Good point.
@TedShifrin It is the same idea I just covered in your text w.r.t getting the derivative of a linear map.
$\|a\|$ renders fine for me on mse
21:53
In this text, $Y$ is treated as a column vector
@D.C.theIII it is a quadratic map...
so $Y Y^T$ is a matrix, right? (a dyad)
YEs. Because all it is is the Least Squares derivation.
Isn’t $Q$ scalar-valued?
ah...I see what you're hinting at copper. Yes $Q$ will be scalar valued
I've been wondering. How does a post make it to "hot network posts" section? I had one question that made it there, and I answered a question that also later made it there. Is it just an algorithm
21:56
if $Y$ is a column, then $Y Y^T$ is a matrix.
Anyhow, I cannot follow your computation at the very beginning.
i am totally confused
if $Y$ is a column then $Q$ is matrix valued, not scalar.
The original "algebra simplification" I wrote?
Yes. The first term should be $Y^\top Y$, DC.
21:59
damn it... I "wrote" the transpose on the wrong vector....switch $Y$ and $Y^t$ around
But the algebra afterwards I don't follow at all.
$Q(a+h)-Q(a) = -2h^\top X^\top Y + h^\top X^\top Xa + a^\top X^\top Xh + h^\top X^\top Xh$.
Expand $Q(\beta +h) - Q(\beta)$. this will have linear terms and one quadratic term. the linear term is the derivative.
then estimate $\|Q(\beta+h)-Q(h)-\text{ linear terms }\|$
@TedShifrin that's what I was expecting to get....maybe I messed up some algebra
Their answer is a bit suspect, too.
what's everybody's favorite letter for a homotopy that's not $H$?
22:02
That's the only letter, @Thor.
$h$
or $\eta$.
If the mappings are $f_0$ and $f_1$, then you could use $f_t$ or $F$.
So $DQ(a) = -2(X^\top Y)^\top + 2a^\top X^\top X$, not what they said.
found my mistake.......I forgot to carry a $t$ from the first step when writing things out.....😭😭
Unless they're using row vectors instead of column vectors.
Remember that $DQ(a)h$ should give the linear terms. Not $h^\top$ in front.
no they use column vectors,
22:06
I need (at least) two homotopies and the maps they are between are kind of not nicely named
give me a second, I'll show you my diagram
trying to get rid of the indices by picking new letters, but it's tough
Oh hell, @Thor. Use $\bar H$ or, if they're totally unrelated, $K$.
nice diagram
math.stackexchange.com/q/4580594/1092912 Am I the only one wondering why there are no bounds?
Yeah, LaTeX cannot typeset that diagram. That's why Mike Spivak (may he rest in peace) invented AMS-LaTeX.
22:07
@Goku check the comments
@D.C.theIII This should be identical to the derivation in my book.
I've considered $K$, but found it awkward cause I already have $k$ and there's no quite clear relation
perhaps I do have to resort to a bar, prime or tilde
@copper.hat Ohh, I knew this didn't seem right
They are writing the derivative as a column, so it's a gradient, not a derivative. Ignorant statisticians.
So they are mistreating the notation?
22:09
yes
Yes.
Where I would have $DF(a)h$, they have $h^\top \nabla F(a)$, but they still write $D$.
Reminds me of SNL: "Jane, you ignorant slut."
Riesz has a lot to answer for
technically, at a later stage, there should also be $s_i$ parallel to the $t_i$, homotopies between those , two new homotopies in place of the $H_j$ that fulfill the same purpose to the $s_i$ as the $H_j$ do to the $t_i$ and then two coherent homotopies of homotopies between those
I would burn the book at the stake.
but I don't think that later stage will be drawn in a diagram form..
22:13
@Thor I'm not sure what you mean by parallel, but perhaps that suggests a bar notation is not bad.
parallel means same domain and codomain
if the paper is not too tough you could use repurpose the book when people are loading up on tp
@TedShifrin whoa, whoa, whoa.....no need to go all Farenheit 451...
Yeah, I think I vote for the bar unless it's going to be confuzling.
People have enough trouble with multivariable calculus without transposing the derivative matrices.
basically, the front and back end of the diagram are fixed and the rest is something of a morphism between the front and back end and then I will consider homotopies between those (however, since the morphism itself involves homotopies, homotopies thereof will involve homotopies of homotopies)
at this point, it's been more of a struggle to try and figure out non-awful notation than it has been to prove the result I'm trying to write down
22:16
Maybe use a subscript for the homotopy of the homotopies. Like $H_u(x,t)$.
Horrid notation is a big impediment to understanding proofs, in my humble experience.
You're not wrong
@D.C.theIII I was thinking more of Salem.
I do make an attempt at streamlining, lowercase letters are continuous maps, upper case letters are homotopies (and the spaces involved, but the difference between those should be clear) and homotopies of homotopies will be greek letters
Oh, Greek letters. I guess if the primary homotopies are $H_0$ and $H_1$, then my notation isn't optimal.
Maybe a superscript? $H^u_t$.
Then some tensor analyst will want to lower indices :D
@D.C.theIII you should get $Q(\beta+h)-Q(\beta) - 2 (X \beta -Y)^T Xh = h^T X^T X h$, and since the term on the rhs is $O(\|h\|^2)$ it is differentiable with $DQ(\beta)h = 2 (X \beta -Y)^T Xh$. If one must use the gradient, it is just $\nabla Q(\beta) = X^T(X \beta -Y)$.
22:21
I do like to write $H_t=H(-,t)$ for homotopies (and similarly for the homotopies of homotopies), but sadly that conflicts with calling the homotopies in the picture $H_1$ and $H_2$, which is why I need to change something
Right. I just realized that. That's why I put the "outer" homotopy as a superscript. Whatever.
I think it's time for me to take my "daily" walk. BBIAB.
have fun
Before I leave, @copper, have you seen this convexity query?
will take a look, thanks @TedShifrin
if I have a markov chain $(X_n)_n$ with values in $S$ with transition matrix $Q$ and initial distribution $\nu$ what can I say about $E_\nu(Q(X_{n+1},y)|F_n)$ for all $y\in S$?
I am thinking about using the simple markov property but I don't see how
22:44
If we had pizza together then it'll probably be really awkward
people behave differently online and in real life
where did that come from?
but all social interactions feel at least a little bit awkward to me
@copper.hat Goku
comes naturally to some, i need practice which covid removed
If we had pizza together you'd see me sitting at a completely separate table away from everyone else, as I do with my own family too. I'm not just good at "eating together"
somehow that sounds a bit sad
22:50
i enjoy company, but don't feel the need to be interacting continually
I'm in the same boat Copper....I'm at a lack of practice.
23:10
@Jakobian it is, because I have very little (if any) social interaction
I think I'm a social anomaly for math geeks.
23:28
Hi everyone! I started a 2nd attempt at a PhD a couple of months ago. The 1st was downgraded to an MPhil due to health & financial issues. I'm concerned the same thing might happen again. I'm studying linear algebraic groups and, so far, I'm learning a lot of algebraic geometry; it's tough. I started strong but old habits, like sleeping in until late, are returning.
I have an assignment due by the end of the month. It's causing stress. I asked for an extension on Friday.
I'm not sure what I'm looking for in the chat here. I guess I just need to vent.
My passion for mathematics is intense and I want to do myself justice in my studies.
I want to make a career of it.
this is a good place to vent
I have a mental health condition. My university knows. I take medication for it.
If, like me, you don't care for much social interaction then study+more than enough sleep should be easy to fit in
@copper.hat I hope so :)
@Goku I'm content socially. I called an old friend earlier today and she cheered me up considerably.
@Shaun lucky you
23:35
I don't use social media though. I think the bad things there outweigh the good.
I don't either.
Algebraic geometry requires a lot of topology, at least for what I'll be using it for. I have never been good at topology.
@Shaun Exercise helps. During the last two quarters of my PhD (during which time my father was diagnosed with lung cancer, my father died, I was in the middle of a messy and acrimonious divorce after 12 years of marriage, oh, and a global pandemic was kicking off), I walked to campus every day (2.5 miles, each way). It helped. A lot.
I have copies of the three "Linear Algebraic Groups" textbooks - yes, they each have that title - and they each start out with a terse treatment of algebraic geometry. I asked my supervisor for guidance. He gave me a selection of exercises from one of them. I've been working on them for the last month.
@XanderHenderson I'm sorry to hear that. Thank you for the advice.
That's why I do Taekwondo and play soccer
23:42
I used to exercise a lot during my undergraduate days: I hit the gym twice a week and exercised before bed each night that wasn't a gym night. Ironically, I had a long stay in hospital (for my mental health) - September 2015 to March 2017 - and I gained weight then. I got out of the habit of exercising and eating healthily.
i'm also a big fan of exercise, i'm pretty sure it's the only way i can sustain not speaking to people for long periods of time
i need exercise to maintain any degree of mental stability
My family are very supportive. They encourage me to lose weight a lot, almost too much really. My current university is associated with a large gym just off campus. I have a discounted membership I haven't used yet. The main barrier is having to wash my gym kit. The local laundrette has exorbitant prices.
i was having issues getting to nearby gyms
because they were not near by
my strategy has been to time caffeine intake energy surges to not mind doing exercise in secluded space
My diagnosis is paranoid schizophrenia. It can render me inactive for a couple of days at a time. Whenever I take the time off from my studies, I feel guilty.
23:51
I don't exercise. I don't talk to other people. I'm fine
11
Normalize not having any human contact
Its normal....right?
@Jakobian the ubermensch
2
And then there's my budget. My 1st PhD attempt wavered my tuition fees and gave me a stipend. This attempt, however, I'm funding for myself using one of those government doctoral loans they have here in the UK.
@Jakobian "Fine" is a matter of judgment.
At least here at Aberdeen, my interests are much more in line with the research I do. The previous attempt always felt like a compromise, not for lack of interest in the area.
23:58
@TedShifrin well yeah. I could do with walks at least but it's too cold
@TedShifrin so, I'm not fine?
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