@Bob Look about a quarter of the way down in this article. In the caption of the pictured model of peaking are both a link to the github for the model (so that you can play with it yourself) and some commentary on what parameters' reasonable ranges might be.
Note that (in case you don't have time to read and reread the article) his point about adjusting fatality rate may seem odd: how can we adjust the fatality rate of the virus? It's not treatments he's talking about; he's talking about fatality of the 5% of cases that are critical sliding from about one-fifth of them (what we're seeing in most places now) to nearly all of them as ICU capacities are exceeded.
I.e. the numbers we have on fatality rates right now come (almost?) exclusively from looking at the fatality of cases that received the very best their medical systems could offer. That gives you 1-2% mortality.
I don't know to what extent #ICU beds is a proxy for a lot of underlying factors: number of ventilators, personnel, ability to turn over cases. I suspect that makeshift hospitals will not be as good as "real" hospitals. Beyond that, I'll leave you to play with the model.
So you might know the history better than most: do you know of a good comparison example where a country (or many countries) effectively decided to enter a recession/depression?
I was trying to talk my mother-in-law off a ledge and used the metaphor: "it's not like we were running a marathon, dropped to the ground, and paramedics are trying to figure out what's wrong. It's like we were running a marathon, we got a phone call that turns out to be critical, we're taking it, we don't know how long it'll be, but after it's done we can get back to our running."
@Bob I'm still going golfing, though. Just going to slip a twenty under the door of the pro shop each time I do. (I live near a course where I've literally never seen more than 4 people on the course at a time, anyway.)
gamestops are california is closed, much to their exec's consternation. "we sell electronics, and people can use that for remote work, so we're totally essential"
Hey everyone. I'm trying to approximate a distribution as a sum of identical unfair dice.
An example of what I mean by "a sum of identical unfair dice" is that I have 64 dice, each of which has a 5% chance of rolling -10, a 30% chance of rolling -1, a 30% chance of 0, a 30% chance of 1, and a 30% chance of 10. I roll all 64 dice and add the resulting values together.
Has there been any research into how to do a good job of approximating a distribution this way?
The reason I'm doing this is I'm trying to implement a lattice model: en.wikipedia.org/wiki/Lattice_model_(finance) Essentially, I'm assuming that the price of a company's stock goes up or down each minute according to an unfair die.
Specifically, a normal-inverse Gaussian distribution with alpha = 0.39, beta = -0.073, mu = 0.54, and delta = 1.9. That particular distribution is a pretty good approximation to the particular set of stock market returns I'm looking at.
Here's a sub-problem I'm thinking about. What properties does a single die need to have for its distribution to have extremely high kurtosis (around 250, say)?
One example is a die with a 0.2% chance of rolling -1, a 99.6% chance of rolling 0, and a 0.2% chance of rolling 1. The kurtosis of this die is (exactly?) 250.
In order to have such a high kurtosis, is it necessary for a die to have faces that are very unlikely to be rolled?
Or can I attain a high kurtosis like that using only faces that are likelier?
@user714630 on the other hand, it is true that the MVT guarantees that the components of the velocity vector, must each vanish at some time. It's just that those times need not coincide.
I forget the Jacobi equation. Let $\gamma : [0, 1] \to M$ be a geodesic in a Riemannian manifold and $X$ be a vector field along $\gamma$ with $X(\gamma(0)) = 0$ such that the variations $\gamma_s : [0, 1] \to M$ defined for $s \in (-\varepsilon, \varepsilon)$ along $X$, i.e., satisfying $\partial_s \gamma_s(t) |_{s = 0} = X(\gamma(t))$ are all geodesics as well.
What was the equation now
So the equation is $\nabla_{\gamma'}^2 X + R(X, \gamma') \gamma' = 0$ but only if I could remember how to obtain this
$R(X, \gamma')\gamma' = \nabla_X \nabla_{\gamma'} \gamma' - \nabla_{\gamma'} \nabla_X \gamma' - \nabla_{[X, \gamma']} \gamma'$, the first term is zero because $\gamma$ is geodesic.
Oh ok this crap. $\nabla_X \gamma' = \nabla_{\gamma'} X + [X, \gamma']$ by the magic torsion vanishing. Take $\nabla_{\gamma'}$ on both sides
I get $R(X, \gamma') \gamma' = -\nabla_{\gamma'}^2 X - \nabla_{\gamma'} [X, \gamma'] - \nabla_{[X, \gamma']} \gamma'$
Dang, some sign mistake
Weird, I don't see the error
I was hoping to apply torsion vanishing again and get some $[[X, \gamma'], \gamma']$ term, oh well. Seems everything is correct.
Ah but I am dumb, $[X, \gamma'] = 0$ I think
$[X, \gamma'] = [\partial_s \gamma_s(t)|_{s = 0}, \partial_t \gamma_s(t)|_{s = 0}]$ and $\partial_s$, $\partial_t$ commutes in $[0, 1] \times (-\varepsilon, \varepsilon)$, the domain of $\gamma_s(t)$. Their pushforwards will also commute.
So $\nabla_{\gamma'}^2 X + R(X, \gamma')\gamma' = 0$ is indeed the Jacobi equation
Lemma:
Let $(X,d)$ be a totally bounded metric space.
Every sequence has a Cauchy subsequence.
My attempt: Let $(x_n)_n$ be any sequence in $X$.
Let $\epsilon=1$. So, we may find $x^1,x^2...,x^m$, for which , $X=\bigcup_{i=1}^mB(x^i,1)$
By the pigeon hole principle, there exists $j$, for which...
One distinct difference between axioms of topology and sigma algebra is the asymmetry between union and intersection; meaning topology is closed under finite intersections sigma-algebra closed under countable union. It is very clear mathematically but is there a way to think; so that we can defin...
Topologicalcafe: yeah, proof looks fine to me, you basically take the intersection of the balls inductively with smaller and smaller radii, then the subsequences are going to converge with d vanishing hence cauchy
You start out at one particular horizontal coordinate, and after walking around in a circle you're at the same coordinate. So you've got x(time 2) = x(time 1)
@Semiclassical Ya know, I learnt to prove that if time period of a planet in a circular motion is $sr^{3/2}$ for $r$ as radius then force is in the form of $c/r^2$.
hence the MVT says there's some time between 1 and 2 where dx/dt = (x(t2)-x(t1)) / (t2-t1) = 0. that is, the horizontal velocity must vanish at some time.
so $T\propto R^{3/2}\implies F\propto R^{-2}$, done
thing is, that's only as short as it is because i'm taking a lot of stuff as 'obvious'
$v=2\pi R/T\propto R/T$ is trivial enough, but $a_c=mv^2/R$ is not obvious at all when you first see it
and neither work if it's not uniform circular motion
@user714630 continuing with my comments above: you can do a similar argument for the vertical coordinate, and therefore conclude that the vertical velocity must be zero at some time
However
Those are two separate MVT arguments. There's no requirement that those two quantities must vanish at the same time. Hence one can't conclude that you were ever 'stopped' on the track; all you can conclude is that you momentarily 'stopped' moving left/right
Another way to see it is that, for the purposes of speed (not horizontal or vertical velocity), what matters is the distance you've walked.
But the distance you've walked keeps increasing with time. You never have "distance at time 1" = "distance at time 2"
Lemma:
Let $(X,d)$ be a totally bounded metric space.
Every sequence has a Cauchy subsequence.
My attempt: Let $(x_n)_n$ be any sequence in $X$.
Let $\epsilon=1$. So, we may find $x^1,x^2...,x^m$, for which , $X=\bigcup_{i=1}^mB(x^i,1)$
By the pigeon hole principle, there exists $j$, for which...
