I will guess: just because the colors go nicely together (black ink on yellow background). Why do women wear red with black?

Statistics: Posted by Alan — 4 minutes ago

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Christy Moore has a song

www.youtube.com/watch?v=8gdtiv-hZTg

He has a variant "The weekend that we went to Amsterdam" but no link here ....

Statistics: Posted by Cuchulainn — 13 minutes ago

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Statistics: Posted by Cuchulainn — 28 minutes ago

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Merkel!

Who do you fancy in the 2:45 at Newmarket? (So I know who not to bet on!)

Gotcha. My great uncle was a stable hand in Newmarket, so be was. He's long gone, so my guess is as good as yours.Who do you fancy in the 2:45 at Newmarket? (So I know who not to bet on!)

quiz; name of horse that won Ascot Gold Cup June 16 1904?

flat or steeplechase?

Does it mean free champagne and hob-nobbing?

Statistics: Posted by Cuchulainn — 31 minutes ago

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You can definitely train for women's handwriting. The little curly hearts above the i's are a dead give-away.

Statistics: Posted by Alan — 50 minutes ago

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Here you go, $100, as predictedMy prediction on vaccine incentives. In the US, by mid-summer, as the vaccine glut develops and vaccination rates plateau around 70%, the Biden administration will try paying hold-outs to get vaccinated. The first offer will be $100. How high will they go?

Statistics: Posted by bearish — 55 minutes ago

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Bill and Melinda Gates to divorce after 27 years of marriage

Lockdown effect

Lockdown effect

Statistics: Posted by tagoma — Today, 6:41 pm

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[$]\frac{1}{\sqrt{2 \pi t}} \exp[-.5 \frac{{Z(t)}^2}{t}] [$]

I found the CDF of the density as

[$]\int_{-\infty}^{Z(t)} \frac{1}{\sqrt{2 \pi t}} \exp[-.5 \frac{{Z(t)}^2}{t}] dZ[$]

I differentiated the above equation in mathematica where I had emphasized the dependence of Z[t] on t and equated it to zero as previously suggested. I am writing the equation as

[$]\frac{d}{dt}[\int_{-\infty}^{Z(t)} \frac{1}{\sqrt{2 \pi t}} \exp[-.5 \frac{{Z(t)}^2}{t}] dZ]=0[$]

After a few manipulations, I got the simple ODE as a result of the above equation as

[$]-0.5 Z(t) + t \frac{dZ(t)}{dt} =0[$]

I solved the above simple ODE in mathematica to get the answer which is also solution for constant CDF lines of brownian motion as

[$]{{Z[t]->t^{0.5} C[1]}}[$]

I am sure it would easily be possible to write the normal form of evolution SDEs in general and then solve them to find an ODE that can be cheaply solved with this using numerical methods(to give constant CDF line solutions) as I am afraid that getting analytical results for all of the SDEs evolution on constant CDF lines would be impossible.

I used mathematica for all of this and I am going to walk the friends through mathematica commands used to find the above solution.

First use the command

1.

D[Integrate[1/Sqrt[2 Pi t] Exp[-.5 (Z[t]^2)/t], {Z[t], -Infinity, Z[t]}], t]

It will give you a relatively complicated solution. I am not writing the solution but I represent it here with 'Solution' in mathematica code.

2.

Simplify[Solution == 0]

The above equation means that we have asked mathematica to equate the result of command 1 with zero and simplify it. (You have to copy the result of command 1 and copy it in command 2 in place of 'solution' words I have written there)

3. As a result of simplification, you will get the following (I have written it in Latex form now.

[$]\frac{e^{-\frac{0.5 Z(t)^2}{t}} \left( t Z'(t)-0.5 Z(t)\right)}{\sqrt{t}}=0[$]

from which you easily retrieve the ODE

[$]t Z'(t)-0.5 Z(t)=0[$]

4. I used the mathematica command to solve this ODE as

DSolve[-0.5 Z[t]+t Z'[t] == 0, Z[t], t]

which gave me the solution to constant CDF lines of brownian motion as

[$]{{Z[t] -> t^{0.5} C[1]}}[$]

In next few days, I will spend time trying to write ODEs for various SDEs and solving them numerically and would be pasting the code here. I have placed the stochastic integrals project for a small hold but many things we will learn here would be useful in solution of stochastic integrals as well as thinking about them led me to this. I will be posting a lot of code in next few days.

I have not done this so I could not be perfectly certain but I believe that you do not need to know the universal density of SDE diffusions as they are mostly not known but I think that we could just have a density representation of instantaneous evolution of SDEs over small time steps and then use that with numerical solution of ODEs to find the solution of SDEs along constant CDF lines. I will keep friends posted as I do more work on this.

Statistics: Posted by Amin — Today, 5:24 pm

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One always hears about what a great job machine learning can do with image recognition. I believe it. Given that, you would think somebody could design a smart fast system to quickly verify signatures and recount mail-in ballots, which are the more problematic ones. I would guess/hope 98% would pass and a couple percent would get kicked out for hand review. Maybe add a fingerprint or something as a backup ID, although there might be issues with that. My point being -- is this really so hard!?

Of course, if a legislature proposed having such a smart fast system, Democrats would complain it must be voter suppression. Or, they'd say it was a tricky way to catch criminals.

Statistics: Posted by Alan — Today, 5:02 pm

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