Interesting thing I've learnt from a local today during a countryside walk. We encountered a black small snake sleeping by the road. It turns out that Scandinavian adders can be blackberry sometimes - an adaptation to better absorb light! Still part of the population preserves the lighter colouring, because it's less visible to predators. Short warm life or long and chilly?

Yeah, I grew up with these guys and can testify to a pretty wide range of colors. I am slightly confused by the blackberry reference, but will assume it involved autocomplete (which, self-referentially, my phone detected about midway through; unlike “self-referentially”, to which it objected vehemently). You would normally be able to make out a zigzag pattern, though.Statistics: Posted by bearish — Yesterday, 9:12 pm

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I would say; no big new features, rather fixes and extra thingies added to C++20.

Statistics: Posted by Cuchulainn — Yesterday, 11:33 am

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The acorn doesn’t fall from the tree.Newton and Euler were the same person??

But Lagrange is closer.

1. Gradient descent (discrete), starting point for ML? [$]k = 1,2,3[$]

2. Continuous realisation of 1. [$]0 \lt t \lt \infty[$]

Statistics: Posted by Cuchulainn — Yesterday, 10:22 am

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Statistics: Posted by tagoma — June 17th, 2021, 8:04 pm

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Newton and Euler were the same person??

The acorn doesn’t fall from the tree.But Lagrange is closer.

Statistics: Posted by Cuchulainn — June 17th, 2021, 7:59 pm

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[$]dX_1 = \mu_1 \, {X_1}^{\beta_1} \, dt + \sum_{n=1}^{N} \, \sigma_{1n} \, {X_1}^{\gamma_1} \, dZ_n[$]

In particular we want to notice that [$]\sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, = \, {\sigma_1}^2 [$]

Please note that after cholesky or SVD, all of the above brownian motions [$]Z_n[$] are orthogonal.

Since I am playing with only one SDE, I will drop the subscript "1" from the bottom of each variable since it is unnecessary in case we are dealing with one SDE only.

expanding the above SDE as we have previously done, we write the equation after applying repeated Ito as

[$]dX = \mu \, {X}^{\beta} \, \int_0^t ds + \, {X}^{\gamma} \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t dZ_n(s) [$]

[$]+(\mu \, \beta {X}^{\beta-1} ) \, (\mu \, {X}^{\beta}) \, \int_0^t \int_0^s dv \, ds[$]

[$]+(\mu \, \beta {X}^{\beta-1} ) \, ({X}^{\gamma} ) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_n(v) ds[$]

[$]+.5 (\mu \, \beta \, (\beta-1) \, {X}^{\beta-2} ) \, ({X}^{2 \gamma} ) \, \sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, \int_0^t \int_0^s dv \, ds[$]

[$]+(\gamma {X}^{\gamma-1} ) \, (\mu \, {X}^{\beta}) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]

[$]+(\gamma {X}^{\gamma-1} ) \, ({X}^{\gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, \sigma_{1m} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_m(v) \, dZ_n(s)[$]

[$]+.5 (\gamma (\gamma-1) \, {X}^{\gamma-2} ) \, ({X}^{2 \gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, {\sigma_{1m}}^2 \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]

since in quadratic variations, [$]\sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, = \, {\sigma_1}^2 [$], we can slightly simplify the above equations by using this identity that changes double summations in quadratic variations to single summations and write the above equation again as

[$]dX = \mu \, {X}^{\beta} \, \int_0^t ds +\, {X}^{\gamma} \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t dZ_n(s) [$]

[$]+(\mu \, \beta {X}^{\beta-1} ) \, (\mu \, {X}^{\beta}) \, \int_0^t \int_0^s dv \, ds[$]

[$]+(\mu \, \beta {X}^{\beta-1} ) \, ({X}^{\gamma} ) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_n(v) ds[$]

[$]+.5 (\mu \, \beta \, (\beta-1) \, {X}^{\beta-2} ) \, ({X}^{2 \gamma} ) \, {\sigma_1}^2 \, \int_0^t \int_0^s dv \, ds[$]

[$]+(\gamma {X}^{\gamma-1} ) \, (\mu \, {X}^{\beta}) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]

[$]+(\gamma {X}^{\gamma-1} ) \, ({X}^{\gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, \sigma_{1m} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_m(v) \, dZ_n(s)[$]

[$]+.5 (\gamma (\gamma-1) \, {X}^{\gamma-2} ) \, ({X}^{2 \gamma}) \, {\sigma_1}^2\, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]

We can solve all of the above integrals analytically.

[$]\int_0^t \int_0^s dZ_n(v) ds= \frac{1}{\sqrt{3}} t Z_n(t)[$]

[$]\int_0^t \int_0^s dZ_n(v) dZ_n(s)= \frac{1}{2} ({Z_n(t)}^2-t)=H_2(Z_n(t))[$]

only difficult integral is

[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]

where we get this integral from Ito Isometry as

[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)[$]

[$]=\int_0^t Z_m(s) dZ_n(s)[$]

and its variance is given as

[$]=\int_0^t {Z_m(s)}^2 ds[$]

[$]=\int_0^t d[{Z_m(s)}^2 s]-\int_0^t s d[{Z_m(s)}^2][$]

[$]= t \, {Z_m(t)}^2 - \int_0^t 2 s \, Z_m(s) \, dZ_m(s)- \int_0^t s \, ds [$]

[$]= t \, H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t^2/2[$]

its representation will be given as

[$] \sqrt{(H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t/2)} \, \sqrt{t} \, N_n[$]

where [$]N_n[$] is standard normal associated with brownian motion [$]Z_n[$]

writing [$]\sqrt{t} \, N_n = Z_n(t)[$] in the above equation we get

[$] \sqrt{(H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t/2)} \, Z_n(t)[$]

[$]= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]

So we can write the integral as

[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]

We can now solve all the integrals and easily simulate a basket option to 2nd expansion order of Ito-Taylor expansion.

Statistics: Posted by Amin — June 17th, 2021, 6:42 pm

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https://en.wikipedia.org/wiki/Newton%27s_method

It is an iterative method and the nasty surprise is that it is the Euler method!

I suppose that the old-timers didn't have recourse to ODE solvers? Same with Cauchy's gradient descent.

Statistics: Posted by Cuchulainn — June 17th, 2021, 3:47 pm

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I suspect his "Goodbye" will be followed by "And another thing..."

I don't think that word means what he thinks it means.he's a bit like Colombo minus the humour.

Statistics: Posted by Cuchulainn — June 17th, 2021, 12:23 pm

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Statistics: Posted by ExSan — June 17th, 2021, 12:23 pm

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God says no mercy for doggists.

Goodbye.

How often does God talk to you Goodbye.

Statistics: Posted by Cuchulainn — June 17th, 2021, 10:56 am

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