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

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 — Today, 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 — Today, 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 — Today, 12:23 pm

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Statistics: Posted by ExSan — Today, 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 — Today, 10:56 am

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Let's skip the zillion fixes and patches for GD. Enough blogs already. One example however is

There is much ado with this and a whole cottage industry has grown around it, e.g. grid search ugh.

GD is really a FD scheme for dissipative gradient ODE (Lagrange, Poincare); the learning rate is the step size in the ODE solver. Equality and inequality constraints are easy (try with GD..)

Statistics: Posted by Cuchulainn — Today, 10:51 am

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www.youtube.com/watch?v=uvPxz5apXN8

Statistics: Posted by Cuchulainn — Today, 9:52 am

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Fantastic story! United we stand, divided we fall: https://fablesofaesop.com/the-four-oxen ... -lion.html

Cute. They weren't Irish oxen.Funny, but I was in Soweto and Stanton, Jo'burg 16 June 2006, the 30th anniversary of the murder of Hector Pieterson in 1976 and the start of the Soweto riots.

Boycott against Apartheid started in Dublin by working class women in 1984. Of course, the middle class liberals came on board when it became trendy to be anti-Apartheid.

https://en.wikipedia.org/wiki/Hector_Pieterson

https://www.youtube.com/watch?v=M6cBDusBiHE

Seamus Heany @2.22. They used to hang poets in Zuid Afrika.

// and it was Bloomsday 16th June

Statistics: Posted by Cuchulainn — Yesterday, 10:56 pm

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Statistics: Posted by katastrofa — Yesterday, 7:40 pm

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Statistics: Posted by tagoma — Yesterday, 7:32 pm

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