Friends, I made some small mistakes in the previous post. I had not taken detailed notes and finessed some equations writing them on the forum. I would have corrected earlier but a fortnightly antipsychotic injection was due and I had to run to take the injection. Anyway here are the corrections which I am sure would have already been obvious to most friends but still I will write them.Friends, I was doing research on ODE like formulas for constant CDF lines of the SDE densities. I realized that only transition SDE is probably not enough to calculate the constant CDF lines and I wanted to explore ways to find a complete density for small steps that takes into account the previous density. I was able to find a simple series formula that could update the entire density for short steps using convolution. It would not work for long steps since I am only using first hermite though I really think that it could be possible to extend this series formula to two hermite polynomials in volatility with a bit of hack( I will share that if that works in my experiments since I still cannot be sure.). Here is the derivation of the new formula.

We are in Bessel/Lamperti format and we are taking a constant volatility [$]\sigma[$]. I write the SDE as

[$]dy(t)= \mu(y) dt + \sigma dz(t)[$]

In general when we expand the SDE using Taylor, we get multiple terms in drift that can have several powers of t and similarly for volatility so I write the above equation in discrete form and use symbols [$]\mu[$] and [$]\sigma[$] to represent sum of all the terms and I have absorbed t or dt in the same symbol representation since it will help keep focus on the main derivation so I write the discrete form of SDE as

[$]y_2=y_1+ \mu(y_1,t) + \sigma(t) Z [$]

Here [$]y_2[$] denotes the SDE variable at time [$]t_2[$] and [$]y_1[$] is the starting density variable of SDE at time [$]t_1[$].

We can write the evolution equation for the density at next step using convolution as

[$]p(y_2) =\int_{-\infty}^{\infty} f(y_1) g(\frac{(y_2-y_1- \mu(y_1,t))}{ \sigma(t) }) dy_1[$]

here [$]f(y_1)[$] is the original density at time [$]t_1[$] and [$]g(.)[$] is a normal density.

We write the above integral equation with normal density explicitly as

[$]p(y_2) =\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} \exp(-.5 \big [\frac{(y_2-y_1- \mu(y_1,t))}{ \sigma(t) } \big ]^2 ) dy_1[$]

Expanding the terms inside the normal exponential, we get

[$]p(y_2) =\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} \exp(-.5 \big [\frac{( {y_2}^2- 2 y_2(y_1+\mu(y_1,t))+ {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

and further re-arrangement of terms gives

[$]p(y_2) =\frac{1}{\sigma \sqrt{2 \pi}} \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) \int_{-\infty}^{\infty} f(y_1) \exp(-.5 \big [\frac{(- 2 y_2(y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]) \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

We can expand the integral as expectation of a series evaluated with respect to density [$]f(y_1)[$]. Below whenever I write expectation symbol E, it means that expectation is taken over the density [$]f(y_1)[$], so we can write the above integral as an expansion as

[$]p(y_2) =\frac{1}{\sigma(t) \sqrt{2 \pi}} \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) E \Big [\exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] )\Big ] *(1 + y_2 E \big [(\frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 }) \big ] [$]

[$]+ \frac{ {y_2}^2}{2} E \big [(\frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 }) \big ]^2 + \frac{ {y_2}^3}{6} E \big [(\frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 }) \big ]^3 + . . .) [$]

please notice that I have taken [$]\frac{1}{\sigma(t) \sqrt{2 \pi}}[$] out of integral since even if we have expanded [$]\sigma(t)[$] to higher order terms in t, its dependence on [$]y_1[$] is totally negligible in Bessel coordinates unless we are extremely close to zero.

So this way we can easily expand the resulting density of SDE to next steps but we will have to keep time steps reasonable. I have not shown this to keep exposition simple, but the powers of time terms would also explicitly appear in the expansion and can be easily differentiated to work with the ODEs etc. We may have to take a lot of terms in the expansion depending upon what the parameters are but it should not be difficult at all.

So eventually we get an expansion for the density of [$]y_2[$] in the form

[$]p(y_2) =C_0 \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) *(1 +C_1 y_2 +C_2 \frac{ {y_2}^2}{2} + C_3 \frac{ {y_2}^3}{6} + . . .) [$]

where all the constants [$]C_0[$],[$]C_1[$],[$]C_2[$] and further are evaluated as expectations over the original density of [$]y_1[$] at starting time.

I take up from the following equation.

[$]p(y_2) =\frac{1}{\sigma \sqrt{2 \pi}} \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) \int_{-\infty}^{\infty} f(y_1) \exp(-.5 \big [\frac{(- 2 y_2(y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]) \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

We expand the first exponential under integral sign and write as

[$]p(y_2) = \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) \int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} \Big[ 1+ y_2 \big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ] [$]

[$]+ \frac{ {y_2}^2}{2} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^2 + \frac{ {y_2}^3}{6} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^3 + . . . \Big] \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

So we get an expansion in the form of series where

[$]p(y_2) = \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) *(C_0 +C_1 y_2 +C_2 \frac{ {y_2}^2}{2} + C_3 \frac{ {y_2}^3}{6} + . . .) [$]

and

[$]C_0=\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}}\exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

[$]=E \Big[\frac{1}{\sigma \sqrt{2 \pi}}\exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) \Big] [$]

and

[$]C_1=\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} \big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ] \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

[$]=E \Big[ \frac{1}{\sigma \sqrt{2 \pi}} \big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ] \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) \Big] [$]

[$]C_2=\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^2 \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

[$]=E \Big[ \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^2 \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) \Big] [$]

[$]C_3=\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^3 \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

[$]=E \Big[ \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^3 \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) \Big] [$]

and nth coefficient is

[$]C_n=\int_{-\infty}^{\infty} f(y_1) \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^n \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) dy_1[$]

[$]=E \Big[ \frac{1}{\sigma \sqrt{2 \pi}} {\big [ -.5 \frac{(- 2 (y_1+\mu(y_1,t)) }{ {\sigma(t)}^2 } \big ]}^n \exp(-.5 \big [\frac{( {(\mu(y_1,t)+{y_1})}^2 ) }{ {\sigma(t)}^2 } \big ] ) \Big] [$]

All these coefficients are expectations over the probability density [$]f(y_1)[$]. If our grid is refined enough, we can easily replace these integrations with summations over the density [$]f(y_1)[$] on the grid. Multiplying Terms in various coefficients are either the same or powers of each other which can easily be calculated once and summed or even integrated if you like in the same one pass over the density.

So we get an expansion in the form of series where

[$]p(y_2) = \exp(-.5 \big [\frac{ {y_2}^2}{ {\sigma(t)}^2 } \big ]) *(C_0 +C_1 y_2 +C_2 \frac{ {y_2}^2}{2} + C_3 \frac{ {y_2}^3}{6} + . . .) [$]

I have already taken the antipsychotic injection shot in my arm and I will lose most facility of thought in a day or so only to come back after 7-10 days if I continue to struggle.