What's the expectation abd variable of the stochstic process

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I could add that it came up in modeling a distribution that can't take values outside of [0,1]. For r=1/2 it's a variation of the stochastic term in the "CIR model" and respecta the bounds I want

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bump

senpaitachi

no takers at all?

I rewards with chocolate

>>8174227

explain briefly the theory behind this (only studied stuff like AR MA and ARMA kind of thing)

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>>8174243
The ARMA model looks like a recurrence scheme where each new term is the previpus one, weighted by a phi and an independent/standalone fluctuation.
A stochastic differential equation scheme has x_i being x_{i-1} plus some function (set zero in the case I ask about) of that point, plus some function weigh multiplied by a fluctuation (dW)

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I have 'a' equal zero and 'b' the quadratic function of X in the OP.
I wonder how the path of X will typically go. The function is not among the basic ones discussed in the books I now looked at.

Sorry for the crappy screenshots

>>8174400
ok tell me if this is bullshit for r=1:
separate variables
dXt/(Xt(1-Xt)) = dWt

when integarated, you get something like ln(Xt/(Xt-1)) = integral(dWt). Is this where you say it follows a normal distribution with std equal to sqrt(t) so N(0,t)

does that make any sense to you?

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>>8174449
Such manipulations don't get you to an exact solution as (when thinking of a discretization) dW_t is not just a small positive quantity like in calculus, but instead a random value (normally distrubuted here, pic related).
I might use your rewriting in an algorithm to solve for X_t, but the stochastic Euler method sure is a more straight forward way to go and I did that (see third pic with the graph).

It's very unlikely that this equation has an exact silution, since I think to know that less complicated equations don't have one.
However, I'm only interested in the variance (or a similar meausre for the typical drift) of X_t anyway!

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Pic related is the full model

>>8174483
I didn't pretend it was an exact solution -_-
my attempt was to show that the random variable Yt = ln(Xt/(1-Xt)) follows a normal distribution of mean 0 and variance t (because if I remember correctly, a random walk over time t has 0 mean and standard deviation sqrt(t))

from there you could probably derive the probability distribution Xt follows from that of Yt

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>>8174507
The expectation <W> of
W=int dW
is zero, as dW is symmetric.

dX is random by definition of the equation. I don't know what to take from the expression when classically integrating dX/(X·(1-X)).

That a process X determines a process f(X), and how, that's the statement of the fundamental theorem for the Ito integral
https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma#Mathematical_formulation_of_It.C3.B4.27s_lemma

The evolution of the probability distribution associated with X evolves according to the Fokker–Planck equation
https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation#One_dimension

>>8171965
https://en.wikipedia.org/wiki/Itô_diffusion

In particular, E(X_t)=EE(X_t|F_0)=EX_0 by the martingale property. So the expectation is just the expectation of the initial condition.

>>8174719
What's F and what about the expected drift?

>>8171965
Ito's lemma:

if Y = f(X)

then dY = df/dX * dX + 1/2 * d^2f / dX^2 * (dX)^2 + df/dt * dt

Apply the formula for f(X) = X - X^2

Note that dW * dW = dt
and df/dt is 0 in our case because the original formula (dX) has no dt coefficient.

>>8174842
>>8174842


F_t=sigma field generated by all X_s with s \leq t

there's no expected drift in this case because there's no dt term in the equation

>>8174957
I don't see what I'm supposed to use
d(X·(1-X))
for, once I've computed it. It it of use to compute the variance of X after a fixed time that passed?

>>8174976
By drift I didn't mean a [math]\mu[/math] but basically, [math] \sqrt{ < X^2 > }[/math], that which comes out to be \sqrt{t} for the classical random walk.

The full equation is in >>8174493 btw.

>>8174976

In general the goal is to make a substitution and get a new equation of the form

dY = a * Y dt + b * Y dZ

or equivalently,
dY/Y = a * dt + b * dZ

(with a and b constants).

Because this implies
Y(t) = Y(0) * e^(a*t + b*Z(t)). (that Y follows Geometric Brownian motion and then you could literally just read off the expected value and variance (expected value is a*t, variance b^2 * t).

Of course that may not be possible in your problem. That would really depend on where it came from.

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>>8175000
>That would really depend on where it came from.
Why? The differential equation is stated (>>>>8174493).

I have no preference on the specifics of dB. The X·(1-X) term is so that the value of X are confined to [0,1], as it's a distribution (of particular car gadgets over the totality of cars)

If you care, I develop the model here
https://axiomsofchoice.org/kfv_._note
(Paragraph 4 and 5 on linear approaches on the non-stochastic care are irrelevant)

>>8174493

The answer will depend on the distribution of X_0.

Let's fix any function f and consider the expression u(t,x)=E(f(X_t)|X_0=x).

Then this function u satisfies a deterministic PDE called Kolmogorov's backward equation which you can read off from the coefficients of your SDE. It's written out here http://math.stackexchange.com/questions/598811/kolmogorov-backward-equation-for-itô-diffusion or in, e.g. Oksendal.

Now let f(x)=x^2 and solve the KBE using your favorite numerical approximation scheme. This gives you E(X_t^2|X_0=x) and then of course
E(X_t^2)=\int E(X_t^2|X_0=x)p(x)dx where p(x) is the density of X_0. etc.