What's the expectation abd variable of the stochstic process giveb by
[math] dX_t = (X_t \, (1-X_t))^r \, dW_t [/math]
?
I'm mostly interested in r equals 1/2 or 1.
What's the expectation abd variable of the stochstic process giveb by
[math] dX_t = (X_t \, (1-X_t))^r \, dW_t [/math]
?
I'm mostly interested in r equals 1/2 or 1.
Other urls found in this thread:
en.wikipedia.org
en.wikipedia.org
en.wikipedia.org
axiomsofchoice.org
math.stackexchange.com
twitter.com
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
bump
senpaitachi
no takers at all?
I rewards with chocolate
explain briefly the theory behind this (only studied stuff like AR MA and ARMA kind of thing)
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)
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
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?
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!
Pic related is the full model
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
The expectation 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
en.wikipedia.org
The evolution of the probability distribution associated with X evolves according to the Fokker–Planck equation
en.wikipedia.org
en.wikipedia.org
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.
What's F and what about the expected drift?
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.
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
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?
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 btw.
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.
>That would really depend on where it came from.
Why? The differential equation is stated (>>).
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
axiomsofchoice.org
(Paragraph 4 and 5 on linear approaches on the non-stochastic care are irrelevant)
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 math.stackexchange.com
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.