What probability density is used in Brownian motion?

Earthland · Dec 4, 2015

Homework Statement

I have a free Brownian particle and its coordinate is given as a function of time:

Displayed equation: x(t) = x(t0) + v(t0)/γ [1 − e^(−γ(t−t0))] + (1/γ) ∫ from t0 to t of [1 − e^(−γ(t−t′))] ζ(t′)/m dt′.

And its first moment, or mean, is given as

Equation for the mean position of a damped Brownian particle: <x(t)> = x(t0) + (v(t0)/gamma)[1 - e^(-gamma(t - t0))].

But what kind of probability density was used to calculate this first moment?

Homework Equations

I know that the first moment is calculated by using probability density

<x> = ∫x*W(x)dx from -infinity to +infinity.

X itself is dependent on t, so is it W(x) or W(x,t) or W(t)? I'm little lost on this one.

The Attempt at a Solution

I did find some kind of density function in regards to Brownian motion

Gaussian heat-kernel density f_t(x) = (1/sqrt(2 pi t)) exp(-x^2/(2t)), a function of position x and time t, shown as an equation.

But I can't see how this would yield the correct result. And still, must I integrate over x or t?

Orodruin · Dec 4, 2015

Just take the expectation value of your first expression (expectation values are linear) keeping in mind that the only actually random variable in it is the driving force ##\zeta(t')##.

What probability density is used in Brownian motion?

Homework Statement

Homework Equations

The Attempt at a Solution

Discussion

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