pde with expectation values

Hi,

I'm looking to solve a differential equation that mixes expectation values and the actual function,

for example:

$\displaystyle f\left( x,t \right) \quad =\quad { Aexp\{ \quad E[\quad \log { (1+\frac { \cfrac { \partial }{ \partial t } f\left( x,t \right) }{ f\left( x,t \right) } )\quad } ]\quad \} \quad }$

where E[ .. ] is the expected value over t (this is an approximated equation, f(x,t) is stochastic)

I'm not expecting to get the actual solution here (and i realize i didn't supply enough details to get a solution), but I need some help on get started- how do you call such differential equations, that involve the function's expectation value? are there some known cases where they are put to use?

thank you

Re: pde with expectation values

Hey dudyu.

The expectation has a well defined definition for both discrete and continuous random variables for E[g(X)] for any function g(X).

These types of equations are called Stochastic Differential Equations or SDE's for short.

If you are taking the expectation with respect to X and t is deterministic, then you can calculate E[g(X,t)] where the g(X,t) is inside the expectation if you have the PDF for X.

I'm speculating that t is not a random variable, but please inform us if that is not correct.

That's basically all there is to it.

Even if you can't get an analytic solution easily, you can use one of the many numerical packages to get a numeric solution in terms of a function of your time variable t (since t is a variable independent from X and is not a random variable) and then you will get some function in terms of t.

Now since you mention that this is an implicit form, you will have to use the use the chain rule of the Fundamental Theorem of Calculus to get rid of the integral.

The Integral for the expectation will look like E[g(X,t)] = Integral (-infinity to infinity) g(X,t)*f(x)dx where f(x) is the PDF for the random variable X (assuming your expectation is with respect to the random variable X).

Re: pde with expectation values

Quote:

Originally Posted by

**chiro** Hey dudyu.

The expectation has a well defined definition for both discrete and continuous random variables for E[g(X)] for any function g(X).

These types of equations are called Stochastic Differential Equations or SDE's for short.

If you are taking the expectation with respect to X and t is deterministic, then you can calculate E[g(X,t)] where the g(X,t) is inside the expectation if you have the PDF for X.

I'm speculating that t is not a random variable, but please inform us if that is not correct.

That's basically all there is to it.

Even if you can't get an analytic solution easily, you can use one of the many numerical packages to get a numeric solution in terms of a function of your time variable t (since t is a variable independent from X and is not a random variable) and then you will get some function in terms of t.

Now since you mention that this is an implicit form, you will have to use the use the chain rule of the Fundamental Theorem of Calculus to get rid of the integral.

The Integral for the expectation will look like E[g(X,t)] = Integral (-infinity to infinity) g(X,t)*f(x)dx where f(x) is the PDF for the random variable X (assuming your expectation is with respect to the random variable X).

Hi, thanks very much for the reply, but I don't really get it- how can I calculate the expectation value if it is dependent on the function that I'm looking for in the first place?

The equation is a relation between a function and its expectation value, and the solution is a function that satisfies this relation..

Re: pde with expectation values

You need to differentiate out the integral. Here is basically the kind of tool used for this sort of thing:

Differentiation under the integral sign - Wikipedia, the free encyclopedia

You'll have to take the logarithm of both sides before you do this and get the integral term as a single linear term, but the above idea is the key catalyst.