## Expected value of integral = integral of expected value?

Hello forumers,

I am studying some of the theory of Brownian motion and stochastic differential equations, and the author of the book I am using (http://www.amazon.com/Numerical-Meth.../dp/0521859719) makes an argument which, despite a week's worth of attempts, I cannot seem to prove.

Though the full argument is somewhat longer, it basically boils down to:

$$\langle \int f(x) dx \rangle=\int \langle f(x) \rangle dx$$

Am I just being thick? The author doesn't substantiate that claim, so I suppose he feels it is rather basic and obvious. But not to me! :(

Does anyone know how to prove (or disprove) that statement?

I wasn't sure if the calculus forum or this forum would be best, so I picked this one first..

Sincerely, my thanks

-clustro
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 Recognitions: Science Advisor The expectation of a random variable can be considered as an integral. As long as the condition for switching the orders of integration holds, then the author's assertion is valid.
 Blog Entries: 1 Recognitions: Homework Help This seems like a weird statement to make. The expected value is going to be a number, as is the integral of a function, and integrating/taking the expected value of a constant is kind of a weird thing to do

Recognitions:
Homework Help

## Expected value of integral = integral of expected value?

$$E_Y \int f(x) dx = \int \int f(x) dx \ g(y) dy =\int\int f(x) g(y) \ dx dy = \int \int f(x)g(y)\ dy dx =\int E_Y f(x)dx$$
If f(x) = a(y) + b(y) x, with Ea(y) = a and Eb(y) = b, EY[a(y) + b(y) x] = a + b x need not be a constant.

 Quote by mathman The expectation of a random variable can be considered as an integral. As long as the condition for switching the orders of integration holds, then the author's assertion is valid.
That's what I was thinking at first. Here is what I tried before:

Symbols:

$$F_R(t)$$ is the random force on a particle at time t. It is the source of Brownian motion.
$$X_R(t)$$ is the displacement of a particle at time t undergoing Brownian motion.
$$t$$ is the time.
$$\Delta t$$ is the time-step.
(The variables with apostrophes are dummy variables).

I left a few constants out to simplify the presentation (all they do is make the units work out.)

$$X_R(t) = \int^{t+\Delta t}_t F_R(t')dt'$$ (1)

The author says that:

$$\langle X_R(t) \rangle = \int^{t+\Delta t}_t \langle F_R(t') \rangle dt'$$ (2)

To get there, I tried doing:

$$\langle X_R(t) \rangle = \int^{t = \infty}_{t = {- \infty}}t'X_R(t')dt'$$ (3)

Substituting (1) into (3)

$$\langle X_R(t) \rangle = \left\langle \int^{t+\Delta t}_t F_R(t') dt' \right\rangle = \int^{t = \infty}_{t = {- \infty}}t' \left [\int^{t'+\Delta t}_{t'} F_R(t'')dt''\right ] dt'$$ (4)

So now we try it the other way, and see if we get the same result as in (4):

$$\langle F_R(t) \rangle = \int^{t = \infty}_{t = -\infty}t' F_R(t')dt'$$ (5)

Substituting (5) into (2):

$$\int^{t+\Delta t}_t \langle F_R(t') \rangle dt' = \int^{t+\Delta t}_t \left [\int^{t' = \infty}_{t' = -\infty} t'' F_R(t'')dt'' \right ] dt'$$ (6)

From what we have just worked out, in order for the author's statement to be true, the right-hand sides of (4) and (6) must be equal:

$$\int^{t = \infty}_{t = {- \infty}}t' \left [\int^{t'+\Delta t}_{t'} F_R(t'')dt''\right ] dt' = \int^{t+\Delta t}_t \left [\int^{t' = \infty}_{t' = -\infty} t'' F_R(t'')dt'' \right ] dt'$$ (7)

I do not see how these two things should be equal. It looks like the integration orders are a flipped, but at the same time it doesn't :(

Looking at EnumaElish's post:

 Quote by EnumaElish $$E_Y \int f(x) dx = \int \int f(x) dx \ g(y) dy =\int\int f(x) g(y) \ dx dy = \int \int f(x)g(y)\ dy dx =\int E_Y f(x)dx$$ If f(x) = a(y) + b(y) x, with Ea(y) = a and Eb(y) = b, EY[a(y) + b(y) x] = a + b x need not be a constant.
It certainly appears similar to what he has posted - its just an integral with the limits switched around. But I am just not seeing it in (7). On the left-hand side of (7), $$t'$$ is outside of the inner integral - I dunno, that just bothers me. I don't thinking moving it to the inner integral changes anything either (...or does it?!) Also, on the right-hand side of (7), the inner integral extends to +/- infinity; what on earth will the outer integral be integrating?!

Any further help in this matter is highly appreciated friends,

-clustro
 Recognitions: Science Advisor The integration limits when you flipped in equation (7) are wrong. Specifically in your first integral, the inner integral goes from t' to t' + ∆t, while your outer integral goes from -∞ to +∞. When you switch, the inner integral will go from t' - ∆t to t', while the outer integral will go from -∞ to +∞.

 Quote by mathman The integration limits when you flipped in equation (7) are wrong. Specifically in your first integral, the inner integral goes from t' to t' + ∆t, while your outer integral goes from -∞ to +∞. When you switch, the inner integral will go from t' - ∆t to t', while the outer integral will go from -∞ to +∞.
I didn't flip anything in (7).

I merely set the right hand sides of (4) and (6) equal to each other.

They should (in theory), be the same, but they didn't come out that way.

How do you even come up with those changes limits of integration you suggest?

The thing is, this problem is not solely switching the order of integration. In the left-hand side of (7), $$t'$$ is outside of the inner integral, whereas in the right-hand side of (7), $$t''$$ is inside the inner integral. I do not see how that situation would ever allow these two integrals to be equal.
 Recognitions: Homework Help Science Advisor Isn't the t' on the left-hand side of (7) a constant with respect to the inner integral, and if so, can't it be placed within the inner integral?

 Quote by EnumaElish Isn't the t' on the left-hand side of (7) a constant with respect to the inner integral, and if so, can't it be placed within the inner integral?
Yes, that is correct, but it won't have the same effect since, you won't be integrating with respect to t'' in that statement. That's probably the main thing that is confusing me.

Another idea I had is to differentiate (7) until we are left with the main "internal" expressions.

$$\int^{t = \infty}_{t = {- \infty}}t' \left [\int^{t'+\Delta t}_{t'} F_R(t'')dt''\right ] dt' = \int^{t+\Delta t}_t \left [\int^{t' = \infty}_{t' = -\infty} t'' F_R(t'')dt'' \right ] dt'$$

Ignoring the limits of integration, differentiating with respect to $$dt'$$ yields:

$$t' \int F_R(t'')dt'' = \int t'' F_R(t'')dt''$$

When differentiate with respect to $$dt''$$, which yields:

$$t' F_R(t'') = t'' F_R(t'')$$

Which shows that $$t' = t''$$.

If these two dummy-variables are equal, what does that get us? I suppose then, it should not matter whether $$t'$$ or $$t''$$ are outside or inside the integral? O_O

Also, EnumaElish, I have a question about the proof you gave. Did you come up with it, or is it from another source? If so, can you provide a link? Also, what does $$E_Y$$ mean? :(

 Quote by clustro Hello forumers, I am studying some of the theory of Brownian motion and stochastic differential equations, and the author of the book I am using (http://www.amazon.com/Numerical-Meth.../dp/0521859719) makes an argument which, despite a week's worth of attempts, I cannot seem to prove. Though the full argument is somewhat longer, it basically boils down to: $$\langle \int f(x) dx \rangle=\int \langle f(x) \rangle dx$$ Am I just being thick? The author doesn't substantiate that claim, so I suppose he feels it is rather basic and obvious. But not to me! :( Does anyone know how to prove (or disprove) that statement? I wasn't sure if the calculus forum or this forum would be best, so I picked this one first.. Sincerely, my thanks -clustro
You are seeing a bunch of mysterious calculations in response to this query, most of which are wrong.

Some one noted that this is apparently related to interchange of the order of integration -- Fubini's theorem, and that is probably the heart of the matter.

However, in order to use Fubini's theorem you need to more explicitly describe your situation in terms of a function of two variables.

This equation

$$\langle \int f(x) dx \rangle=\int \langle f(x) \rangle dx$$

does not make sense unless and until you have described what you mean by $$\int f(x) dx$$ as a random variable. As it stands it seems to be simply a number.

What one would expect is that, with a more clear definition of what $$\int f(x) dx$$ means and what expectation means that what you have is something more like

$$\langle \int f(x) dx \rangle=\int f(x,y)dxdy = \int f(x,y)dydx=\int \langle f(x) \rangle dx$$

This of course requires that the conditions for the application of Fubini's theorem apply -- that the integrals exist for the absolute value of f in place of f.

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Homework Help
 Quote by DrRocket You are seeing a bunch of mysterious calculations in response to this query, most of which are wrong.
Do you mind pointing out specific errors that you see?

 Quote by EnumaElish $$E_Y \int f(x) dx = \int \int f(x) dx \ g(y) dy =\int\int f(x) g(y) \ dx dy = \int \int f(x)g(y)\ dy dx =\int E_Y f(x)dx$$ If f(x) = a(y) + b(y) x, with Ea(y) = a and Eb(y) = b, EY[a(y) + b(y) x] = a + b x need not be a constant.
 Quote by EnumaElish Do you mind pointing out specific errors that you see?
The errors are conceptual, but here is one specific example based on the above. Since $$f$$ is treated as a funtion of $$x$$ along the calculation should read

$$E_Y \int f(x) dx = \int \int f(x) dx \ g(y) dy = \int f(x) dx \int g(y) dy = \int f(x) dx$$

which doesn't shed much light on things.

The problem is that, as I said, one need to make clear how one is thinking of $$\int f(x) dx$$ as a random variable. That is probably what you meant to imply by introducing the variable y, but to do that you need to show the dependence of f on not only x but also on y.

One that is done then one can apply the Fubini theorem and the rest is pretty easy. The difficult here lies in an incompletely or erroneously specified problem and not in the attempted solutions, which fail because the problem is not properly formulated.

Now if $$f$$were a function of both $$x$$ and$$y$$ and , you have
$$E_Y \int f(x,y) dx = \int \int f(x,y) dx \ g(y) dy =\int\int f(x,y) g(y) \ dx dy = \int \int f(x,y)g(y) dy dx =\int E_Y f(x, \centerdot)dx$$

And this is nothing but an application of Fubini's theorem.

Depending on one's background it might be easier to see this if we use general measures, say $$\mu$$ and $$\nu$$. Here $$\nu$$ is the probability measure and $$/mu$$ is whatever measue is used to integrate $$f$$, which one suspects is probably actually $$\nu$$ although the OP was not clear on that point. In this case the above calculation becomes

$$E_Y \int f(x,y) d \mu(x) = \int \int f(x,y) d \mu(x) d \nu (y)= \int\int f(x,y) d \nu (y) d \mu (x) =\int E_Y f(x, \centerdot) d \mu (x)$$

I you are not familiar with the general theory of measure and integration then simply refer to the earlier calculation.

Recognitions:
 Quote by clustro That's what I was thinking at first. Here is what I tried before: Symbols: $$F_R(t)$$ is the random force on a particle at time t. It is the source of Brownian motion. $$X_R(t)$$ is the displacement of a particle at time t undergoing Brownian motion. $$t$$ is the time. $$\Delta t$$ is the time-step. (The variables with apostrophes are dummy variables). I left a few constants out to simplify the presentation (all they do is make the units work out.) $$X_R(t) = \int^{t+\Delta t}_t F_R(t')dt'$$ (1) The author says that: $$\langle X_R(t) \rangle = \int^{t+\Delta t}_t \langle F_R(t') \rangle dt'$$ (2) To get there, I tried doing: $$\langle X_R(t) \rangle = \int^{t = \infty}_{t = {- \infty}}t'X_R(t')dt'$$ (3) Substituting (1) into (3) $$\langle X_R(t) \rangle = \left\langle \int^{t+\Delta t}_t F_R(t') dt' \right\rangle = \int^{t = \infty}_{t = {- \infty}}t' \left [\int^{t'+\Delta t}_{t'} F_R(t'')dt''\right ] dt'$$ (4) So now we try it the other way, and see if we get the same result as in (4): $$\langle F_R(t) \rangle = \int^{t = \infty}_{t = -\infty}t' F_R(t')dt'$$ (5) Substituting (5) into (2): $$\int^{t+\Delta t}_t \langle F_R(t') \rangle dt' = \int^{t+\Delta t}_t \left [\int^{t' = \infty}_{t' = -\infty} t'' F_R(t'')dt'' \right ] dt'$$ (6) From what we have just worked out, in order for the author's statement to be true, the right-hand sides of (4) and (6) must be equal: $$\int^{t = \infty}_{t = {- \infty}}t' \left [\int^{t'+\Delta t}_{t'} F_R(t'')dt''\right ] dt' = \int^{t+\Delta t}_t \left [\int^{t' = \infty}_{t' = -\infty} t'' F_R(t'')dt'' \right ] dt'$$ (7) I do not see how these two things should be equal. It looks like the integration orders are a flipped, but at the same time it doesn't :( Looking at EnumaElish's post: It certainly appears similar to what he has posted - its just an integral with the limits switched around. But I am just not seeing it in (7). On the left-hand side of (7), $$t'$$ is outside of the inner integral - I dunno, that just bothers me. I don't thinking moving it to the inner integral changes anything either (...or does it?!) Also, on the right-hand side of (7), the inner integral extends to +/- infinity; what on earth will the outer integral be integrating?! Any further help in this matter is highly appreciated friends, -clustro
I was a little careless before. On closer examination, I believe the problem lies in equation (3). I have no idea what you were thinking, but it is not an accurate description of $$\langle X_R(t) \rangle$$.

Recognitions:
Homework Help
 Quote by DrRocket The errors are conceptual, but here is one specific example based on the above. Since $$f$$ is treated as a funtion of $$x$$ along the calculation should read $$E_Y \int f(x) dx = \int \int f(x) dx \ g(y) dy = \int f(x) dx \int g(y) dy = \int f(x) dx$$ which doesn't shed much light on things. The problem is that, as I said, one need to make clear how one is thinking of $$\int f(x) dx$$ as a random variable. That is probably what you meant to imply by introducing the variable y, but to do that you need to show the dependence of f on not only x but also on y. One that is done then one can apply the Fubini theorem and the rest is pretty easy. The difficult here lies in an incompletely or erroneously specified problem and not in the attempted solutions, which fail because the problem is not properly formulated. Now if $$f$$were a function of both $$x$$ and$$y$$ and , you have $$E_Y \int f(x,y) dx = \int \int f(x,y) dx \ g(y) dy =\int\int f(x,y) g(y) \ dx dy = \int \int f(x,y)g(y) dy dx =\int E_Y f(x, \centerdot)dx$$ And this is nothing but an application of Fubini's theorem. Depending on one's background it might be easier to see this if we use general measures, say $$\mu$$ and $$\nu$$. Here $$\nu$$ is the probability measure and $$/mu$$ is whatever measue is used to integrate $$f$$, which one suspects is probably actually $$\nu$$ although the OP was not clear on that point. In this case the above calculation becomes $$E_Y \int f(x,y) d \mu(x) = \int \int f(x,y) d \mu(x) d \nu (y)= \int\int f(x,y) d \nu (y) d \mu (x) =\int E_Y f(x, \centerdot) d \mu (x)$$ I you are not familiar with the general theory of measure and integration then simply refer to the earlier calculation.
 Recognitions: Homework Help Science Advisor clustro, when you write $$\langle X_R(t) \rangle = \int^{t = \infty}_{t = {- \infty}}t' X_R(t')dt'$$ should that be $$\langle X_R(t) \rangle = \int^{t = \infty}_{t = {- \infty}}t' dX_R(t')$$ instead? (using DrRocket's tip)
 Well, upon further reading into Dr. Rocket's comment, it appears I have been treading water. Ever since I can remember, I thought the $$\langle f(x) \rangle$$ bracket operators were just an alternative notation for $$E(f(x))$$. Apparently it seems, that the bracket operators mean either the "time-average" of the function or the "ensemble average" of the function - which is distinctly different from the expected value of a function. I am not familiar really with these concepts, so I'm going to have to do some more reading so I can speak intelligently on the subject. Sorry guys :[