I Calculating the expected value of the square of an integral of Brownian Motion

AI Thread Summary
The discussion revolves around calculating the expected value of the square of an integral of standard one-dimensional Brownian motion, specifically E[(1/T ∫_0^T W_t dt)^2]. Participants express confusion over the simplification of terms and the meaning of T, questioning whether it represents a stopping time or simply an upper limit of integration. Clarifications are made regarding the notation used for the expectations operator, which some find inconsistent. A link to the Itô isometry is shared as a potential resource for solving the problem. Ultimately, the original poster reports that they have resolved their query.
JohanL
Messages
154
Reaction score
0
For a standard one-dimensional Brownian motion W(t), calculate:

$$E\bigg[\Big(\frac{1}{T}\int\limits_0^TW_t\, dt\Big)^2\bigg]$$I can't figure out how the middle term simplifies.

$$
\mathsf E\left(\int_0^T W_t\mathrm dt\right)^2 = \mathsf E\left[T^2W_T^2\right] - 2T\mathsf E\left[W_T\int_0^T t\mathrm dW_t\right]+\mathsf E\left(\int_0^T t\mathrm dW_t\right)^2
$$
$$
= T^3- 2T\int_0^Tt\mathrm dt+\int_0^Tt^2\mathrm dt
$$

i.e. why is

$$
\mathsf 2T\mathsf E\left[W_T\int_0^T t\mathrm dW_t\right]
= 2T\int_0^Tt\mathrm dt
$$
?
 
Physics news on Phys.org
will you tell us what ##T## is? The standard usage of a capital letter ##T## would be for a stopping time (i.e. a random variable), but this seems to contradict other equations... you also haven't consistently used brackets for the expectations operator, which makes this not so easy to read.
 
StoneTemplePython said:
will you tell us what ##T## is? The standard usage of a capital letter ##T## would be for a stopping time (i.e. a random variable), but this seems to contradict other equations.

I found the exercise and solution online. They don't say anything about T.

Im guessing its just the upper limit of integration and not a stopping time if you say it contradicts the other equations.

StoneTemplePython said:
you also haven't consistently used brackets for the expectations operator, which makes this not so easy to read.

Sorry about that. Seems like i can't edit now?
 
Last edited:
Last edited:
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top