Expectation of a product of Brownian Motions

[jamesa00789]

Let Bt1, Bt2 and Bt3 be standard Brownian motions with ~N(0,1).

Then what is E[Bt1.Bt2.Bt3] ?

Any help would be much appreciated.

 

[chiro]

Let Bt1, Bt2 and Bt3 be standard Brownian motions with ~N(0,1).

Then what is E[Bt1.Bt2.Bt3] ?

Any help would be much appreciated.

Hey jamesa00789 and welcome to the forums.

What are the conditions for each BM? Are they independent? Do they refer to different intervals for the same process? Maybe some overlap in intervals?

If they are truly independent you can use the property that E[XY] = E[X]E[Y] and take it from there.

 

[jamesa00789]

Yes they are of the same standard brownian motion at different time intervals.

 

[chiro]

Yes they are of the same standard brownian motion at different time intervals.

If they are are at non-overlapping intervals, then use the definition of the Brownian motion. If they are over-lapping, then decompose it into processes that are non-overlapping and take care of parts that are overlapping.

Using this, the fact that E[XY] = E[X]E[Y], and the definition of BM, what do you get?

 

[blue_raver22]

I would like to clarify that Brownian motion is a random process, meaning that the values it takes at any given time are unpredictable. Therefore, the expectation of a product of Brownian motions cannot be determined with certainty. However, we can use mathematical tools to estimate the expected value.

The expected value of a product of Brownian motions can be calculated using the properties of normal distributions. Since Bt1, Bt2, and Bt3 are standard Brownian motions, they all have a mean of 0 and a variance of 1. This means that their product would follow a normal distribution with a mean of 0 and a variance of 1.

Using the properties of normal distributions, we can calculate the expected value of the product by multiplying the means of each Brownian motion. In this case, the expected value would be 0, since 0 multiplied by any number is still 0.

In summary, the expected value of Bt1.Bt2.Bt3 is 0, but it is important to note that this is an estimated value and not a certain one. The actual value could vary due to the random nature of Brownian motion.