Proving a process is Brownian Motion

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SUMMARY

The process \{X(t)\}_{t\geq 0}, defined as X(t)=\rho B_1(t)+\sqrt{1-\rho^2}B_2(t) where \rho\in(0,1) and B_1(t), B_2(t) are independent standard Brownian motions, is indeed a standard Brownian motion. The key steps to prove this include demonstrating that X(0)=0 and that increments such as X(t_2)-X(t_1) are normally distributed with mean 0 and variance t_2-t_1. Additionally, establishing the independence of increments requires showing that the covariance of X(t_3)-X(t_2) and X(t_2)-X(t_1) is zero, which is valid under the Gaussian assumption.

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Homework Statement


Is the process \{X(t)\}_{t\geq 0}, where X(t)=\rho B_1(t)+\sqrt{1-\rho^2}B_2(t) Standard Brownian Motion?

Where \rho\in(0,1), \ B_1(t) and B_2(t) are independent standard brownian motions

Homework Equations


The Attempt at a Solution



Obviously X(0)=0. Now let 0\leq t_1<t_2<t_3. Then, I can show that X(t_2)-X(t_1)\sim\mathcal{N}(0, t_2-t_1). My problem is showing that X(t_3)-X(t_2) and X(t_2)-X(t_1) are independent. I can show their covariance is 0, but that only implies independence if the process is Gaussian, which I have not shown. Any help would be much appreciated!
 
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IniquiTrance said:

Homework Statement


Is the process \{X(t)\}_{t\geq 0}, where X(t)=\rho B_1(t)+\sqrt{1-\rho^2}B_2(t) Standard Brownian Motion?

Where \rho\in(0,1), \ B_1(t) and B_2(t) are independent standard brownian motions


Homework Equations





The Attempt at a Solution



Obviously X(0)=0. Now let 0\leq t_1<t_2<t_3. Then, I can show that X(t_2)-X(t_1)\sim\mathcal{N}(0, t_2-t_1). My problem is showing that X(t_3)-X(t_2) and X(t_2)-X(t_1) are independent. I can show their covariance is 0, but that only implies independence if the process is Gaussian, which I have not shown. Any help would be much appreciated!

For i = 1,2 and j = 1,2, are the ##B_i(t_3)-B_i(t_2)## independent of the ##B_j(t_2)-B_j(t_1)##?
 
Ray Vickson said:
For i = 1,2 and j = 1,2, are the ##B_i(t_3)-B_i(t_2)## independent of the ##B_j(t_2)-B_j(t_1)##?

Hi Ray, yes they are.
 
Not sure how to proceed.
 
Are you implying that if A\perp B\perp C \perp D, then A+B \perp C+D, where \perp means independent?
 
IniquiTrance said:
Are you implying that if A\perp B\perp C \perp D, then A+B \perp C+D, where \perp means independent?

More than that: if A\perp B\perp C \perp D, then f(A,B) \perp g(C,D) for any (measurable) functions f(.,.) and g(.,.).
 
Ray Vickson said:
More than that: if A\perp B\perp C \perp D, then f(A,B) \perp g(C,D) for any (measurable) functions f(.,.) and g(.,.).

Thanks. How can I go about proving that?
 
IniquiTrance said:
Thanks. How can I go about proving that?

It depends on what you know already. Anyway, all you need is the result for A+B and C+D, and you can use any number of standard tools, but first you need to know what those tools are. Do you know about characteristic functions? What can you say about the density (or cdf) of a set of independent random variables? How can you test whether two random variables (A+B and C+D in this case) are independent?

All of these things can be found in books and articles---many freely available on-line---so I will leave it to you to search for the answers if you do not know them already. If you do know them already, just think about how you would use them.
 

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