Is (Wt)2 a Brownian Motion? Analyzing the Distribution of Vt+s - Vs

[Firepanda]

I need to show that (W_t)² is a brownian motion

So let V_t = (W_t)²

I need to first show that V_t+s - V_s ~ N(0,t)

V_t+s - V_s = (W_t+s)² - (W_s)² = (W_t+s + W_s)(W_t+s - W_s)

(W_t+s - W_s) ~ N(0,t)

But is (W_t+s + W_s) ~ N(0,t)?

If it is what happens when I multiply two RV's that are normally distributed together? What can I say about the variance of the new distribution?

Thanks

 

[Ray Vickson]

I need to show that (W_t)² is a brownian motion

So let V_t = (W_t)²

I need to first show that V_t+s - V_s ~ N(0,t)

V_t+s - V_s = (W_t+s)² - (W_s)² = (W_t+s + W_s)(W_t+s - W_s)

(W_t+s - W_s) ~ N(0,t)

But is (W_t+s + W_s) ~ N(0,t)?

If it is what happens when I multiply two RV's that are normally distributed together? What can I say about the variance of the new distribution?

Thanks

You are being asked to show something that is FALSE: W_t^2 is not a Brownian motion, unless you are using an unusual definition of Brownian motion. The square of a normal random variable is essentially a chi-squared random variable with parameter 1; that is N(0,1)^2 = Chi^2(1).

RGV

 

[Firepanda]

Hi Ray

Thanks for responding!

Yeah sorry, the question was show if it is or is not a Brownian Motion, for some odd reason I assumed it was because the rest of the examples were

How do I show it is not a brownian motion then, which property of a BM does not hold?
Thanks

 

[Ray Vickson]

Hi Ray

Thanks for responding!

Yeah sorry, the question was show if it is or is not a Brownian Motion, for some odd reason I assumed it was because the rest of the examples were

How do I show it is not a brownian motion then, which property of a BM does not hold?
Thanks

What properties of BM are you trying to verify or deny? I have seen alternative definitions of BM; which one are you using?

RGV

 

[Firepanda]

These:

Also in the question I should note that W is a standard brownian motion

 

[Ray Vickson]

These:

Also in the question I should note that W is a standard brownian motion

Does property (i) hold?

RGV

 

[Firepanda]

That's what I was attempting in my original post.

So my brownian motion V is not normally distributed? Not sure how to show that (apart from my earlier attempt)