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

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SUMMARY

The discussion centers on the assertion that (Wt)² is a Brownian motion, which is established as false. The user, RGV, seeks to demonstrate that Vt = (Wt)² does not satisfy the properties of a Brownian motion. Key points include the transformation of the standard Brownian motion W into Vt and the realization that the distribution of Vt is not normally distributed, as it resembles a chi-squared distribution instead. The conversation emphasizes the need to identify which properties of Brownian motion are violated by Vt.

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

So let Vt = (Wt)2

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

Vt+s - Vs = (Wt+s)2 - (Ws)2 = (Wt+s + Ws)(Wt+s - Ws)

(Wt+s - Ws) ~ N(0,t)

But is (Wt+s + Ws) ~ 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
 
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Firepanda said:
I need to show that (Wt)2 is a brownian motion

So let Vt = (Wt)2

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

Vt+s - Vs = (Wt+s)2 - (Ws)2 = (Wt+s + Ws)(Wt+s - Ws)

(Wt+s - Ws) ~ N(0,t)

But is (Wt+s + Ws) ~ 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
 
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
 
Firepanda said:
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
 
These:

2q1updw.png


Also in the question I should note that W is a standard brownian motion
 
Firepanda said:
These:

2q1updw.png


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

Does property (i) hold?

RGV
 
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)
 

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