Prove the following is standard Brownian from this expression of non-standard Brownian motion

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I have attempted the following question but fear I may have sort of perhaps made my solution fit the answer for convenience sake. I have the following:

Y is Brownian motion with variance parameter $sigma^2$. Prove the process X with
$$
X(t) = Yleft(fractsigma^2right)
$$
is a standard Brownian motion

My attempt is as follows:

We know: $ mboxVar(Y(t)) = E[Y(t+s) - Y(s)^2] = sigma^2t$

This is a known result. We now try to prove $mboxVar(X(t))=t$
$$
begineqnarray*
mboxVar(X(t)) & = & E[X(t+s) - X(s)^2] \
& = & Eleft[leftY(fract+ssigma^2) - Y(fracssigma^2)right^2right] \
&=& Eleft[ left frac1sigmaY(t+s) - frac1sigmaY(s) right^2 right] \
&=& Eleft[frac1sigma^2 left Y(t+s) - Y(s) right^2 right]\
&=& frac1sigma^2 Eleft[ left Y(t+s) - Y(s) right^2 right] \
&=& frac1sigma^2( sigma^2 t)\
&=& t
endeqnarray*
$$
Thus we prove it has variance of a standard Brownian process. This is the only difficult part I struggled in proving and think I may have perfomed steps that aren't true such as the third step. The rest of the requirement for standard Brownian motion were simple to prove. Please may I ask for your guidance and assistance in understanding how to prove this. I also messed around with the idea of fidgeting with the transition function but...did not progress far.

stochastic-processes normal-distribution brownian-motion

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asked Jul 28 at 20:57

Yes

203

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up vote
1
down vote

favorite

I have attempted the following question but fear I may have sort of perhaps made my solution fit the answer for convenience sake. I have the following:

Y is Brownian motion with variance parameter $sigma^2$. Prove the process X with
$$
X(t) = Yleft(fractsigma^2right)
$$
is a standard Brownian motion

My attempt is as follows:

We know: $ mboxVar(Y(t)) = E[Y(t+s) - Y(s)^2] = sigma^2t$

This is a known result. We now try to prove $mboxVar(X(t))=t$
$$
begineqnarray*
mboxVar(X(t)) & = & E[X(t+s) - X(s)^2] \
& = & Eleft[leftY(fract+ssigma^2) - Y(fracssigma^2)right^2right] \
&=& Eleft[ left frac1sigmaY(t+s) - frac1sigmaY(s) right^2 right] \
&=& Eleft[frac1sigma^2 left Y(t+s) - Y(s) right^2 right]\
&=& frac1sigma^2 Eleft[ left Y(t+s) - Y(s) right^2 right] \
&=& frac1sigma^2( sigma^2 t)\
&=& t
endeqnarray*
$$
Thus we prove it has variance of a standard Brownian process. This is the only difficult part I struggled in proving and think I may have perfomed steps that aren't true such as the third step. The rest of the requirement for standard Brownian motion were simple to prove. Please may I ask for your guidance and assistance in understanding how to prove this. I also messed around with the idea of fidgeting with the transition function but...did not progress far.

stochastic-processes normal-distribution brownian-motion

share|cite|improve this question

asked Jul 28 at 20:57

Yes

203

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have attempted the following question but fear I may have sort of perhaps made my solution fit the answer for convenience sake. I have the following:

Y is Brownian motion with variance parameter $sigma^2$. Prove the process X with
$$
X(t) = Yleft(fractsigma^2right)
$$
is a standard Brownian motion

My attempt is as follows:

We know: $ mboxVar(Y(t)) = E[Y(t+s) - Y(s)^2] = sigma^2t$

This is a known result. We now try to prove $mboxVar(X(t))=t$
$$
begineqnarray*
mboxVar(X(t)) & = & E[X(t+s) - X(s)^2] \
& = & Eleft[leftY(fract+ssigma^2) - Y(fracssigma^2)right^2right] \
&=& Eleft[ left frac1sigmaY(t+s) - frac1sigmaY(s) right^2 right] \
&=& Eleft[frac1sigma^2 left Y(t+s) - Y(s) right^2 right]\
&=& frac1sigma^2 Eleft[ left Y(t+s) - Y(s) right^2 right] \
&=& frac1sigma^2( sigma^2 t)\
&=& t
endeqnarray*
$$
Thus we prove it has variance of a standard Brownian process. This is the only difficult part I struggled in proving and think I may have perfomed steps that aren't true such as the third step. The rest of the requirement for standard Brownian motion were simple to prove. Please may I ask for your guidance and assistance in understanding how to prove this. I also messed around with the idea of fidgeting with the transition function but...did not progress far.

stochastic-processes normal-distribution brownian-motion

share|cite|improve this question

asked Jul 28 at 20:57

Yes

203

I have attempted the following question but fear I may have sort of perhaps made my solution fit the answer for convenience sake. I have the following:

Y is Brownian motion with variance parameter $sigma^2$. Prove the process X with
$$
X(t) = Yleft(fractsigma^2right)
$$
is a standard Brownian motion

My attempt is as follows:

We know: $ mboxVar(Y(t)) = E[Y(t+s) - Y(s)^2] = sigma^2t$

This is a known result. We now try to prove $mboxVar(X(t))=t$
$$
begineqnarray*
mboxVar(X(t)) & = & E[X(t+s) - X(s)^2] \
& = & Eleft[leftY(fract+ssigma^2) - Y(fracssigma^2)right^2right] \
&=& Eleft[ left frac1sigmaY(t+s) - frac1sigmaY(s) right^2 right] \
&=& Eleft[frac1sigma^2 left Y(t+s) - Y(s) right^2 right]\
&=& frac1sigma^2 Eleft[ left Y(t+s) - Y(s) right^2 right] \
&=& frac1sigma^2( sigma^2 t)\
&=& t
endeqnarray*
$$
Thus we prove it has variance of a standard Brownian process. This is the only difficult part I struggled in proving and think I may have perfomed steps that aren't true such as the third step. The rest of the requirement for standard Brownian motion were simple to prove. Please may I ask for your guidance and assistance in understanding how to prove this. I also messed around with the idea of fidgeting with the transition function but...did not progress far.

stochastic-processes normal-distribution brownian-motion

share|cite|improve this question

asked Jul 28 at 20:57

Yes

203

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 20:57

Yes

203

asked Jul 28 at 20:57

Yes

203

asked Jul 28 at 20:57

Yes

203

203

add a comment | 

add a comment | 

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accepted

Your th ird step can be justified as follows: it is enough to show that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$. We know that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $Y(frac t sigma^2)$ and $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$ has the same distribution as $frac Y(t) sigma$. But $Y(t)$ is normal so $Y(frac t sigma^2)$ has the same distribution as $frac Y(t) sigma$.

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answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
  – Yes
  Jul 29 at 9:37
  

  
  
  
  

add a comment | 

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Your th ird step can be justified as follows: it is enough to show that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$. We know that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $Y(frac t sigma^2)$ and $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$ has the same distribution as $frac Y(t) sigma$. But $Y(t)$ is normal so $Y(frac t sigma^2)$ has the same distribution as $frac Y(t) sigma$.

share|cite|improve this answer

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
  – Yes
  Jul 29 at 9:37
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Your th ird step can be justified as follows: it is enough to show that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$. We know that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $Y(frac t sigma^2)$ and $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$ has the same distribution as $frac Y(t) sigma$. But $Y(t)$ is normal so $Y(frac t sigma^2)$ has the same distribution as $frac Y(t) sigma$.

share|cite|improve this answer

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
  – Yes
  Jul 29 at 9:37
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Your th ird step can be justified as follows: it is enough to show that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$. We know that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $Y(frac t sigma^2)$ and $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$ has the same distribution as $frac Y(t) sigma$. But $Y(t)$ is normal so $Y(frac t sigma^2)$ has the same distribution as $frac Y(t) sigma$.

share|cite|improve this answer

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

Your th ird step can be justified as follows: it is enough to show that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$. We know that $Y(frac t+s sigma^2)-Y(frac s sigma^2)$ has the same distribution as $Y(frac t sigma^2)$ and $frac 1 sigma Y( t+s)-frac 1 sigma Y(s)$ has the same distribution as $frac Y(t) sigma$. But $Y(t)$ is normal so $Y(frac t sigma^2)$ has the same distribution as $frac Y(t) sigma$.

share|cite|improve this answer

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

answered Jul 28 at 23:27

Kavi Rama Murthy

19.8k2829

19.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
  – Yes
  Jul 29 at 9:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
  – Yes
  Jul 29 at 9:37
  

  
  
  
  

Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
– Yes
Jul 29 at 9:37

Thank you very much. This was very well explained and useful. I am now able to complete further activities like this in the future.
– Yes
Jul 29 at 9:37

add a comment | 

 

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