Mixing
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Riemann Integral of Brownian Motion
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I am taking an intro to financial mathematics class and came across the above equation. Unfortunately, I am completely unable to decipher the meaning. I am assuming that Y subscript i is your random variable (could be wrong), however, I can't see how they arrived at the following expansion. Do they successive Y terms represent random values the variable could take at each time step? Also, normally in a Riemann sum (at least at the level I have studied calculus) you end up with a 1/n term, however, here I see (n - 1), (n - 2), etc. and I would like to know how they arrived at this result. Either a qualitative explanation or more complete proof of this explanation would be much appreciated!
random-variables brownian-motion
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
asked Jul 15 at 9:54
Reti2roll
91
add a comment |Â
up vote
1
down vote
favorite
I am taking an intro to financial mathematics class and came across the above equation. Unfortunately, I am completely unable to decipher the meaning. I am assuming that Y subscript i is your random variable (could be wrong), however, I can't see how they arrived at the following expansion. Do they successive Y terms represent random values the variable could take at each time step? Also, normally in a Riemann sum (at least at the level I have studied calculus) you end up with a 1/n term, however, here I see (n - 1), (n - 2), etc. and I would like to know how they arrived at this result. Either a qualitative explanation or more complete proof of this explanation would be much appreciated!
random-variables brownian-motion
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
asked Jul 15 at 9:54
Reti2roll
91
Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am taking an intro to financial mathematics class and came across the above equation. Unfortunately, I am completely unable to decipher the meaning. I am assuming that Y subscript i is your random variable (could be wrong), however, I can't see how they arrived at the following expansion. Do they successive Y terms represent random values the variable could take at each time step? Also, normally in a Riemann sum (at least at the level I have studied calculus) you end up with a 1/n term, however, here I see (n - 1), (n - 2), etc. and I would like to know how they arrived at this result. Either a qualitative explanation or more complete proof of this explanation would be much appreciated!
random-variables brownian-motion
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
asked Jul 15 at 9:54
Reti2roll
91
I am taking an intro to financial mathematics class and came across the above equation. Unfortunately, I am completely unable to decipher the meaning. I am assuming that Y subscript i is your random variable (could be wrong), however, I can't see how they arrived at the following expansion. Do they successive Y terms represent random values the variable could take at each time step? Also, normally in a Riemann sum (at least at the level I have studied calculus) you end up with a 1/n term, however, here I see (n - 1), (n - 2), etc. and I would like to know how they arrived at this result. Either a qualitative explanation or more complete proof of this explanation would be much appreciated!
random-variables brownian-motion
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
asked Jul 15 at 9:54
Reti2roll
91
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
edited Jul 15 at 9:55
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 15 at 9:54
Reti2roll
91
asked Jul 15 at 9:54
Reti2roll
91
asked Jul 15 at 9:54
Reti2roll
91
91
Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16
add a comment |Â
Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16
Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16
add a comment |Â
1 Answer
1
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0
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If $Y_i$ represent a stochastic process, where each $Y_i$ is a normally distributed random variable, then what the notes are saying is that $Y_i$'s are not independent, but the random variables $Z_i$, obtained from their 'differences' or increments are independent, i.e. $Z_i=Y_i+1-Y_i$ are independent.
So in general, given the importance of independence, one tries to 'handle' problems regarding $Y_i$'s by somehow `transforming' them to problems regarding the increments.
For example, $Y_2=Y_1+(Y_2-Y_1)$ and $Y_1$ and $Y_2-Y_1$ are independent.
$Y_i $ as you say is the random variable for the process at time $i $.
The sum given in the equation is just a convenient, algebraic identity, and you can verift that it holds.
If you are not familiar with thw definition of the Rieman integral, it may be a good idea to look this up.
answered Jul 15 at 12:00
AnyAD
1,451611
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $Y_i$ represent a stochastic process, where each $Y_i$ is a normally distributed random variable, then what the notes are saying is that $Y_i$'s are not independent, but the random variables $Z_i$, obtained from their 'differences' or increments are independent, i.e. $Z_i=Y_i+1-Y_i$ are independent.
So in general, given the importance of independence, one tries to 'handle' problems regarding $Y_i$'s by somehow `transforming' them to problems regarding the increments.
For example, $Y_2=Y_1+(Y_2-Y_1)$ and $Y_1$ and $Y_2-Y_1$ are independent.
$Y_i $ as you say is the random variable for the process at time $i $.
The sum given in the equation is just a convenient, algebraic identity, and you can verift that it holds.
If you are not familiar with thw definition of the Rieman integral, it may be a good idea to look this up.
answered Jul 15 at 12:00
AnyAD
1,451611
add a comment |Â
up vote
0
down vote
If $Y_i$ represent a stochastic process, where each $Y_i$ is a normally distributed random variable, then what the notes are saying is that $Y_i$'s are not independent, but the random variables $Z_i$, obtained from their 'differences' or increments are independent, i.e. $Z_i=Y_i+1-Y_i$ are independent.
So in general, given the importance of independence, one tries to 'handle' problems regarding $Y_i$'s by somehow `transforming' them to problems regarding the increments.
For example, $Y_2=Y_1+(Y_2-Y_1)$ and $Y_1$ and $Y_2-Y_1$ are independent.
$Y_i $ as you say is the random variable for the process at time $i $.
The sum given in the equation is just a convenient, algebraic identity, and you can verift that it holds.
If you are not familiar with thw definition of the Rieman integral, it may be a good idea to look this up.
answered Jul 15 at 12:00
AnyAD
1,451611
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $Y_i$ represent a stochastic process, where each $Y_i$ is a normally distributed random variable, then what the notes are saying is that $Y_i$'s are not independent, but the random variables $Z_i$, obtained from their 'differences' or increments are independent, i.e. $Z_i=Y_i+1-Y_i$ are independent.
So in general, given the importance of independence, one tries to 'handle' problems regarding $Y_i$'s by somehow `transforming' them to problems regarding the increments.
For example, $Y_2=Y_1+(Y_2-Y_1)$ and $Y_1$ and $Y_2-Y_1$ are independent.
$Y_i $ as you say is the random variable for the process at time $i $.
The sum given in the equation is just a convenient, algebraic identity, and you can verift that it holds.
If you are not familiar with thw definition of the Rieman integral, it may be a good idea to look this up.
answered Jul 15 at 12:00
AnyAD
1,451611
If $Y_i$ represent a stochastic process, where each $Y_i$ is a normally distributed random variable, then what the notes are saying is that $Y_i$'s are not independent, but the random variables $Z_i$, obtained from their 'differences' or increments are independent, i.e. $Z_i=Y_i+1-Y_i$ are independent.
So in general, given the importance of independence, one tries to 'handle' problems regarding $Y_i$'s by somehow `transforming' them to problems regarding the increments.
For example, $Y_2=Y_1+(Y_2-Y_1)$ and $Y_1$ and $Y_2-Y_1$ are independent.
$Y_i $ as you say is the random variable for the process at time $i $.
The sum given in the equation is just a convenient, algebraic identity, and you can verift that it holds.
If you are not familiar with thw definition of the Rieman integral, it may be a good idea to look this up.
answered Jul 15 at 12:00
AnyAD
1,451611
edited Jul 15 at 17:07
answered Jul 15 at 12:00
AnyAD
1,451611
answered Jul 15 at 12:00
AnyAD
1,451611
answered Jul 15 at 12:00
AnyAD
1,451611
1,451611
add a comment |Â
add a comment |Â
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Did you try to compute the RHS, say, for $n=5$?
– Did
Jul 15 at 10:10
Reading on the characterisation of the Wiener process here en.wikipedia.org/wiki/Wiener_process or in a textbook may help.
– AnyAD
Jul 15 at 12:16