measurable functions commute with conditional expectation
Let $W_t = W (t)$ be a (one-dimensional) Wiener process, and fix an admissible filtration $mathbb{F}.$
An adapted process $V_t$ is called elementary if it has the form
$$V_t = sum_{j = 0}^{k} xi_j chi_{(tj ,tj+1]}(t)$$
where $0 = t_0 < t_1 <···< t_k = 1,$ and for each index j the random variable $xi_j$ is measurable relative to $mathcal{F}_{t_j}.$
Define the Ito Integral of a simple process as follows:
$$I_{t}(V) = int_{0}^{t}V_{s}dW_{s} = sum_{j=0}^{k}xi_j(W_{t_{j+1}} - W_{t_{j}})$$
In some proofs of the properties of the Ito Integral, I see the following type of argument: (For simplicity, I will use a process with only one jump.)
begin{align*}
mathbb{E}left(I_{t}(V)right)&= mathbb{E}(xi_j(W_{t_{j+1}} - W_{t_{j}}))\
&= mathbb{E}left[mathbb{E}left(xi_j(W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]\
&=mathbb{E}xi_{j}left[mathbb{E}left((W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]
end{align*}
I have been staring at this for a bit, but I am unsure where this string of equalities comes from. I am guessing that the second inequality comes from the Tower Property for conditional expectations? Also where the heck does the third equality come from? If we write this out we've
begin{align*}
int_{Omega}I_{t}(V)dmu &= int_{Omega}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu\
&= int_{Omega}int_{mathcal{F}_{t_{i+1}}}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu dmu \
&=int_{Omega}xi_{j}int_{mathcal{F}_{t_{i+1}}}(W_{t_{j+1}} - W_{t_{j}})dmu dmu
end{align*}
where $(Omega,mu)$ is the background measure space. I still don't see why we can pull out the measurable function like that. Are my definitions wrong? Please help.
stochastic-processes brownian-motion conditional-expectation
asked Jan 4 at 22:34
IndexIndex
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Let $W_t = W (t)$ be a (one-dimensional) Wiener process, and fix an admissible filtration $mathbb{F}.$
An adapted process $V_t$ is called elementary if it has the form
$$V_t = sum_{j = 0}^{k} xi_j chi_{(tj ,tj+1]}(t)$$
where $0 = t_0 < t_1 <···< t_k = 1,$ and for each index j the random variable $xi_j$ is measurable relative to $mathcal{F}_{t_j}.$
Define the Ito Integral of a simple process as follows:
$$I_{t}(V) = int_{0}^{t}V_{s}dW_{s} = sum_{j=0}^{k}xi_j(W_{t_{j+1}} - W_{t_{j}})$$
In some proofs of the properties of the Ito Integral, I see the following type of argument: (For simplicity, I will use a process with only one jump.)
begin{align*}
mathbb{E}left(I_{t}(V)right)&= mathbb{E}(xi_j(W_{t_{j+1}} - W_{t_{j}}))\
&= mathbb{E}left[mathbb{E}left(xi_j(W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]\
&=mathbb{E}xi_{j}left[mathbb{E}left((W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]
end{align*}
I have been staring at this for a bit, but I am unsure where this string of equalities comes from. I am guessing that the second inequality comes from the Tower Property for conditional expectations? Also where the heck does the third equality come from? If we write this out we've
begin{align*}
int_{Omega}I_{t}(V)dmu &= int_{Omega}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu\
&= int_{Omega}int_{mathcal{F}_{t_{i+1}}}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu dmu \
&=int_{Omega}xi_{j}int_{mathcal{F}_{t_{i+1}}}(W_{t_{j+1}} - W_{t_{j}})dmu dmu
end{align*}
where $(Omega,mu)$ is the background measure space. I still don't see why we can pull out the measurable function like that. Are my definitions wrong? Please help.
stochastic-processes brownian-motion conditional-expectation
asked Jan 4 at 22:34
IndexIndex
63
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
1
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03
|
show 1 more comment
Let $W_t = W (t)$ be a (one-dimensional) Wiener process, and fix an admissible filtration $mathbb{F}.$
An adapted process $V_t$ is called elementary if it has the form
$$V_t = sum_{j = 0}^{k} xi_j chi_{(tj ,tj+1]}(t)$$
where $0 = t_0 < t_1 <···< t_k = 1,$ and for each index j the random variable $xi_j$ is measurable relative to $mathcal{F}_{t_j}.$
Define the Ito Integral of a simple process as follows:
$$I_{t}(V) = int_{0}^{t}V_{s}dW_{s} = sum_{j=0}^{k}xi_j(W_{t_{j+1}} - W_{t_{j}})$$
In some proofs of the properties of the Ito Integral, I see the following type of argument: (For simplicity, I will use a process with only one jump.)
begin{align*}
mathbb{E}left(I_{t}(V)right)&= mathbb{E}(xi_j(W_{t_{j+1}} - W_{t_{j}}))\
&= mathbb{E}left[mathbb{E}left(xi_j(W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]\
&=mathbb{E}xi_{j}left[mathbb{E}left((W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]
end{align*}
I have been staring at this for a bit, but I am unsure where this string of equalities comes from. I am guessing that the second inequality comes from the Tower Property for conditional expectations? Also where the heck does the third equality come from? If we write this out we've
begin{align*}
int_{Omega}I_{t}(V)dmu &= int_{Omega}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu\
&= int_{Omega}int_{mathcal{F}_{t_{i+1}}}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu dmu \
&=int_{Omega}xi_{j}int_{mathcal{F}_{t_{i+1}}}(W_{t_{j+1}} - W_{t_{j}})dmu dmu
end{align*}
where $(Omega,mu)$ is the background measure space. I still don't see why we can pull out the measurable function like that. Are my definitions wrong? Please help.
stochastic-processes brownian-motion conditional-expectation
asked Jan 4 at 22:34
IndexIndex
63
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $W_t = W (t)$ be a (one-dimensional) Wiener process, and fix an admissible filtration $mathbb{F}.$
An adapted process $V_t$ is called elementary if it has the form
$$V_t = sum_{j = 0}^{k} xi_j chi_{(tj ,tj+1]}(t)$$
where $0 = t_0 < t_1 <···< t_k = 1,$ and for each index j the random variable $xi_j$ is measurable relative to $mathcal{F}_{t_j}.$
Define the Ito Integral of a simple process as follows:
$$I_{t}(V) = int_{0}^{t}V_{s}dW_{s} = sum_{j=0}^{k}xi_j(W_{t_{j+1}} - W_{t_{j}})$$
In some proofs of the properties of the Ito Integral, I see the following type of argument: (For simplicity, I will use a process with only one jump.)
begin{align*}
mathbb{E}left(I_{t}(V)right)&= mathbb{E}(xi_j(W_{t_{j+1}} - W_{t_{j}}))\
&= mathbb{E}left[mathbb{E}left(xi_j(W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]\
&=mathbb{E}xi_{j}left[mathbb{E}left((W_{t_{j+1}} - W_{t_{j}})| mathcal{F}_{t_{i+1}}right)right]
end{align*}
I have been staring at this for a bit, but I am unsure where this string of equalities comes from. I am guessing that the second inequality comes from the Tower Property for conditional expectations? Also where the heck does the third equality come from? If we write this out we've
begin{align*}
int_{Omega}I_{t}(V)dmu &= int_{Omega}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu\
&= int_{Omega}int_{mathcal{F}_{t_{i+1}}}xi_j(W_{t_{j+1}} - W_{t_{j}})dmu dmu \
&=int_{Omega}xi_{j}int_{mathcal{F}_{t_{i+1}}}(W_{t_{j+1}} - W_{t_{j}})dmu dmu
end{align*}
where $(Omega,mu)$ is the background measure space. I still don't see why we can pull out the measurable function like that. Are my definitions wrong? Please help.
stochastic-processes brownian-motion conditional-expectation
stochastic-processes brownian-motion conditional-expectation
asked Jan 4 at 22:34
IndexIndex
63
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 22:34
IndexIndex
63
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 5 at 1:30
Index
asked Jan 4 at 22:34
IndexIndex
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Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 22:34
IndexIndex
63
asked Jan 4 at 22:34
IndexIndex
63
63
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Index is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
1
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03
|
show 1 more comment
$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
1
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03
$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
1
1
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03
|
show 1 more comment
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$mathcal F_{t_{j+1}}supset mathcal F_{t_j}$ so if $xi_j$ is measurable with $mathcal F_{t_j}$ it is measurable with respect to $mathcal F_{t_{j+1}}$ and hence we can pull it out as a known factor - no tower property involved.
– Math1000
Jan 4 at 22:44
1
What do you mean by integral over a sigma algebra? I think you should take a set in the sigma algebra and integrate over it.
– Kavi Rama Murthy
Jan 4 at 23:42
@KaviRamaMurthy I guess I don't really know how to define conditional expectation as an integral. Is the correct definition
– Index
Jan 5 at 0:48
@Index Conditional expectation is a random variable in general, not a number. It is not an integral.
– Kavi Rama Murthy
Jan 5 at 0:49
@Math1000 I'm not sure what you mean by a "known factor". Like, if it is measurable with respect to a sub sigma algebra, we can think of it as a constant?
– Index
Jan 5 at 1:03