Conditional expectation of integral of Ornstein-Uhlenbeck processCan I apply the Girsanov theorem to an Ornstein-Uhlenbeck process?Maximum Likelihood Estimation of an Ornstein-Uhlenbeck processOrnstein - Uhlenbeck ProcessOrnstein-Uhlenbeck process: Markov, but not martingale?Quadratic variation of the Ornstein-Uhlenbeck processConditional expectation of Ornstein-Uhlenbeck processSimulation Ornstein-Uhlenbeck continuousOrnstein-Uhlenbeck and StationarityInterpreting equation from article with Ornstein-Uhlenbeck processVariance of convolution between filter $A$ and Ornstein-Uhlenbeck process $x_t$
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Conditional expectation of integral of Ornstein-Uhlenbeck process
Can I apply the Girsanov theorem to an Ornstein-Uhlenbeck process?Maximum Likelihood Estimation of an Ornstein-Uhlenbeck processOrnstein - Uhlenbeck ProcessOrnstein-Uhlenbeck process: Markov, but not martingale?Quadratic variation of the Ornstein-Uhlenbeck processConditional expectation of Ornstein-Uhlenbeck processSimulation Ornstein-Uhlenbeck continuousOrnstein-Uhlenbeck and StationarityInterpreting equation from article with Ornstein-Uhlenbeck processVariance of convolution between filter $A$ and Ornstein-Uhlenbeck process $x_t$
$begingroup$
Given that $X(t)$ is an Ornstein-Uhlenbeck process with $X(0) = x_0$, which is a Markov process, but not a Martingale, how could I go forward if I would like to calculate
$E[int_0^T X(s)ds | mathcalF_t]$?
I have been twisting my brain for hours, but can't seem to find any reasonable approach.
stochastic-processes conditional-expectation martingales markov-process stochastic-analysis
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
$endgroup$
add a comment |
$begingroup$
Given that $X(t)$ is an Ornstein-Uhlenbeck process with $X(0) = x_0$, which is a Markov process, but not a Martingale, how could I go forward if I would like to calculate
$E[int_0^T X(s)ds | mathcalF_t]$?
I have been twisting my brain for hours, but can't seem to find any reasonable approach.
stochastic-processes conditional-expectation martingales markov-process stochastic-analysis
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
$endgroup$
$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02
add a comment |
$begingroup$
Given that $X(t)$ is an Ornstein-Uhlenbeck process with $X(0) = x_0$, which is a Markov process, but not a Martingale, how could I go forward if I would like to calculate
$E[int_0^T X(s)ds | mathcalF_t]$?
I have been twisting my brain for hours, but can't seem to find any reasonable approach.
stochastic-processes conditional-expectation martingales markov-process stochastic-analysis
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
$endgroup$
Given that $X(t)$ is an Ornstein-Uhlenbeck process with $X(0) = x_0$, which is a Markov process, but not a Martingale, how could I go forward if I would like to calculate
$E[int_0^T X(s)ds | mathcalF_t]$?
I have been twisting my brain for hours, but can't seem to find any reasonable approach.
stochastic-processes conditional-expectation martingales markov-process stochastic-analysis
stochastic-processes conditional-expectation martingales markov-process stochastic-analysis
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
edited Mar 12 at 19:16
Ken Klark
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
asked Mar 12 at 19:10
Ken KlarkKen Klark
91
91
$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02
add a comment |
$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02
$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02
$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will try to reply to your question with some assumptions as you were not precise in your description.
Let $(Omega,mathcalF,lbracemathcalF_trbrace_tinmathbbR_+,P)$ be a filtered probability space where we defined a $lbracemathcalF_trbrace-$brownian motion $lbraceW_trbrace_tinmathbbR_+$ starting from 0. We suppose that the process $X$ satisfies the following SDE:
beginequation
dX_t = theta(mu-X_t)dt + sigma dW_t
endequation
where $sigma >0, theta >0, mu in mathbbR$ and $X_0 in L^2$ and independant to $W$. We can show that it exists a unique strong solution to this SDE and is given by:
beginequation*
forall tin mathbbR_+, quad X_t = e^-theta tX_0 + mu(1+e^-theta t) +sigmaint_0^te^-theta(t-s)dW_t
endequation*
The problem was to compute: for $T>t$, $Eleft[int_0^TX_sds|mathcalF_tright]$
beginalign*
Eleft[int_0^TX_sds|mathcalF_tright] &= Eleft[int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s) +sigmaint_0^se^-theta(s-u)dW_uright)ds|mathcalF_tright] \
&=int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_0^se^-theta(s-u)dW_uds|mathcalF_tright] quad \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_u^Te^-theta(s-u)dsdW_u|mathcalF_tright] textFubini stochastic \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tphi(u)dW_u|mathcalF_tright] textwhere phi(u) := int_u^Te^-theta(s-u)ds \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma int_0^tphi(u)dW_u quad texta.s.
endalign*
The last equality holds because the function $phi$ is $L^2(mathbbR_+, dt)$. Thus the stochastic integral is a Wiener integral (which is a martingale).
answered Mar 13 at 1:08
SesameSesame
515
$endgroup$
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
I will try to reply to your question with some assumptions as you were not precise in your description.
Let $(Omega,mathcalF,lbracemathcalF_trbrace_tinmathbbR_+,P)$ be a filtered probability space where we defined a $lbracemathcalF_trbrace-$brownian motion $lbraceW_trbrace_tinmathbbR_+$ starting from 0. We suppose that the process $X$ satisfies the following SDE:
beginequation
dX_t = theta(mu-X_t)dt + sigma dW_t
endequation
where $sigma >0, theta >0, mu in mathbbR$ and $X_0 in L^2$ and independant to $W$. We can show that it exists a unique strong solution to this SDE and is given by:
beginequation*
forall tin mathbbR_+, quad X_t = e^-theta tX_0 + mu(1+e^-theta t) +sigmaint_0^te^-theta(t-s)dW_t
endequation*
The problem was to compute: for $T>t$, $Eleft[int_0^TX_sds|mathcalF_tright]$
beginalign*
Eleft[int_0^TX_sds|mathcalF_tright] &= Eleft[int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s) +sigmaint_0^se^-theta(s-u)dW_uright)ds|mathcalF_tright] \
&=int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_0^se^-theta(s-u)dW_uds|mathcalF_tright] quad \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_u^Te^-theta(s-u)dsdW_u|mathcalF_tright] textFubini stochastic \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tphi(u)dW_u|mathcalF_tright] textwhere phi(u) := int_u^Te^-theta(s-u)ds \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma int_0^tphi(u)dW_u quad texta.s.
endalign*
The last equality holds because the function $phi$ is $L^2(mathbbR_+, dt)$. Thus the stochastic integral is a Wiener integral (which is a martingale).
answered Mar 13 at 1:08
SesameSesame
515
$endgroup$
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
add a comment |
$begingroup$
I will try to reply to your question with some assumptions as you were not precise in your description.
Let $(Omega,mathcalF,lbracemathcalF_trbrace_tinmathbbR_+,P)$ be a filtered probability space where we defined a $lbracemathcalF_trbrace-$brownian motion $lbraceW_trbrace_tinmathbbR_+$ starting from 0. We suppose that the process $X$ satisfies the following SDE:
beginequation
dX_t = theta(mu-X_t)dt + sigma dW_t
endequation
where $sigma >0, theta >0, mu in mathbbR$ and $X_0 in L^2$ and independant to $W$. We can show that it exists a unique strong solution to this SDE and is given by:
beginequation*
forall tin mathbbR_+, quad X_t = e^-theta tX_0 + mu(1+e^-theta t) +sigmaint_0^te^-theta(t-s)dW_t
endequation*
The problem was to compute: for $T>t$, $Eleft[int_0^TX_sds|mathcalF_tright]$
beginalign*
Eleft[int_0^TX_sds|mathcalF_tright] &= Eleft[int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s) +sigmaint_0^se^-theta(s-u)dW_uright)ds|mathcalF_tright] \
&=int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_0^se^-theta(s-u)dW_uds|mathcalF_tright] quad \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_u^Te^-theta(s-u)dsdW_u|mathcalF_tright] textFubini stochastic \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tphi(u)dW_u|mathcalF_tright] textwhere phi(u) := int_u^Te^-theta(s-u)ds \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma int_0^tphi(u)dW_u quad texta.s.
endalign*
The last equality holds because the function $phi$ is $L^2(mathbbR_+, dt)$. Thus the stochastic integral is a Wiener integral (which is a martingale).
answered Mar 13 at 1:08
SesameSesame
515
$endgroup$
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
add a comment |
$begingroup$
I will try to reply to your question with some assumptions as you were not precise in your description.
Let $(Omega,mathcalF,lbracemathcalF_trbrace_tinmathbbR_+,P)$ be a filtered probability space where we defined a $lbracemathcalF_trbrace-$brownian motion $lbraceW_trbrace_tinmathbbR_+$ starting from 0. We suppose that the process $X$ satisfies the following SDE:
beginequation
dX_t = theta(mu-X_t)dt + sigma dW_t
endequation
where $sigma >0, theta >0, mu in mathbbR$ and $X_0 in L^2$ and independant to $W$. We can show that it exists a unique strong solution to this SDE and is given by:
beginequation*
forall tin mathbbR_+, quad X_t = e^-theta tX_0 + mu(1+e^-theta t) +sigmaint_0^te^-theta(t-s)dW_t
endequation*
The problem was to compute: for $T>t$, $Eleft[int_0^TX_sds|mathcalF_tright]$
beginalign*
Eleft[int_0^TX_sds|mathcalF_tright] &= Eleft[int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s) +sigmaint_0^se^-theta(s-u)dW_uright)ds|mathcalF_tright] \
&=int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_0^se^-theta(s-u)dW_uds|mathcalF_tright] quad \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_u^Te^-theta(s-u)dsdW_u|mathcalF_tright] textFubini stochastic \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tphi(u)dW_u|mathcalF_tright] textwhere phi(u) := int_u^Te^-theta(s-u)ds \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma int_0^tphi(u)dW_u quad texta.s.
endalign*
The last equality holds because the function $phi$ is $L^2(mathbbR_+, dt)$. Thus the stochastic integral is a Wiener integral (which is a martingale).
answered Mar 13 at 1:08
SesameSesame
515
$endgroup$
I will try to reply to your question with some assumptions as you were not precise in your description.
Let $(Omega,mathcalF,lbracemathcalF_trbrace_tinmathbbR_+,P)$ be a filtered probability space where we defined a $lbracemathcalF_trbrace-$brownian motion $lbraceW_trbrace_tinmathbbR_+$ starting from 0. We suppose that the process $X$ satisfies the following SDE:
beginequation
dX_t = theta(mu-X_t)dt + sigma dW_t
endequation
where $sigma >0, theta >0, mu in mathbbR$ and $X_0 in L^2$ and independant to $W$. We can show that it exists a unique strong solution to this SDE and is given by:
beginequation*
forall tin mathbbR_+, quad X_t = e^-theta tX_0 + mu(1+e^-theta t) +sigmaint_0^te^-theta(t-s)dW_t
endequation*
The problem was to compute: for $T>t$, $Eleft[int_0^TX_sds|mathcalF_tright]$
beginalign*
Eleft[int_0^TX_sds|mathcalF_tright] &= Eleft[int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s) +sigmaint_0^se^-theta(s-u)dW_uright)ds|mathcalF_tright] \
&=int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_0^se^-theta(s-u)dW_uds|mathcalF_tright] quad \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tint_u^Te^-theta(s-u)dsdW_u|mathcalF_tright] textFubini stochastic \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma Eleft[int_0^Tphi(u)dW_u|mathcalF_tright] textwhere phi(u) := int_u^Te^-theta(s-u)ds \
&= int_0^Tleft(e^-theta sX_0 + mu(1+e^-theta s)right)ds + sigma int_0^tphi(u)dW_u quad texta.s.
endalign*
The last equality holds because the function $phi$ is $L^2(mathbbR_+, dt)$. Thus the stochastic integral is a Wiener integral (which is a martingale).
answered Mar 13 at 1:08
SesameSesame
515
answered Mar 13 at 1:08
SesameSesame
515
answered Mar 13 at 1:08
SesameSesame
515
answered Mar 13 at 1:08
SesameSesame
515
515
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
add a comment |
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
$begingroup$
Thank you so much. I'm sorry I wasn't precise enough. I just assumed the standard assumptions and that they would be familiar to anyone answering
$endgroup$
– Ken Klark
Mar 13 at 8:13
add a comment |
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$begingroup$
What is $mathcalF_t$?
$endgroup$
– d.k.o.
Mar 12 at 21:02