Show that $E(X|mathcal{F}_tau)=sumlimits_{ninmathbb{N}}E(X|mathcal{F}_n)mathbf{1}_{{tau=n}}$
$begingroup$
If $mathbf{E}X<infty$ and $tau$ is a stopping time, then $$mathbf{E}(X|mathcal{F}_tau)=sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}.$$
My attempt: First assume that $X$ is nonnegative. The general case will follow directly from nonnegative case.
Let $Ain mathcal{F}_tau.$ Then $Acap {tau=n}in mathcal{F}_n$ for every $ninmathbb{N}.$ Therefore, begin{align*}
int_{A}sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}&=sum_{ninmathbb{N}}int_{A}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}=sum_{ninmathbb{N}}int_{Acap {tau=n}}mathbf{E}(X|mathcal{F}_n)dmathbf{P}\&=sum_{ninmathbb{N}}int_{Acap {tau=n}}Xdmathbf{P}=int_A Xdmathbf{P}
end{align*}
where the first equality follows from monotone convergence theorem, third follows from the definition of the conditional expectation.
Question: (1) To prove that $sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}$ is indeed the conditional expectation of $X$ wrt $mathcal{F}_tau$, I have to prove that it is $mathcal{F}_tau$-measurable. How should I do so?
(2) Since the stopping time can take infinite value, my calculation of the integration above holds only when $tau<infty$ almost surely. Is the condition $tau<infty$ a.s. necessary here?
Thanks in advance!
probability-theory stochastic-processes conditional-expectation stopping-times
$endgroup$
add a comment |
$begingroup$
If $mathbf{E}X<infty$ and $tau$ is a stopping time, then $$mathbf{E}(X|mathcal{F}_tau)=sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}.$$
My attempt: First assume that $X$ is nonnegative. The general case will follow directly from nonnegative case.
Let $Ain mathcal{F}_tau.$ Then $Acap {tau=n}in mathcal{F}_n$ for every $ninmathbb{N}.$ Therefore, begin{align*}
int_{A}sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}&=sum_{ninmathbb{N}}int_{A}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}=sum_{ninmathbb{N}}int_{Acap {tau=n}}mathbf{E}(X|mathcal{F}_n)dmathbf{P}\&=sum_{ninmathbb{N}}int_{Acap {tau=n}}Xdmathbf{P}=int_A Xdmathbf{P}
end{align*}
where the first equality follows from monotone convergence theorem, third follows from the definition of the conditional expectation.
Question: (1) To prove that $sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}$ is indeed the conditional expectation of $X$ wrt $mathcal{F}_tau$, I have to prove that it is $mathcal{F}_tau$-measurable. How should I do so?
(2) Since the stopping time can take infinite value, my calculation of the integration above holds only when $tau<infty$ almost surely. Is the condition $tau<infty$ a.s. necessary here?
Thanks in advance!
probability-theory stochastic-processes conditional-expectation stopping-times
$endgroup$
add a comment |
$begingroup$
If $mathbf{E}X<infty$ and $tau$ is a stopping time, then $$mathbf{E}(X|mathcal{F}_tau)=sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}.$$
My attempt: First assume that $X$ is nonnegative. The general case will follow directly from nonnegative case.
Let $Ain mathcal{F}_tau.$ Then $Acap {tau=n}in mathcal{F}_n$ for every $ninmathbb{N}.$ Therefore, begin{align*}
int_{A}sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}&=sum_{ninmathbb{N}}int_{A}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}=sum_{ninmathbb{N}}int_{Acap {tau=n}}mathbf{E}(X|mathcal{F}_n)dmathbf{P}\&=sum_{ninmathbb{N}}int_{Acap {tau=n}}Xdmathbf{P}=int_A Xdmathbf{P}
end{align*}
where the first equality follows from monotone convergence theorem, third follows from the definition of the conditional expectation.
Question: (1) To prove that $sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}$ is indeed the conditional expectation of $X$ wrt $mathcal{F}_tau$, I have to prove that it is $mathcal{F}_tau$-measurable. How should I do so?
(2) Since the stopping time can take infinite value, my calculation of the integration above holds only when $tau<infty$ almost surely. Is the condition $tau<infty$ a.s. necessary here?
Thanks in advance!
probability-theory stochastic-processes conditional-expectation stopping-times
$endgroup$
If $mathbf{E}X<infty$ and $tau$ is a stopping time, then $$mathbf{E}(X|mathcal{F}_tau)=sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}.$$
My attempt: First assume that $X$ is nonnegative. The general case will follow directly from nonnegative case.
Let $Ain mathcal{F}_tau.$ Then $Acap {tau=n}in mathcal{F}_n$ for every $ninmathbb{N}.$ Therefore, begin{align*}
int_{A}sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}&=sum_{ninmathbb{N}}int_{A}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}dmathbf{P}=sum_{ninmathbb{N}}int_{Acap {tau=n}}mathbf{E}(X|mathcal{F}_n)dmathbf{P}\&=sum_{ninmathbb{N}}int_{Acap {tau=n}}Xdmathbf{P}=int_A Xdmathbf{P}
end{align*}
where the first equality follows from monotone convergence theorem, third follows from the definition of the conditional expectation.
Question: (1) To prove that $sum_{ninmathbb{N}}mathbf{E}(X|mathcal{F}_n)mathbf{1}_{{tau=n}}$ is indeed the conditional expectation of $X$ wrt $mathcal{F}_tau$, I have to prove that it is $mathcal{F}_tau$-measurable. How should I do so?
(2) Since the stopping time can take infinite value, my calculation of the integration above holds only when $tau<infty$ almost surely. Is the condition $tau<infty$ a.s. necessary here?
Thanks in advance!
probability-theory stochastic-processes conditional-expectation stopping-times
probability-theory stochastic-processes conditional-expectation stopping-times
edited Nov 27 '18 at 17:20
Did
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asked Nov 27 '18 at 14:51
bellcirclebellcircle
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2 Answers
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$begingroup$
(1) Recall $mathscr{F}_tau$ is, by definition, the set of events satisfying $mathrm{E} cap {tau = n} in mathscr{F}_n.$ Then, all you need to do is to show that $underbrace{left{ sumlimits_{nin mathbf{N}} mathbf{E}(X mid mathscr{F}_n) mathbf{1}_{{tau=n}} in mathrm{A} right}}_{mathrm{E}} cap {tau = m} in mathscr{F}_m.$ The intersection on the left side becomes ${mathbf{E}(X mid mathscr{F}_m) in mathrm{A}}cap{tau=m},$ which clearly belongs to $mathscr{F}_m.$ Q.E.D.
(2) If $tau = infty$ with positive probability, you would need to add it in the sum and make sense of the case $n = infty$ everywhere.
$endgroup$
add a comment |
$begingroup$
It suffices to prove that for each $n$ and each $mathcal F_n$-measurable random variable $Y$, the random variable $Ymathbf 1_{{tau=n}}$ is $mathcal{F}_{tau}$-measurable. By an approximation by simple function argument, it suffices to prove it in the most restrictive case where $Y$ is the indicator function of an $mathcal{F}_{n}$-measurable set, say $B$. This can be done by checking the definition, by proving that $Bcap {tau=n}cap {tau=k}$ is an element of $mathcal F_n$ for all $k$. This intersection is empty for $kneq n$ and for $k=n$, $mathcal{F}_{n}$-measurability of $Bcap {tau=n}$ is guaranteed by the fact that $tau$ is a stopping time.
We have to extend the filtration to $t=+infty$ by taking $mathcal{F}_{+infty}$ as the $sigma$-algebra generated by all the $mathcal{F}_{t}$, add the term $mathbb Eleft[Xmid mathcal F_{+infty}right]mathbf 1_{{tau=+infty}}$ and change the definition of $mathcal{F}_{tau}$ as
$$
mathcal{F}_{tau}=left{Ain mathcal{F}_{+infty}mid forall k, Acap {tau=k}in mathcal F_kright}.
$$
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
(1) Recall $mathscr{F}_tau$ is, by definition, the set of events satisfying $mathrm{E} cap {tau = n} in mathscr{F}_n.$ Then, all you need to do is to show that $underbrace{left{ sumlimits_{nin mathbf{N}} mathbf{E}(X mid mathscr{F}_n) mathbf{1}_{{tau=n}} in mathrm{A} right}}_{mathrm{E}} cap {tau = m} in mathscr{F}_m.$ The intersection on the left side becomes ${mathbf{E}(X mid mathscr{F}_m) in mathrm{A}}cap{tau=m},$ which clearly belongs to $mathscr{F}_m.$ Q.E.D.
(2) If $tau = infty$ with positive probability, you would need to add it in the sum and make sense of the case $n = infty$ everywhere.
$endgroup$
add a comment |
$begingroup$
(1) Recall $mathscr{F}_tau$ is, by definition, the set of events satisfying $mathrm{E} cap {tau = n} in mathscr{F}_n.$ Then, all you need to do is to show that $underbrace{left{ sumlimits_{nin mathbf{N}} mathbf{E}(X mid mathscr{F}_n) mathbf{1}_{{tau=n}} in mathrm{A} right}}_{mathrm{E}} cap {tau = m} in mathscr{F}_m.$ The intersection on the left side becomes ${mathbf{E}(X mid mathscr{F}_m) in mathrm{A}}cap{tau=m},$ which clearly belongs to $mathscr{F}_m.$ Q.E.D.
(2) If $tau = infty$ with positive probability, you would need to add it in the sum and make sense of the case $n = infty$ everywhere.
$endgroup$
add a comment |
$begingroup$
(1) Recall $mathscr{F}_tau$ is, by definition, the set of events satisfying $mathrm{E} cap {tau = n} in mathscr{F}_n.$ Then, all you need to do is to show that $underbrace{left{ sumlimits_{nin mathbf{N}} mathbf{E}(X mid mathscr{F}_n) mathbf{1}_{{tau=n}} in mathrm{A} right}}_{mathrm{E}} cap {tau = m} in mathscr{F}_m.$ The intersection on the left side becomes ${mathbf{E}(X mid mathscr{F}_m) in mathrm{A}}cap{tau=m},$ which clearly belongs to $mathscr{F}_m.$ Q.E.D.
(2) If $tau = infty$ with positive probability, you would need to add it in the sum and make sense of the case $n = infty$ everywhere.
$endgroup$
(1) Recall $mathscr{F}_tau$ is, by definition, the set of events satisfying $mathrm{E} cap {tau = n} in mathscr{F}_n.$ Then, all you need to do is to show that $underbrace{left{ sumlimits_{nin mathbf{N}} mathbf{E}(X mid mathscr{F}_n) mathbf{1}_{{tau=n}} in mathrm{A} right}}_{mathrm{E}} cap {tau = m} in mathscr{F}_m.$ The intersection on the left side becomes ${mathbf{E}(X mid mathscr{F}_m) in mathrm{A}}cap{tau=m},$ which clearly belongs to $mathscr{F}_m.$ Q.E.D.
(2) If $tau = infty$ with positive probability, you would need to add it in the sum and make sense of the case $n = infty$ everywhere.
answered Nov 27 '18 at 17:27
Will M.Will M.
2,505315
2,505315
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$begingroup$
It suffices to prove that for each $n$ and each $mathcal F_n$-measurable random variable $Y$, the random variable $Ymathbf 1_{{tau=n}}$ is $mathcal{F}_{tau}$-measurable. By an approximation by simple function argument, it suffices to prove it in the most restrictive case where $Y$ is the indicator function of an $mathcal{F}_{n}$-measurable set, say $B$. This can be done by checking the definition, by proving that $Bcap {tau=n}cap {tau=k}$ is an element of $mathcal F_n$ for all $k$. This intersection is empty for $kneq n$ and for $k=n$, $mathcal{F}_{n}$-measurability of $Bcap {tau=n}$ is guaranteed by the fact that $tau$ is a stopping time.
We have to extend the filtration to $t=+infty$ by taking $mathcal{F}_{+infty}$ as the $sigma$-algebra generated by all the $mathcal{F}_{t}$, add the term $mathbb Eleft[Xmid mathcal F_{+infty}right]mathbf 1_{{tau=+infty}}$ and change the definition of $mathcal{F}_{tau}$ as
$$
mathcal{F}_{tau}=left{Ain mathcal{F}_{+infty}mid forall k, Acap {tau=k}in mathcal F_kright}.
$$
$endgroup$
add a comment |
$begingroup$
It suffices to prove that for each $n$ and each $mathcal F_n$-measurable random variable $Y$, the random variable $Ymathbf 1_{{tau=n}}$ is $mathcal{F}_{tau}$-measurable. By an approximation by simple function argument, it suffices to prove it in the most restrictive case where $Y$ is the indicator function of an $mathcal{F}_{n}$-measurable set, say $B$. This can be done by checking the definition, by proving that $Bcap {tau=n}cap {tau=k}$ is an element of $mathcal F_n$ for all $k$. This intersection is empty for $kneq n$ and for $k=n$, $mathcal{F}_{n}$-measurability of $Bcap {tau=n}$ is guaranteed by the fact that $tau$ is a stopping time.
We have to extend the filtration to $t=+infty$ by taking $mathcal{F}_{+infty}$ as the $sigma$-algebra generated by all the $mathcal{F}_{t}$, add the term $mathbb Eleft[Xmid mathcal F_{+infty}right]mathbf 1_{{tau=+infty}}$ and change the definition of $mathcal{F}_{tau}$ as
$$
mathcal{F}_{tau}=left{Ain mathcal{F}_{+infty}mid forall k, Acap {tau=k}in mathcal F_kright}.
$$
$endgroup$
add a comment |
$begingroup$
It suffices to prove that for each $n$ and each $mathcal F_n$-measurable random variable $Y$, the random variable $Ymathbf 1_{{tau=n}}$ is $mathcal{F}_{tau}$-measurable. By an approximation by simple function argument, it suffices to prove it in the most restrictive case where $Y$ is the indicator function of an $mathcal{F}_{n}$-measurable set, say $B$. This can be done by checking the definition, by proving that $Bcap {tau=n}cap {tau=k}$ is an element of $mathcal F_n$ for all $k$. This intersection is empty for $kneq n$ and for $k=n$, $mathcal{F}_{n}$-measurability of $Bcap {tau=n}$ is guaranteed by the fact that $tau$ is a stopping time.
We have to extend the filtration to $t=+infty$ by taking $mathcal{F}_{+infty}$ as the $sigma$-algebra generated by all the $mathcal{F}_{t}$, add the term $mathbb Eleft[Xmid mathcal F_{+infty}right]mathbf 1_{{tau=+infty}}$ and change the definition of $mathcal{F}_{tau}$ as
$$
mathcal{F}_{tau}=left{Ain mathcal{F}_{+infty}mid forall k, Acap {tau=k}in mathcal F_kright}.
$$
$endgroup$
It suffices to prove that for each $n$ and each $mathcal F_n$-measurable random variable $Y$, the random variable $Ymathbf 1_{{tau=n}}$ is $mathcal{F}_{tau}$-measurable. By an approximation by simple function argument, it suffices to prove it in the most restrictive case where $Y$ is the indicator function of an $mathcal{F}_{n}$-measurable set, say $B$. This can be done by checking the definition, by proving that $Bcap {tau=n}cap {tau=k}$ is an element of $mathcal F_n$ for all $k$. This intersection is empty for $kneq n$ and for $k=n$, $mathcal{F}_{n}$-measurability of $Bcap {tau=n}$ is guaranteed by the fact that $tau$ is a stopping time.
We have to extend the filtration to $t=+infty$ by taking $mathcal{F}_{+infty}$ as the $sigma$-algebra generated by all the $mathcal{F}_{t}$, add the term $mathbb Eleft[Xmid mathcal F_{+infty}right]mathbf 1_{{tau=+infty}}$ and change the definition of $mathcal{F}_{tau}$ as
$$
mathcal{F}_{tau}=left{Ain mathcal{F}_{+infty}mid forall k, Acap {tau=k}in mathcal F_kright}.
$$
answered Nov 27 '18 at 15:56
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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