Mixing
On calculating sigma algebras generated by specific functions.
$begingroup$
I started (again!) with the intention to build an interesting example of a computation of conditional expectation with respect to $sigma(X) $ when $X $ is not a step function. My first example, choosing $Omega = [-1, 1], $ ${cal F} = {cal B}(Omega), $ and $P = (1/2)lambda $ with
$X: Omega mapsto R: omega mapsto omega^2 $ or $|omega | $ worked fine, since $sigma(X) = {A in {cal B}(Omega): A= -A}. $
When I tried to have a function that is not symmetric, say
$Y(omega) = -omegacdot I_{[-1, 0]}(omega) + 2omega cdot I_{[0, 1]}(omega), $ not such luck. Now, I can see the overall structure of $sigma(Y), $ but cannot really write it down in an elegant manner: I see that intervals like $(a, b) $ with $a= -1/2*b $ should be part of $sigma(Y), $ if $b > 0, $ but I was hoping to get an elegant expression like in the case of $sigma(X) $ above. Is that possible?
Thank you.
Maurice
probability-theory conditional-expectation
asked Jan 29 at 17:59
MauriceMaurice
45849
$endgroup$
|
show 4 more comments
$begingroup$
I started (again!) with the intention to build an interesting example of a computation of conditional expectation with respect to $sigma(X) $ when $X $ is not a step function. My first example, choosing $Omega = [-1, 1], $ ${cal F} = {cal B}(Omega), $ and $P = (1/2)lambda $ with
$X: Omega mapsto R: omega mapsto omega^2 $ or $|omega | $ worked fine, since $sigma(X) = {A in {cal B}(Omega): A= -A}. $
When I tried to have a function that is not symmetric, say
$Y(omega) = -omegacdot I_{[-1, 0]}(omega) + 2omega cdot I_{[0, 1]}(omega), $ not such luck. Now, I can see the overall structure of $sigma(Y), $ but cannot really write it down in an elegant manner: I see that intervals like $(a, b) $ with $a= -1/2*b $ should be part of $sigma(Y), $ if $b > 0, $ but I was hoping to get an elegant expression like in the case of $sigma(X) $ above. Is that possible?
Thank you.
Maurice
probability-theory conditional-expectation
asked Jan 29 at 17:59
MauriceMaurice
45849
$endgroup$
$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14
|
show 4 more comments
$begingroup$
I started (again!) with the intention to build an interesting example of a computation of conditional expectation with respect to $sigma(X) $ when $X $ is not a step function. My first example, choosing $Omega = [-1, 1], $ ${cal F} = {cal B}(Omega), $ and $P = (1/2)lambda $ with
$X: Omega mapsto R: omega mapsto omega^2 $ or $|omega | $ worked fine, since $sigma(X) = {A in {cal B}(Omega): A= -A}. $
When I tried to have a function that is not symmetric, say
$Y(omega) = -omegacdot I_{[-1, 0]}(omega) + 2omega cdot I_{[0, 1]}(omega), $ not such luck. Now, I can see the overall structure of $sigma(Y), $ but cannot really write it down in an elegant manner: I see that intervals like $(a, b) $ with $a= -1/2*b $ should be part of $sigma(Y), $ if $b > 0, $ but I was hoping to get an elegant expression like in the case of $sigma(X) $ above. Is that possible?
Thank you.
Maurice
probability-theory conditional-expectation
asked Jan 29 at 17:59
MauriceMaurice
45849
$endgroup$
I started (again!) with the intention to build an interesting example of a computation of conditional expectation with respect to $sigma(X) $ when $X $ is not a step function. My first example, choosing $Omega = [-1, 1], $ ${cal F} = {cal B}(Omega), $ and $P = (1/2)lambda $ with
$X: Omega mapsto R: omega mapsto omega^2 $ or $|omega | $ worked fine, since $sigma(X) = {A in {cal B}(Omega): A= -A}. $
When I tried to have a function that is not symmetric, say
$Y(omega) = -omegacdot I_{[-1, 0]}(omega) + 2omega cdot I_{[0, 1]}(omega), $ not such luck. Now, I can see the overall structure of $sigma(Y), $ but cannot really write it down in an elegant manner: I see that intervals like $(a, b) $ with $a= -1/2*b $ should be part of $sigma(Y), $ if $b > 0, $ but I was hoping to get an elegant expression like in the case of $sigma(X) $ above. Is that possible?
Thank you.
Maurice
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Jan 29 at 17:59
MauriceMaurice
45849
asked Jan 29 at 17:59
MauriceMaurice
45849
edited Jan 29 at 19:23
Maurice
asked Jan 29 at 17:59
MauriceMaurice
45849
asked Jan 29 at 17:59
MauriceMaurice
45849
asked Jan 29 at 17:59
MauriceMaurice
45849
45849
$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14
|
show 4 more comments
$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14
$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First, for any $Binmathcal{B}(mathbb{R})$,
$$
X^{-1}(B)=X^{-1}(Bcap[0,1])cup X^{-1}(Bcap(1,2])equiv C_1cup C_2.
$$
Notice that $C_1cap C_2=emptyset$, $C_2=emptyset$ or $C_2subseteq(1/2,1]$ and $C_1$ is "symmetric" around $0$ in the following sense: if $C_1^+equiv C_1cap [0,1/2]$ and $C_1^-equiv C_1cap [-1,0)$, then $C_1^-=-2C_1^+$.
For an integrable random variable $Y$, the conditional expectation $mathsf{E}[Ymid sigma(X)]=Z(omega)$, where
begin{align}
Z(omega)&:=Y(omega)1_{(1/2,1]}(omega)+frac{Y(omega)+Y(-2omega)}{2}1_{[0,1/2]}(omega)\
&quad+frac{Y(omega)+Y(-omega/2)}{2}1_{[-1,0)}(omega).
end{align}
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
First, for any $Binmathcal{B}(mathbb{R})$,
$$
X^{-1}(B)=X^{-1}(Bcap[0,1])cup X^{-1}(Bcap(1,2])equiv C_1cup C_2.
$$
Notice that $C_1cap C_2=emptyset$, $C_2=emptyset$ or $C_2subseteq(1/2,1]$ and $C_1$ is "symmetric" around $0$ in the following sense: if $C_1^+equiv C_1cap [0,1/2]$ and $C_1^-equiv C_1cap [-1,0)$, then $C_1^-=-2C_1^+$.
For an integrable random variable $Y$, the conditional expectation $mathsf{E}[Ymid sigma(X)]=Z(omega)$, where
begin{align}
Z(omega)&:=Y(omega)1_{(1/2,1]}(omega)+frac{Y(omega)+Y(-2omega)}{2}1_{[0,1/2]}(omega)\
&quad+frac{Y(omega)+Y(-omega/2)}{2}1_{[-1,0)}(omega).
end{align}
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
add a comment |
$begingroup$
First, for any $Binmathcal{B}(mathbb{R})$,
$$
X^{-1}(B)=X^{-1}(Bcap[0,1])cup X^{-1}(Bcap(1,2])equiv C_1cup C_2.
$$
Notice that $C_1cap C_2=emptyset$, $C_2=emptyset$ or $C_2subseteq(1/2,1]$ and $C_1$ is "symmetric" around $0$ in the following sense: if $C_1^+equiv C_1cap [0,1/2]$ and $C_1^-equiv C_1cap [-1,0)$, then $C_1^-=-2C_1^+$.
For an integrable random variable $Y$, the conditional expectation $mathsf{E}[Ymid sigma(X)]=Z(omega)$, where
begin{align}
Z(omega)&:=Y(omega)1_{(1/2,1]}(omega)+frac{Y(omega)+Y(-2omega)}{2}1_{[0,1/2]}(omega)\
&quad+frac{Y(omega)+Y(-omega/2)}{2}1_{[-1,0)}(omega).
end{align}
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
add a comment |
$begingroup$
First, for any $Binmathcal{B}(mathbb{R})$,
$$
X^{-1}(B)=X^{-1}(Bcap[0,1])cup X^{-1}(Bcap(1,2])equiv C_1cup C_2.
$$
Notice that $C_1cap C_2=emptyset$, $C_2=emptyset$ or $C_2subseteq(1/2,1]$ and $C_1$ is "symmetric" around $0$ in the following sense: if $C_1^+equiv C_1cap [0,1/2]$ and $C_1^-equiv C_1cap [-1,0)$, then $C_1^-=-2C_1^+$.
For an integrable random variable $Y$, the conditional expectation $mathsf{E}[Ymid sigma(X)]=Z(omega)$, where
begin{align}
Z(omega)&:=Y(omega)1_{(1/2,1]}(omega)+frac{Y(omega)+Y(-2omega)}{2}1_{[0,1/2]}(omega)\
&quad+frac{Y(omega)+Y(-omega/2)}{2}1_{[-1,0)}(omega).
end{align}
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
$endgroup$
First, for any $Binmathcal{B}(mathbb{R})$,
$$
X^{-1}(B)=X^{-1}(Bcap[0,1])cup X^{-1}(Bcap(1,2])equiv C_1cup C_2.
$$
Notice that $C_1cap C_2=emptyset$, $C_2=emptyset$ or $C_2subseteq(1/2,1]$ and $C_1$ is "symmetric" around $0$ in the following sense: if $C_1^+equiv C_1cap [0,1/2]$ and $C_1^-equiv C_1cap [-1,0)$, then $C_1^-=-2C_1^+$.
For an integrable random variable $Y$, the conditional expectation $mathsf{E}[Ymid sigma(X)]=Z(omega)$, where
begin{align}
Z(omega)&:=Y(omega)1_{(1/2,1]}(omega)+frac{Y(omega)+Y(-2omega)}{2}1_{[0,1/2]}(omega)\
&quad+frac{Y(omega)+Y(-omega/2)}{2}1_{[-1,0)}(omega).
end{align}
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
edited Jan 29 at 23:24
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
answered Jan 29 at 22:48
d.k.o.d.k.o.
10.5k630
10.5k630
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
add a comment |
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
$begingroup$
Thank you very much... Much appreciated... Now I get the structure so that it could be used to condition and guess the form of $E[Y| sigma(X)].
$endgroup$
– Maurice
Jan 30 at 2:04
add a comment |
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$begingroup$
$sigma(Y)=mathcal{F}$...
$endgroup$
– d.k.o.
Jan 29 at 18:29
$begingroup$
I am not sure if the comment above was an attempt to answer or just some humor...
$endgroup$
– Maurice
Jan 29 at 18:51
$begingroup$
Here is the joke: ${[a,b]:-1le a<ble 1}subset sigma(X)$.
$endgroup$
– d.k.o.
Jan 29 at 18:56
$begingroup$
@ d.k.o. I was hoping for an equality more than an inclusion. Something like $sigma(X) = {A: {cal B}(Omega): .....}. $
$endgroup$
– Maurice
Jan 29 at 19:09
$begingroup$
$sigma(Y)$ is the Borel $sigma$-algebra on $[0,1]$. Isn't the latter elegant enough?
$endgroup$
– d.k.o.
Jan 29 at 19:14