Is the sigma algebra generated by $X$ random variable and its square equal to the sigma algebra generated by...
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I would like to understand which relationships hold among the sigma algebras $sigma(X, X^2)$, $sigma(X)$ and $sigma(X^2)$, where X is a random variable. I would expect that $sigma(X, X^2)=sigma(X)$. If this is true, then is $X^2$ $sigma(X)$-measurable?
Moreover, I would like to know if there exist more sigma algebras w.r.t. a process is adapted. I am thinking that if $X_i$ are i.i.d. than $S_n=sum_{i=1}^n X_i$ is $mathcal{F}_n:=sigma(S_n)$-measurable but also $mathcal{G}_n:=sigma(X_1, X_2, dots, X_n)$-measurable. What changes if I take ${mathcal{F}_n}$ instead of ${mathcal{G}_n}$ as a filtration for the process ${S_n}$? Because for both the process is a martingale!
probability measure-theory stochastic-processes martingales
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
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show 1 more comment
$begingroup$
I would like to understand which relationships hold among the sigma algebras $sigma(X, X^2)$, $sigma(X)$ and $sigma(X^2)$, where X is a random variable. I would expect that $sigma(X, X^2)=sigma(X)$. If this is true, then is $X^2$ $sigma(X)$-measurable?
Moreover, I would like to know if there exist more sigma algebras w.r.t. a process is adapted. I am thinking that if $X_i$ are i.i.d. than $S_n=sum_{i=1}^n X_i$ is $mathcal{F}_n:=sigma(S_n)$-measurable but also $mathcal{G}_n:=sigma(X_1, X_2, dots, X_n)$-measurable. What changes if I take ${mathcal{F}_n}$ instead of ${mathcal{G}_n}$ as a filtration for the process ${S_n}$? Because for both the process is a martingale!
probability measure-theory stochastic-processes martingales
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
$endgroup$
1
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
1
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
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@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00
|
show 1 more comment
$begingroup$
I would like to understand which relationships hold among the sigma algebras $sigma(X, X^2)$, $sigma(X)$ and $sigma(X^2)$, where X is a random variable. I would expect that $sigma(X, X^2)=sigma(X)$. If this is true, then is $X^2$ $sigma(X)$-measurable?
Moreover, I would like to know if there exist more sigma algebras w.r.t. a process is adapted. I am thinking that if $X_i$ are i.i.d. than $S_n=sum_{i=1}^n X_i$ is $mathcal{F}_n:=sigma(S_n)$-measurable but also $mathcal{G}_n:=sigma(X_1, X_2, dots, X_n)$-measurable. What changes if I take ${mathcal{F}_n}$ instead of ${mathcal{G}_n}$ as a filtration for the process ${S_n}$? Because for both the process is a martingale!
probability measure-theory stochastic-processes martingales
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
$endgroup$
I would like to understand which relationships hold among the sigma algebras $sigma(X, X^2)$, $sigma(X)$ and $sigma(X^2)$, where X is a random variable. I would expect that $sigma(X, X^2)=sigma(X)$. If this is true, then is $X^2$ $sigma(X)$-measurable?
Moreover, I would like to know if there exist more sigma algebras w.r.t. a process is adapted. I am thinking that if $X_i$ are i.i.d. than $S_n=sum_{i=1}^n X_i$ is $mathcal{F}_n:=sigma(S_n)$-measurable but also $mathcal{G}_n:=sigma(X_1, X_2, dots, X_n)$-measurable. What changes if I take ${mathcal{F}_n}$ instead of ${mathcal{G}_n}$ as a filtration for the process ${S_n}$? Because for both the process is a martingale!
probability measure-theory stochastic-processes martingales
probability measure-theory stochastic-processes martingales
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
edited Dec 17 '18 at 19:11
PiccoloPaolo
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
asked Dec 17 '18 at 19:01
PiccoloPaoloPiccoloPaolo
413
413
1
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
1
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
$begingroup$
@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00
|
show 1 more comment
1
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
1
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
$begingroup$
@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00
1
1
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
1
1
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
$begingroup$
@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00
|
show 1 more comment
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1
$begingroup$
"I would expect that $sigma(X, X^2)=sigma(X)$." True. "If this is true, then is $X^2$ $sigma(X)$measurable?" It is. "Because for both the process is a martingale!" No, even for centered i.i.d. $(X_n)$, the process $(S_n)$ is not an $(mathcal F_n)$-martingale in general, only a $(mathcal G_n)$-martingale, for example $S_{n-1}$ is not even $mathcal F_n$-measurable in general.
$endgroup$
– Did
Dec 17 '18 at 19:13
$begingroup$
Just work it out. For example $X^2leq y=(Xleq sqrt{y})cap (Xgeq -sqrt{y})$ is clearly $sigma(X)$ measurable for all $y$.
$endgroup$
– Alex R.
Dec 17 '18 at 19:15
1
$begingroup$
If $f:mathbb Rtomathbb R$ is a Borel-measurable function and $X$ is a random variable then $Y:=f(X)$ is a random variable measurable wrt $sigma(X)$. This because for every Borelset $B$ we have: $$sigma(Y)ni{Yin B}={Xin f^{-1}(B)}insigma(X)$$You can apply this on function $xmapsto x^2$ and conclude that $sigma(X^2)subseteqsigma(X)$ and consequently $sigma(X,X^2)=sigma(X)$.
$endgroup$
– drhab
Dec 17 '18 at 19:17
$begingroup$
@Did If the $X_i$ are centered, then $S_n$ is $mathcal{F}_n$-measurable and $mathbb{E}[S_n|mathcal{F}_{n-1}]=S_{n-1}$ because $X_n$ is indipendent w.r.t. $mathcal{F}_{n-1}$ and its expected value is 0, and $S_{n-1}$ is $mathcal{F}_{n-1}$-measurable. Why do you say that $S_n$ is not a $mathcal{F}_n$ -martingale?
$endgroup$
– PiccoloPaolo
Dec 17 '18 at 19:24
$begingroup$
Because, in general, $S_{n-1}$ is not $mathcal F_n$-measurable (as already mentioned). The trouble comes from the fact that the sequence $(mathcal F_n)$ is not (in general) a filtration.
$endgroup$
– Did
Dec 17 '18 at 20:00