intuition expectation adapted processes
$begingroup$
Let $X=(X_n)_{ninmathbb{N}}$ be a sequence of independent identically distributed random variables satisfying $mathbb{E}X_1^2=1$. Let $mathbb{F}=(mathcal{F}_n)_{ninmathbb{N}}$ the filtration generated by X. (Or $mathcal{F}_n=sigma(X_1,X_2,dots,X_n)$.)
Why is $mathbb{E}[X_n^2|mathcal{F}_{n-1}]=1$?
If it follows from independence of $X_n$ and $mathcal{F}_{n-1}$, what is the intuition behind it? Or does it follow from the fact that $X$ is iid? I would like a detailed answer so I can really understand what is happening here and be able to recognise similar situations in the future.
probability-theory conditional-expectation
asked Dec 23 '18 at 14:11
Lou95Lou95
659
$endgroup$
add a comment |
$begingroup$
Let $X=(X_n)_{ninmathbb{N}}$ be a sequence of independent identically distributed random variables satisfying $mathbb{E}X_1^2=1$. Let $mathbb{F}=(mathcal{F}_n)_{ninmathbb{N}}$ the filtration generated by X. (Or $mathcal{F}_n=sigma(X_1,X_2,dots,X_n)$.)
Why is $mathbb{E}[X_n^2|mathcal{F}_{n-1}]=1$?
If it follows from independence of $X_n$ and $mathcal{F}_{n-1}$, what is the intuition behind it? Or does it follow from the fact that $X$ is iid? I would like a detailed answer so I can really understand what is happening here and be able to recognise similar situations in the future.
probability-theory conditional-expectation
asked Dec 23 '18 at 14:11
Lou95Lou95
659
$endgroup$
add a comment |
$begingroup$
Let $X=(X_n)_{ninmathbb{N}}$ be a sequence of independent identically distributed random variables satisfying $mathbb{E}X_1^2=1$. Let $mathbb{F}=(mathcal{F}_n)_{ninmathbb{N}}$ the filtration generated by X. (Or $mathcal{F}_n=sigma(X_1,X_2,dots,X_n)$.)
Why is $mathbb{E}[X_n^2|mathcal{F}_{n-1}]=1$?
If it follows from independence of $X_n$ and $mathcal{F}_{n-1}$, what is the intuition behind it? Or does it follow from the fact that $X$ is iid? I would like a detailed answer so I can really understand what is happening here and be able to recognise similar situations in the future.
probability-theory conditional-expectation
asked Dec 23 '18 at 14:11
Lou95Lou95
659
$endgroup$
Let $X=(X_n)_{ninmathbb{N}}$ be a sequence of independent identically distributed random variables satisfying $mathbb{E}X_1^2=1$. Let $mathbb{F}=(mathcal{F}_n)_{ninmathbb{N}}$ the filtration generated by X. (Or $mathcal{F}_n=sigma(X_1,X_2,dots,X_n)$.)
Why is $mathbb{E}[X_n^2|mathcal{F}_{n-1}]=1$?
If it follows from independence of $X_n$ and $mathcal{F}_{n-1}$, what is the intuition behind it? Or does it follow from the fact that $X$ is iid? I would like a detailed answer so I can really understand what is happening here and be able to recognise similar situations in the future.
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Dec 23 '18 at 14:11
Lou95Lou95
659
asked Dec 23 '18 at 14:11
Lou95Lou95
659
asked Dec 23 '18 at 14:11
Lou95Lou95
659
asked Dec 23 '18 at 14:11
Lou95Lou95
659
asked Dec 23 '18 at 14:11
Lou95Lou95
659
659
add a comment |
add a comment |
1 Answer
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$begingroup$
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=mathbb E[Xmid Y]$ (also denoted as $mathbb E[Xmid sigma(Y)]$) which is a random variable that is characterized by:
$Z$ is measurable wrt $sigma(Y)$.
$mathbb EXmathbf1_A=mathbb EZmathbf1_A$ for every $Ainsigma(Y)$
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:mathbb Rtomathbb R$.
The second bullet implies that $mathbb EXg(Y)=mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=mathbb E[Xmid Y]$ the constant function that is prescribed by: $omegamapstomathbb EX$.
As a constant function of course it is measurable wrt $sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $mathbf1_A$ are independent so that: $$mathbb EXmathbf1_A=mathbb EXcdotmathbb Emathbf1_A=mathbb E[[mathbb EX]mathbf1_A]$$
So in your case we get:$$mathbb{E}[X_n^2|mathcal{F}_{n-1}]=mathbb{E}[X_n^2]=1$$
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
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add a comment |
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$begingroup$
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=mathbb E[Xmid Y]$ (also denoted as $mathbb E[Xmid sigma(Y)]$) which is a random variable that is characterized by:
$Z$ is measurable wrt $sigma(Y)$.
$mathbb EXmathbf1_A=mathbb EZmathbf1_A$ for every $Ainsigma(Y)$
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:mathbb Rtomathbb R$.
The second bullet implies that $mathbb EXg(Y)=mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=mathbb E[Xmid Y]$ the constant function that is prescribed by: $omegamapstomathbb EX$.
As a constant function of course it is measurable wrt $sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $mathbf1_A$ are independent so that: $$mathbb EXmathbf1_A=mathbb EXcdotmathbb Emathbf1_A=mathbb E[[mathbb EX]mathbf1_A]$$
So in your case we get:$$mathbb{E}[X_n^2|mathcal{F}_{n-1}]=mathbb{E}[X_n^2]=1$$
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=mathbb E[Xmid Y]$ (also denoted as $mathbb E[Xmid sigma(Y)]$) which is a random variable that is characterized by:
$Z$ is measurable wrt $sigma(Y)$.
$mathbb EXmathbf1_A=mathbb EZmathbf1_A$ for every $Ainsigma(Y)$
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:mathbb Rtomathbb R$.
The second bullet implies that $mathbb EXg(Y)=mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=mathbb E[Xmid Y]$ the constant function that is prescribed by: $omegamapstomathbb EX$.
As a constant function of course it is measurable wrt $sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $mathbf1_A$ are independent so that: $$mathbb EXmathbf1_A=mathbb EXcdotmathbb Emathbf1_A=mathbb E[[mathbb EX]mathbf1_A]$$
So in your case we get:$$mathbb{E}[X_n^2|mathcal{F}_{n-1}]=mathbb{E}[X_n^2]=1$$
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=mathbb E[Xmid Y]$ (also denoted as $mathbb E[Xmid sigma(Y)]$) which is a random variable that is characterized by:
$Z$ is measurable wrt $sigma(Y)$.
$mathbb EXmathbf1_A=mathbb EZmathbf1_A$ for every $Ainsigma(Y)$
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:mathbb Rtomathbb R$.
The second bullet implies that $mathbb EXg(Y)=mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=mathbb E[Xmid Y]$ the constant function that is prescribed by: $omegamapstomathbb EX$.
As a constant function of course it is measurable wrt $sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $mathbf1_A$ are independent so that: $$mathbb EXmathbf1_A=mathbb EXcdotmathbb Emathbf1_A=mathbb E[[mathbb EX]mathbf1_A]$$
So in your case we get:$$mathbb{E}[X_n^2|mathcal{F}_{n-1}]=mathbb{E}[X_n^2]=1$$
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
$endgroup$
It follows from independence.
If $X$ and $Y$ are random variables then we can go searching for $Z=mathbb E[Xmid Y]$ (also denoted as $mathbb E[Xmid sigma(Y)]$) which is a random variable that is characterized by:
$Z$ is measurable wrt $sigma(Y)$.
$mathbb EXmathbf1_A=mathbb EZmathbf1_A$ for every $Ainsigma(Y)$
The first bullet can be restated as: $Z=f(Y)$ for some Borel-measurable function $f:mathbb Rtomathbb R$.
The second bullet implies that $mathbb EXg(Y)=mathbb EZg(Y)$ for every Borel-measurable function $g$.
In the special case where $X$ and $Y$ are independent we can take for $Z=mathbb E[Xmid Y]$ the constant function that is prescribed by: $omegamapstomathbb EX$.
As a constant function of course it is measurable wrt $sigma(Y)$ (so the first bullet is satisfied).
Concerning the second bullet note that in this case $X$ and $mathbf1_A$ are independent so that: $$mathbb EXmathbf1_A=mathbb EXcdotmathbb Emathbf1_A=mathbb E[[mathbb EX]mathbf1_A]$$
So in your case we get:$$mathbb{E}[X_n^2|mathcal{F}_{n-1}]=mathbb{E}[X_n^2]=1$$
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
edited Dec 23 '18 at 19:33
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
answered Dec 23 '18 at 14:29
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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