Intuition of sub sigma-algebra definition
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I am having trouble understanding the sub σ-algebra definition on Wikipedia. I understand the following:
Let $X$ be a set, and let $A,B$ be σ-algebras on $X$. Then $B$ is said to be a sub-σ-algebra of $A$ iff $mathcal B subseteq mathcal A$.
For example, let $X=[0,1]$ and let $A$ be the σ-algebra generated by the sets $[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]$.
Then the σ-algebra $B$ generated by the sets $[0,1/2), [1/2,1]$ is a sub σ-algebra of $A$. Is this correct?
And now regarding the wikipedia definition that I have trouble with:
Formally, since you need to use subsets of Ω, this is codified as the σ-algebra
${displaystyle {mathcal {G}}_{n}={Atimes {H,T}^{infty }:Asubset {H,T}^{n}}.}$
Observe that then ${displaystyle {mathcal {G}}_{1}subset {mathcal {G}}_{2}subset {mathcal {G}}_{3}subset cdots subset {mathcal {G}}_{infty },}$
where ${displaystyle {mathcal {G}}_{infty }}$ is the smallest σ-algebra containing all the others.
I don't understand the last observation. My reasoning is that ${mathcal {G}}_{1}$ (for example a sequence of H, T starting with H) has more elements than ${mathcal {G}}_{2}$ (a sequence of H, T starting with HH).
measure-theory intuition
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
$endgroup$
add a comment |
$begingroup$
I am having trouble understanding the sub σ-algebra definition on Wikipedia. I understand the following:
Let $X$ be a set, and let $A,B$ be σ-algebras on $X$. Then $B$ is said to be a sub-σ-algebra of $A$ iff $mathcal B subseteq mathcal A$.
For example, let $X=[0,1]$ and let $A$ be the σ-algebra generated by the sets $[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]$.
Then the σ-algebra $B$ generated by the sets $[0,1/2), [1/2,1]$ is a sub σ-algebra of $A$. Is this correct?
And now regarding the wikipedia definition that I have trouble with:
Formally, since you need to use subsets of Ω, this is codified as the σ-algebra
${displaystyle {mathcal {G}}_{n}={Atimes {H,T}^{infty }:Asubset {H,T}^{n}}.}$
Observe that then ${displaystyle {mathcal {G}}_{1}subset {mathcal {G}}_{2}subset {mathcal {G}}_{3}subset cdots subset {mathcal {G}}_{infty },}$
where ${displaystyle {mathcal {G}}_{infty }}$ is the smallest σ-algebra containing all the others.
I don't understand the last observation. My reasoning is that ${mathcal {G}}_{1}$ (for example a sequence of H, T starting with H) has more elements than ${mathcal {G}}_{2}$ (a sequence of H, T starting with HH).
measure-theory intuition
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
$endgroup$
$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12
add a comment |
$begingroup$
I am having trouble understanding the sub σ-algebra definition on Wikipedia. I understand the following:
Let $X$ be a set, and let $A,B$ be σ-algebras on $X$. Then $B$ is said to be a sub-σ-algebra of $A$ iff $mathcal B subseteq mathcal A$.
For example, let $X=[0,1]$ and let $A$ be the σ-algebra generated by the sets $[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]$.
Then the σ-algebra $B$ generated by the sets $[0,1/2), [1/2,1]$ is a sub σ-algebra of $A$. Is this correct?
And now regarding the wikipedia definition that I have trouble with:
Formally, since you need to use subsets of Ω, this is codified as the σ-algebra
${displaystyle {mathcal {G}}_{n}={Atimes {H,T}^{infty }:Asubset {H,T}^{n}}.}$
Observe that then ${displaystyle {mathcal {G}}_{1}subset {mathcal {G}}_{2}subset {mathcal {G}}_{3}subset cdots subset {mathcal {G}}_{infty },}$
where ${displaystyle {mathcal {G}}_{infty }}$ is the smallest σ-algebra containing all the others.
I don't understand the last observation. My reasoning is that ${mathcal {G}}_{1}$ (for example a sequence of H, T starting with H) has more elements than ${mathcal {G}}_{2}$ (a sequence of H, T starting with HH).
measure-theory intuition
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
$endgroup$
I am having trouble understanding the sub σ-algebra definition on Wikipedia. I understand the following:
Let $X$ be a set, and let $A,B$ be σ-algebras on $X$. Then $B$ is said to be a sub-σ-algebra of $A$ iff $mathcal B subseteq mathcal A$.
For example, let $X=[0,1]$ and let $A$ be the σ-algebra generated by the sets $[0,1/4), [1/4,1/2), [1/2,3/4), [3/4,1]$.
Then the σ-algebra $B$ generated by the sets $[0,1/2), [1/2,1]$ is a sub σ-algebra of $A$. Is this correct?
And now regarding the wikipedia definition that I have trouble with:
Formally, since you need to use subsets of Ω, this is codified as the σ-algebra
${displaystyle {mathcal {G}}_{n}={Atimes {H,T}^{infty }:Asubset {H,T}^{n}}.}$
Observe that then ${displaystyle {mathcal {G}}_{1}subset {mathcal {G}}_{2}subset {mathcal {G}}_{3}subset cdots subset {mathcal {G}}_{infty },}$
where ${displaystyle {mathcal {G}}_{infty }}$ is the smallest σ-algebra containing all the others.
I don't understand the last observation. My reasoning is that ${mathcal {G}}_{1}$ (for example a sequence of H, T starting with H) has more elements than ${mathcal {G}}_{2}$ (a sequence of H, T starting with HH).
measure-theory intuition
measure-theory intuition
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
edited Dec 7 '18 at 12:04
GNUSupporter 8964民主女神 地下教會
12.7k72445
12.7k72445
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
asked Dec 7 '18 at 11:54
Jorge GarciaJorge Garcia
214
214
$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12
add a comment |
$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12
$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12
add a comment |
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$begingroup$
In fact, $mathcal G_n$ has exactly $2^n$ elements, since that is how many sets $A$ there are.
$endgroup$
– GEdgar
Dec 7 '18 at 12:23
$begingroup$
$mathcal{G}_n$ is set of set of string. Not set of string.
$endgroup$
– Lee.HW
Dec 7 '18 at 13:12