$A_i=S_itimesOmega ,S_isubseteq{0,1}^{n_i} ,n_ige1$, $i=1,2$; show $A_1cup A_2$ and $A_1cap A_2$ are of the...
$begingroup$
[Random walk with the actual infinite sample space, $Omega={0,1}^{mathbb{N}}$]
If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{0,1}^{n_i}$ for some $n_ige1$, $i=1,2$, then show that $A_1cup A_2$ and $A_1cap A_2$ are again of the same form $(S_0timesOmega)$.
My approach:
To prove: $(A_1cap A_2)=S_3timesOmega$ for some $S_3subseteq{0,1}^{n_3}$, $n_3ge1$
Without loss of generality assume $n_1>n_2$
$S_2subseteq{0,1}^{n_2}$
Then, we get some $S_3$ such that-
$S_3=S_2times{0,1}^{(n_1-n_2)}subseteq{0,1}^{(n_2)+(n_1-n_2)}={0,1}^{n_1}$
$A_2=S_2timesOmega$ where $S_2subseteq{0,1}^{n_2}$
Also, $A_2=S_3timesOmega$ where $S_3subseteq{0,1}^{n_1}$ $_{...(1)}$
$A_1=S_1timesOmega$ where $S_1subseteq{0,1}^{n_1}$ $_{...(2)}$
From $(1)$ and $(2)$, we get that-
$A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{0,1}^{n_1}$
[since, $S_1,S_3subseteq{0,1}^{n_1}$]
Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?
probability probability-theory random-walk
asked Jan 4 at 3:44
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
[Random walk with the actual infinite sample space, $Omega={0,1}^{mathbb{N}}$]
If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{0,1}^{n_i}$ for some $n_ige1$, $i=1,2$, then show that $A_1cup A_2$ and $A_1cap A_2$ are again of the same form $(S_0timesOmega)$.
My approach:
To prove: $(A_1cap A_2)=S_3timesOmega$ for some $S_3subseteq{0,1}^{n_3}$, $n_3ge1$
Without loss of generality assume $n_1>n_2$
$S_2subseteq{0,1}^{n_2}$
Then, we get some $S_3$ such that-
$S_3=S_2times{0,1}^{(n_1-n_2)}subseteq{0,1}^{(n_2)+(n_1-n_2)}={0,1}^{n_1}$
$A_2=S_2timesOmega$ where $S_2subseteq{0,1}^{n_2}$
Also, $A_2=S_3timesOmega$ where $S_3subseteq{0,1}^{n_1}$ $_{...(1)}$
$A_1=S_1timesOmega$ where $S_1subseteq{0,1}^{n_1}$ $_{...(2)}$
From $(1)$ and $(2)$, we get that-
$A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{0,1}^{n_1}$
[since, $S_1,S_3subseteq{0,1}^{n_1}$]
Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?
probability probability-theory random-walk
asked Jan 4 at 3:44
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
[Random walk with the actual infinite sample space, $Omega={0,1}^{mathbb{N}}$]
If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{0,1}^{n_i}$ for some $n_ige1$, $i=1,2$, then show that $A_1cup A_2$ and $A_1cap A_2$ are again of the same form $(S_0timesOmega)$.
My approach:
To prove: $(A_1cap A_2)=S_3timesOmega$ for some $S_3subseteq{0,1}^{n_3}$, $n_3ge1$
Without loss of generality assume $n_1>n_2$
$S_2subseteq{0,1}^{n_2}$
Then, we get some $S_3$ such that-
$S_3=S_2times{0,1}^{(n_1-n_2)}subseteq{0,1}^{(n_2)+(n_1-n_2)}={0,1}^{n_1}$
$A_2=S_2timesOmega$ where $S_2subseteq{0,1}^{n_2}$
Also, $A_2=S_3timesOmega$ where $S_3subseteq{0,1}^{n_1}$ $_{...(1)}$
$A_1=S_1timesOmega$ where $S_1subseteq{0,1}^{n_1}$ $_{...(2)}$
From $(1)$ and $(2)$, we get that-
$A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{0,1}^{n_1}$
[since, $S_1,S_3subseteq{0,1}^{n_1}$]
Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?
probability probability-theory random-walk
asked Jan 4 at 3:44
Za IraZa Ira
161115
$endgroup$
[Random walk with the actual infinite sample space, $Omega={0,1}^{mathbb{N}}$]
If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{0,1}^{n_i}$ for some $n_ige1$, $i=1,2$, then show that $A_1cup A_2$ and $A_1cap A_2$ are again of the same form $(S_0timesOmega)$.
My approach:
To prove: $(A_1cap A_2)=S_3timesOmega$ for some $S_3subseteq{0,1}^{n_3}$, $n_3ge1$
Without loss of generality assume $n_1>n_2$
$S_2subseteq{0,1}^{n_2}$
Then, we get some $S_3$ such that-
$S_3=S_2times{0,1}^{(n_1-n_2)}subseteq{0,1}^{(n_2)+(n_1-n_2)}={0,1}^{n_1}$
$A_2=S_2timesOmega$ where $S_2subseteq{0,1}^{n_2}$
Also, $A_2=S_3timesOmega$ where $S_3subseteq{0,1}^{n_1}$ $_{...(1)}$
$A_1=S_1timesOmega$ where $S_1subseteq{0,1}^{n_1}$ $_{...(2)}$
From $(1)$ and $(2)$, we get that-
$A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{0,1}^{n_1}$
[since, $S_1,S_3subseteq{0,1}^{n_1}$]
Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?
probability probability-theory random-walk
probability probability-theory random-walk
asked Jan 4 at 3:44
Za IraZa Ira
161115
asked Jan 4 at 3:44
Za IraZa Ira
161115
edited Jan 16 at 7:01
Za Ira
asked Jan 4 at 3:44
Za IraZa Ira
161115
asked Jan 4 at 3:44
Za IraZa Ira
161115
asked Jan 4 at 3:44
Za IraZa Ira
161115
161115
add a comment |
add a comment |
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