$A_i=S_itimesOmega ,S_isubseteq{+1,-1}^{n_i} ,n_ige1$, $i=1,2$; show $A_1cup A_2$ and $A_1cap A_2$ are of the...












0














[Random walk with the actual infinite sample space, $Omega={+1,-1}^{mathbb{N}}$]




If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{+1,-1}^{n_i}$ 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 $S_3subseteq{+1,-1}^{n_3}$, $n_3ge1$


Without loss of generality assume $n_1>n_2$

$S_2subseteq{+1,-1}^{n_2}$

Then, we get some $S_3$ such that-
$S_3=S_2times{+1,-1}^{(n_1-n_2)}subseteq{+1,-1}^{(n_2)+(n_1-n_2)}={+1,-1}^{n_1}$
$A_2=S_2timesOmega$ where $S_2subseteq{+1,-1}^{n_2}$

Also, $A_2=S_3timesOmega$ where $S_3subseteq{+1,-1}^{n_1}$ $_{...(1)}$

$A_1=S_1timesOmega$ where $S_1subseteq{+1,-1}^{n_1}$ $_{...(2)}$


From $(1)$ and $(2)$, we get that-
$A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{+1,-1}^{n_1}$

[since, $S_1,S_3subseteq{+1,-1}^{n_1}$]




Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?







probability probability-theory random-walk







share|cite|improve this question





























    0














    [Random walk with the actual infinite sample space, $Omega={+1,-1}^{mathbb{N}}$]




    If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{+1,-1}^{n_i}$ 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 $S_3subseteq{+1,-1}^{n_3}$, $n_3ge1$


    Without loss of generality assume $n_1>n_2$

    $S_2subseteq{+1,-1}^{n_2}$

    Then, we get some $S_3$ such that-
    $S_3=S_2times{+1,-1}^{(n_1-n_2)}subseteq{+1,-1}^{(n_2)+(n_1-n_2)}={+1,-1}^{n_1}$
    $A_2=S_2timesOmega$ where $S_2subseteq{+1,-1}^{n_2}$

    Also, $A_2=S_3timesOmega$ where $S_3subseteq{+1,-1}^{n_1}$ $_{...(1)}$

    $A_1=S_1timesOmega$ where $S_1subseteq{+1,-1}^{n_1}$ $_{...(2)}$


    From $(1)$ and $(2)$, we get that-
    $A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{+1,-1}^{n_1}$

    [since, $S_1,S_3subseteq{+1,-1}^{n_1}$]




    Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?







    probability probability-theory random-walk







    share|cite|improve this question



























      0












      0








      0







      [Random walk with the actual infinite sample space, $Omega={+1,-1}^{mathbb{N}}$]




      If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{+1,-1}^{n_i}$ 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 $S_3subseteq{+1,-1}^{n_3}$, $n_3ge1$


      Without loss of generality assume $n_1>n_2$

      $S_2subseteq{+1,-1}^{n_2}$

      Then, we get some $S_3$ such that-
      $S_3=S_2times{+1,-1}^{(n_1-n_2)}subseteq{+1,-1}^{(n_2)+(n_1-n_2)}={+1,-1}^{n_1}$
      $A_2=S_2timesOmega$ where $S_2subseteq{+1,-1}^{n_2}$

      Also, $A_2=S_3timesOmega$ where $S_3subseteq{+1,-1}^{n_1}$ $_{...(1)}$

      $A_1=S_1timesOmega$ where $S_1subseteq{+1,-1}^{n_1}$ $_{...(2)}$


      From $(1)$ and $(2)$, we get that-
      $A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{+1,-1}^{n_1}$

      [since, $S_1,S_3subseteq{+1,-1}^{n_1}$]




      Is my approach correct? Can I prove the same for $A_1cup A_2$ similarly?







      probability probability-theory random-walk







      share|cite|improve this question















      [Random walk with the actual infinite sample space, $Omega={+1,-1}^{mathbb{N}}$]




      If $A_1=S_1timesOmega$ and $A_2=S_2timesOmega$ where $S_isubseteq{+1,-1}^{n_i}$ 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 $S_3subseteq{+1,-1}^{n_3}$, $n_3ge1$


      Without loss of generality assume $n_1>n_2$

      $S_2subseteq{+1,-1}^{n_2}$

      Then, we get some $S_3$ such that-
      $S_3=S_2times{+1,-1}^{(n_1-n_2)}subseteq{+1,-1}^{(n_2)+(n_1-n_2)}={+1,-1}^{n_1}$
      $A_2=S_2timesOmega$ where $S_2subseteq{+1,-1}^{n_2}$

      Also, $A_2=S_3timesOmega$ where $S_3subseteq{+1,-1}^{n_1}$ $_{...(1)}$

      $A_1=S_1timesOmega$ where $S_1subseteq{+1,-1}^{n_1}$ $_{...(2)}$


      From $(1)$ and $(2)$, we get that-
      $A_1cap A_2=(S_1cap S_3)timesOmega=S_0timesOmega$ where $S_0subseteq{+1,-1}^{n_1}$

      [since, $S_1,S_3subseteq{+1,-1}^{n_1}$]




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 3:52







      Za Ira

















      asked Jan 4 at 3:44









      Za IraZa Ira

      147113




      147113






















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