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Is independence indpendent on measure?

  1. May 25, 2012 #1
    Is it possible that 2 sigma-algebras could be independent under one measure but not independent under another?

    Many thanks.
  2. jcsd
  3. May 26, 2012 #2


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    Hey tunaaa and welcome to the forums.

    I don't know much about measure theory so maybe you could give us the definition of independence for a measure. I've heard about decomposing measures into orthogonal parts but I don't think this is what you are asking about.
  4. May 26, 2012 #3
    He probably means independence wrt a probability measure. This is defined as follows: take a probability space [itex](\Omega,\mathcal{F},P)[/itex] and take [itex]\mathcal{B}_1[/itex] and [itex]\mathcal{B}_2[/itex] sigma-algebra's which are part of [itex]\mathcal{F}[/itex]. They are independent if for every [itex]B_1\in \mathcal{B}_1[/itex] and [itex]B_2\in \mathcal{B}_2[/itex] holds that

    [tex]P(B_1\cap B_2)=P(B_1)P(B_2)[/tex]

    Like the definition suggests, the measure P is critical here. If we have another measure on [itex](\Omega,\mathcal{F}[/itex], then independence must not hold.

    For example, look at [itex](\Omega,\mathcal{F})=(\mathbb{N},\mathcal{P} (\mathbb{N}))[/itex] and take [itex]P_1[/itex] uniquely defined by [itex]P_1(\{0\})=1[/itex]. Further, take [itex]P_2[/itex] uniquely define by [itex]P_2(\{0\})=P_2(\{1\})=1/2[/itex].

    Then {0} and {1} (which generate sigma-algebras) are independent for [itex]P_1[/itex], but dependent for [itex]P_2[/itex].
  5. May 26, 2012 #4
    Many thanks
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