I am trying to understand the following equality involving probability measures in Wikipedia: ‖ μ...

rzfansubs87

rzfansubs87

Answered

2022-07-10

I am trying to understand the following equality involving probability measures in Wikipedia:
μ ν = | μ ν | ( X ) = 2 sup { | μ ( A ) ν ( A ) | : A Σ }
where the total variation norm and the total variation of a measure | | are defined in the some articles. The article has been flagged for not citing sources or offering a proof.
I have found a proof of a similar result for discrete probability distributions (1). But I have not been able to adapt it to continuous probability distributions, which is the result above.

Answer & Explanation

Valeria Wolfe

Valeria Wolfe

Expert

2022-07-11Added 11 answers

First we need to show μ ν ( ) is a (finite) signed measure. I am not sure if that is trivial or requires some clever trick. Then, we let B be the measurable set E D + using the Hahn decomposition ( D + D ) of X under μ ν s.t. μ ν is positive on D + . It is then easy to prove that for all A Σ, μ ( A ) ν ( A ) μ ( B ) ν ( B ) and ν ( B c ) μ ( B c ) = μ ( B ) ν ( B ). Then you go on to say that
sup A Σ | μ ( A ) ν ( A ) | μ ( B ) ν ( B )
and, since B Σ,
sup A Σ | μ ( A ) ν ( A ) | μ ( B ) ν ( B ) .
From this, it is possible to conclude
sup A Σ | μ ( A ) ν ( A ) | = μ ( B ) ν ( B ) = 1 2 ( μ ( B ) ν ( B ) + ν ( B c ) μ ( B c ) ) = 1 2 ( μ ( A D + ) ν ( A D + ) + ν ( A D ) μ ( A D ) ) = 1 2 ( ( μ ν ) ( A D + ) + ( μ ν ) ( A D ) ) = | μ ν | ( A ) .

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