rzfansubs87

Answered

2022-07-10

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

$\Vert \mu -\nu \Vert =|\mu -\nu |(X)=2sup\{|\mu (A)-\nu (A)|:A\in \mathrm{\Sigma}\}$

where the total variation norm $\Vert \cdot \Vert $ and the total variation of a measure $|\cdot |$ 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.

$\Vert \mu -\nu \Vert =|\mu -\nu |(X)=2sup\{|\mu (A)-\nu (A)|:A\in \mathrm{\Sigma}\}$

where the total variation norm $\Vert \cdot \Vert $ and the total variation of a measure $|\cdot |$ 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

Expert

2022-07-11Added 11 answers

First we need to show $\mu -\nu (\cdot )$ 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 $E\cap {D}^{+}$ using the Hahn decomposition (${D}^{+}\bigsqcup {D}^{-}$) of $X$ under $\mu -\nu $ s.t. $\mu -\nu $ is positive on ${D}^{+}$. It is then easy to prove that for all $A\in \mathrm{\Sigma}$, $\mu (A)-\nu (A)\le \mu (B)-\nu (B)$ and $\le \nu ({B}^{c})-\mu ({B}^{c})=\mu (B)-\nu (B)$. Then you go on to say that

$\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|\le \mu (B)-\nu (B)$

and, since $B\in \mathrm{\Sigma}$,

$\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|\ge \mu (B)-\nu (B).$

From this, it is possible to conclude

$\begin{array}{rl}\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|& =\mu (B)-\nu (B)\\ & =\frac{1}{2}(\mu (B)-\nu (B)+\nu ({B}^{c})-\mu ({B}^{c}))\\ & =\frac{1}{2}(\mu (A\cup {D}^{+})-\nu (A\cup {D}^{+})+\nu (A\cup {D}^{-})-\mu (A\cup {D}^{-}))\\ & =\frac{1}{2}((\mu -\nu )(A\cup {D}^{+})+(\mu -\nu )(A\cup {D}^{-}))\\ & =|\mu -\nu |(A).\end{array}$

$\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|\le \mu (B)-\nu (B)$

and, since $B\in \mathrm{\Sigma}$,

$\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|\ge \mu (B)-\nu (B).$

From this, it is possible to conclude

$\begin{array}{rl}\underset{A\in \mathrm{\Sigma}}{sup}|\mu (A)-\nu (A)|& =\mu (B)-\nu (B)\\ & =\frac{1}{2}(\mu (B)-\nu (B)+\nu ({B}^{c})-\mu ({B}^{c}))\\ & =\frac{1}{2}(\mu (A\cup {D}^{+})-\nu (A\cup {D}^{+})+\nu (A\cup {D}^{-})-\mu (A\cup {D}^{-}))\\ & =\frac{1}{2}((\mu -\nu )(A\cup {D}^{+})+(\mu -\nu )(A\cup {D}^{-}))\\ & =|\mu -\nu |(A).\end{array}$

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