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world:totalvariation


2017/10/07 23:39 ·

# The total variation norm between two probability measures

Let $\mu, \nu \in \meas{1}(\Xset)$ where $(\Xset,\Xsigma)$. The total variation norm between $\mu$ and $\nu$ is defined by $$\tv{\mu-\nu}=\sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|=\gamma(|f-g|)$$ where $\mu\preceq \gamma$, $\nu \preceq \gamma$ and $\mu= f \cdot \gamma$, $\nu= g \cdot \gamma$.

# Proof

Indeed for all $h \in \mcf(\Xset)$, $$|\mu(h)-\nu(h)|=|\gamma( (f-g)h)| \leq \supnorm{h} \gamma|f-g|$$ showing that $\sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|\leq \gamma(|f-g|)$

Moreover, $$\gamma(|f-g|)=\gamma( (f-g)\ \underbrace{\sg(f-g)}_{h})=\mu(h) -\nu(h)$$ where $h=\sg(f-g)$ and since $\supnorm{h}=1$, we thus have $\gamma(|f-g|) \leq \sup_{h \in \mcf(\Xset), \supnorm{h}\leq 1} |\mu(h)-\nu(h)|$. The proof is completed.

world/totalvariation.txt · Last modified: 2022/03/16 07:40 (external edit)