6.5. Total Variation: Scheffe’s Theorem

Remark

\((\mathbb{R}, \mathcal{R})\)에서의 probability measures \(\mu_n\) and \(\mu\)에 대해 아래의 조건은 \((3\implies 2\implies 1)\)이다 (progressively stronger forms);

1. \(\mu_n\leadsto \mu\).

2. \(\mu(A)\rightarrow \mu(A)\) for all \(A\in \mathcal{R}\).

3. \(\sup_{A\in \mathcal{R}}|\mu_n(A)-\mu(A)|\rightarrow 0\).

• \((2\implies 1)\): Portmanteau의 마지막 condition \((\lim_n \mu_n(A)=\mu(A)\) for all \(A\in \mathcal{R}\) with \(\mu(\partial A)=0)\).

• \((3\implies 2)\): absolute convergence \(\implies\) convergence.

• The convergence in 3 is called Convergence in total variation norm: \[ ||\mu-\nu||_{\text{TV}}=\sup_{A\in\mathcal{R}}|\mu(A)-\nu(A)|. \] Scheffe의 Theorem은 Convergence in total variation norm의 sufficient condition을 알려준다.

Scheffe’s theorem을 위해 필요

Lemma

If \(\mu\) and \(\nu\) are probability measures, with densities \(f\) and \(g\) w.r.t some dominating measure \(\gamma\), then

\[ \sup_{A\in \mathcal{F}}|\mu(A)-\nu(A)|=\frac{1}{2}\int|f-g|d\gamma. \]

• Dominating measure는 \(\mu,\nu \ll \gamma\)를 의미한다. 즉 이는 Radon Nidodym에 의해 \(\mu(A)=\int_A f \mbox{ }d\gamma\), \(\nu(A)=\int_A g \mbox{ }d\gamma\)를 의미한다.

• \(\int(f-g)\mbox{ }d\gamma=\mu(\Omega)-\nu(\Omega)=0\)

  \(\implies \int_A(f-g)\mbox{ }d\gamma+ \int_{A^c}(f-g)\mbox{ }d\gamma=0\)

  \(\implies |\int_A(f-g)\mbox{ }d\gamma|=|\int_{A^c}(f-g)\mbox{ }d\gamma|\).

• Thus, for any \(A\in \mathcal{F}\),

\[\begin{eqnarray*} |\mu(A)-\nu(A)|&=&\left| \int_A (f-g)d\gamma \right|\\ &=& \frac{1}{2}\left| \int_A (f-g)d\gamma \right|+ \frac{1}{2}\left| \int_{A^c} (f-g)d\gamma \right|\\ &\le&\frac{1}{2}\int_A \left| f-g\right|d\gamma + \frac{1}{2} \int_{A^c} \left|f-g\right|d\gamma \\ &=&\frac{1}{2}\int \left| f-g\right|d\gamma. \end{eqnarray*}\]

• Let \(A=[f\ge g]=\{\omega:f(\omega)\ge g(\omega)\}\). Then, \(A\in \mathcal{F}\), and as above,

\[\begin{eqnarray*} |\mu(A)-\nu(A)|&=& \frac{1}{2}\left| \int_A (f-g)d\gamma \right|+ \frac{1}{2}\left| \int_{A^c} (f-g)d\gamma \right|\\ &=&\frac{1}{2}\left|\int_A (f-g)d\gamma\right| + \frac{1}{2}\left|\int_{A^c} (g-f)d\gamma\right| \\ &=&\frac{1}{2}\int_A \left| f-g\right|d\gamma+\frac{1}{2}\int_{A^c} \left| f-g\right|d\gamma= \frac{1}{2}\int \left| f-g\right|d\gamma. \end{eqnarray*}\] 즉, \(\sup_{A\in \mathcal{F}}|\mu(A)-\nu(A)|=\frac{1}{2}\int|f-g|d\gamma\)이다.

중요하다

Theorem (Scheffe’s Theorem)

\(\mu_n,n\ge 1\) and \(\mu\) are probability measures having densities \(f_n\) and \(f\), respectively, w.r.t some dominating measure \(\gamma\). If \[ f_n\rightarrow f\mbox{ }\mbox{ }\mbox{ } \gamma\mbox{-a.e. }\mbox{ as }n\rightarrow \infty, \] then \(\mu_n\) converges to \(\mu\) in total variation norm, i.e., \[ \sup_{A\in\mathcal{R}}|\mu_n(A)-\mu(A)|\rightarrow 0. \]

• 증명: \(f_n\rightarrow f\) \(\gamma\)-a.e.\(\implies 0\le (f-f_n)^+\le |f-f_n|\rightarrow 0\) \(\gamma\)-a.e.이다.

  즉, \((f-f_n)^+\rightarrow 0\) \(\gamma\)-a.e.이다.

  또한 \(f,f_n\ge 0\)이기 때문에 \((f-f_n)^+\le f\in L^1\)이다.

  때문에 D.C.T에 의해 \(\int (f-f_n)^+ d\gamma\rightarrow 0\)이다.

• 또한 \[ 0\stackrel{\text{prob}=1}{=}\int (f-f_n)d\gamma= \int (f-f_n)^+d\gamma- \int (f-f_n)^-d\gamma\implies \int (f-f_n)^+d\gamma= \int (f-f_n)^-d\gamma \] 이다.

• 그러므로 \[ \int |f-f_n|d\gamma= \int (f-f_n)^+d\gamma+\int (f-f_n)^-d\gamma= 2\int (f-f_n)^+d\gamma\rightarrow 0. \] 때문에 위의 lemma에 따르면 \[ \sup_{A\in\mathcal{R}}|\mu_n(A)-\mu(A)|=\frac{1}{2}\int|f_n-f|d\gamma \rightarrow 0. \]

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