world:pinsker
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world:pinsker [2022/03/16 07:40] 127.0.0.1 external edit |
world:pinsker [2025/07/08 16:52] (current) rdouc [The proof] |
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| If $0\leq f\leq 1$, then $\Var_{\mu}(f)=\PE[f^2]-\PE^2[f]\leq \PE[f]-\PE^2[f]=\PE[f](1-\PE[f]) \leq 1/4$ and we get | If $0\leq f\leq 1$, then $\Var_{\mu}(f)=\PE[f^2]-\PE^2[f]\leq \PE[f]-\PE^2[f]=\PE[f](1-\PE[f]) \leq 1/4$ and we get | ||
| $$ | $$ | ||
| - | \rmd_{tv}(\sigma,\pi) \leq KL(\sigma|| \pi)/2 | + | \rmd_{tv}^2(\sigma,\pi) \leq KL(\sigma|| \pi)/2 |
| $$ | $$ | ||
| ===== The proof ===== | ===== The proof ===== | ||
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| </WRAP> | </WRAP> | ||
| - | Indeed, set $h(y)=\frac{2}{3}(2y+1)(-\log y+y-1)-(y-1)^2$ and check that $h(1)=h'(1)=0$ and $h''(y)=2(x-1)^2/(3y^2)>0$ for all $y \geq 0$. Therefore, $h$ is a convex function on $\rset^+$ and attains its minimum at $y=1$. Therefore, $h(y) \geq 0$ for all $y\geq 0$, which proves $(\star)$. Then, using $(\star)$ with $y=\frac{\pi(x)}{\sigma(x)}$, and taking $\bar f=f-\mu(f)$ where $\mu=2\pi/3 +\sigma/3$, | + | Indeed, set $h(y)=\frac{2}{3}(2y+1)(-\log y+y-1)-(y-1)^2$ and check that $h(1)=h'(1)=0$ and $h''(y)=2(y-1)^2/(3y^2)>0$ for all $y \geq 0$. Therefore, $h$ is a convex function on $\rset^+$ and attains its minimum at $y=1$. Therefore, $h(y) \geq 0$ for all $y\geq 0$, which proves $(\star)$. Then, using $(\star)$ with $y=\frac{\pi(x)}{\sigma(x)}$, and taking $\bar f=f-\mu(f)$ where $\mu=2\pi/3 +\sigma/3$, |
| \begin{align*} | \begin{align*} | ||
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| \left(\int |\bar f(x)|\ |\sigma(x)-\pi(x)|\rmd x\right)^2 \\ | \left(\int |\bar f(x)|\ |\sigma(x)-\pi(x)|\rmd x\right)^2 \\ | ||
| &=\left(\int |\bar f(x)|\ \sigma(x)\ |\frac{\pi(x)}{\sigma(x)}-1|\rmd x\right)^2 \\ | &=\left(\int |\bar f(x)|\ \sigma(x)\ |\frac{\pi(x)}{\sigma(x)}-1|\rmd x\right)^2 \\ | ||
| - | &=\left(\int \sigma(\rmd x) \sqrt{\frac{2\bar f^2(x)}{3}\left(2\frac{\pi(x)}{\sigma(x)}+1\right)} \sqrt{\left( -\log \frac{\pi(x)}{\sigma(x)} | + | &\leq\left(\int \sigma(\rmd x) \sqrt{\frac{2\bar f^2(x)}{3}\left(2\frac{\pi(x)}{\sigma(x)}+1\right)} \sqrt{\left( -\log \frac{\pi(x)}{\sigma(x)} |
| +\frac{\pi(x)}{\sigma(x)}-1\right)}\right)^2 \end{align*} And we conclude by applying Holder's inequality: $\left(\int \sigma(\rmd x) \sqrt{a(x)}\sqrt{b(x)} \right)^2\leq \int \sigma(\rmd x) a(x) \times \int \sigma(\rmd x) b(x)$ $\eproof$ | +\frac{\pi(x)}{\sigma(x)}-1\right)}\right)^2 \end{align*} And we conclude by applying Holder's inequality: $\left(\int \sigma(\rmd x) \sqrt{a(x)}\sqrt{b(x)} \right)^2\leq \int \sigma(\rmd x) a(x) \times \int \sigma(\rmd x) b(x)$ $\eproof$ | ||
world/pinsker.1647412823.txt.gz · Last modified: 2022/03/16 07:40 by 127.0.0.1
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