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world:pinsker
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world:pinsker [2025/07/08 16:44] rdouc [What?] |
world:pinsker [2025/07/08 16:52] (current) rdouc [The proof] |
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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.1751985844.txt.gz · Last modified: 2025/07/08 16:44 by rdouc
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