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Consider a random variable $X$ in $(\Omega, \mathcal{F}, \mathbb{P})$. Let $p,q$ be two densities with respect to a measure $\mu$ in $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra in $\mathbb{R}$. Could you help me to show that $$ \left(\int_{-\infty}^\infty \sqrt{p}\sqrt{q}d\mu\right)^2\le 2 \int_{-\infty}^\infty\min\{p,q\}d\mu $$ from van der Vaart "Asymptotic Statistics" proof of Lemma 14.31.

My attempt:

\begin{align}\left(\int_{-\infty}^\infty \sqrt{p}\sqrt{q}d\mu\right)^2&\le \left(\int_{-\infty}^\infty \min\{\sqrt{p},\sqrt{q}\}\left(\sqrt{p}+\sqrt{q}\right)d\mu\right)^2\\[0.2cm]& \le \int_{-\infty}^{\infty}\left(\sqrt{p}+\sqrt{q}\right)^2d\mu \int_{-\infty}^{\infty}\min\left\{\sqrt{p},\sqrt{q}\right\}^2d\mu\\[0.2cm]&=\left(2+2\int_{-\infty}^{\infty}\sqrt{p}\sqrt{q}d\mu\right)\int_{-\infty}^{\infty}\min\{p,q\}d\mu=\dots?\end{align}

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Let

\begin{align}\left( \int_{-\infty}^{\infty} \sqrt{p}\sqrt{q} \;d\mu \right)^2 &= \left( \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})\cdot \max(\sqrt{p},\sqrt{q}) \;d\mu \right)^2 \\ &\leq \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})^2 \;d\mu \cdot \int_{-\infty}^{\infty} \max(\sqrt{p},\sqrt{q})^2 \;d\mu \\ &\leq \int_{-\infty}^{\infty} \min(p,q)\, d\mu \cdot \int_{-\infty}^{\infty}(p + q) \; d\mu \\ &= 2\int_{-\infty}^{\infty} \min(p,q)\, d\mu. \end{align}

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    $\begingroup$ Maybe the last step is an equality? thanks $\endgroup$ – user299158 Feb 5 '16 at 15:55
  • $\begingroup$ @user299158 true, edited it. $\endgroup$ – user159517 Feb 5 '16 at 18:11

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