Gaussian Pointcare Inequality Theorem

[[concept]]

Topics

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Statement

Gaussian Pointcare Inquality Theorem (GIP)

has PI where (independent of !)

Proof

By the tensorization lemma, it suffices to do . We will use the central limit theorem and Efron-Stein.

Introduce . Then we have

\text{Var}_{g\sim {\cal N}(0, 1)}(f(g)) &\overset{\text{(CLT)}}\approx \text{Var}\left( f\left( \frac{1}{\sqrt{ k }}\sum_{i=1}^k z_{k} \right) \right) \\ &\overset{(ES)}\leq \frac{1}{2} \sum_{i=1}^k \mathbb{E}_{}\left[\left( f\left( \underbrace{ \frac{1}{\sqrt{ k }}\sum_{j=1}^k z_{j} }_{ \frac{1}{\sqrt{ k }}\sum_{j\neq i}z_{j} +\frac{z_{i}}{\sqrt{ k }} }\right) -f\left( \underbrace{ \frac{1}{\sqrt{ k }} \sum_{j\neq i}z_{j} }_{ S_{i} } + \frac{z_{i}}{\sqrt{ k }} \right) \right)^2\right] \\ &= \frac{1}{4} \sum_{i=1}^k \mathbb{E}_{}\left[\left( f\left( S_{i}+\frac{1}{\sqrt{ k }} \right)-f\left( S_{i}-\frac{1}{\sqrt{ k }} \right) \right)^2\right] \\ &\approx \frac{1}{4} \sum_{i=1}^k \mathbb{E}_{}\left[\left( f'(S_{i}) \frac{2}{\sqrt{ k }} \right)^2\right] \\ &= \mathbb{E}_{}\left[(f'(S_{1}))^2\right] \\ &\approx \mathbb{E}_{g \sim {\cal N}(0,1)}\left[(f'(g))^2\right] \end{align}$$

Corollaries

Corollary

Proof

We can view as and construct where

\sqrt{ 2 }g_{1,1} & \dots & & \\ \vdots & \ddots & g_{i,j} & & \\ & g_{i,j} & \ddots \\ & & & \sqrt{ 2 }g_{\frac{d(d+1)}{2}} \end{bmatrix}$$ Then define $f(g)=\lambda_{1}(X(g))$. This is Lipschitz with $L=\sqrt{ 2 }$ and we have $$\begin{align} \text{Var}_{X \sim GOE(GPI)}(\lambda_{1}(X)) &= \text{Var}_{g \sim {\cal N}(0, I_{\underbrace{ N }_{ d(d+1)/2 }})}(f(g)) \\ \lvert f(g)-f(h) \rvert &= \lvert \lambda_{1}(X(g)) \rvert -\lvert \lambda_{1}(X(h)) \rvert \\ &\leq \lvert \lvert X(g) - X(h) \rvert \rvert _{\text{op}} \\ &\leq \lvert \lvert X(g) - X(h) \rvert \rvert _{F} \\ &\leq \sqrt{ 2 } \lvert \lvert g-h \rvert \rvert \end{align}$$ $$\tag*{$\blacksquare$}$$

Corollary

References

References

See Also

Mentions

Mentions

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