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Suppose $0\leq X\leq 1.$ Suppose we are given that $\mathrm{Var}(X)\leq a$ where $a$ is some small constant.

What are the best upper bounds we can provide on $\mathrm{Var}(f(X))$ if

a) $f:[0,1]\mapsto\mathbb{R}$ is a Lipschitz function with Lipschitz constant $L$, such as say, $f(x) = x^2$ which has Lipschitz constant $2.$

b) $f:[0,1]\mapsto\mathbb{R}$ is not Lipschitz but is a Hölder continuous function such as say $f(x) = \sqrt{x}.$

I am interested in upper bounds that go to zero as $a$ goes to zero.

Hedonist
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1 Answers1

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In any cases, we utilize the following inequality:

$$ \operatorname{Var}(f(X)) \leq \Bbb{E}[(f(X) - f(\Bbb{E}X))^2]. $$

This is easily proved from the inequality $\operatorname{Var}(Y) \leq \Bbb{E}[Y^2]$ with $Y = f(X) - f(\Bbb{E}X)$. Now assume that

$$|f(x) - f(y)| \leq C|x - y|^{\alpha} \qquad \forall x, y \in [0, 1]$$

for some $\alpha \in (0, 1]$ and $C \in (0, \infty)$. Then

$$ \operatorname{Var}(f(X)) \leq C^2 \Bbb{E}[|X - \Bbb{E}X|^{2\alpha}]. $$

From the Jensen's inequality, we have

$$ \Bbb{E}[|X - \Bbb{E}X|^{2\alpha}] \leq \Bbb{E}[|X - \Bbb{E}X|^{2}]^{\alpha} = \operatorname{Var}(X)^{\alpha}. $$

Consequently we have

$$\operatorname{Var}(f(X)) \leq C^2 \operatorname{Var}(X)^{\alpha}. $$

Sangchul Lee
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  • Can we generalise this result for functions $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ in order to take similar result.Like $V(f(X,Y))\leq C^{2}(V(X)+V(Y))$ – Jonathan1234 Dec 20 '17 at 19:29