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Let $X$ be a standard normal random variable. Then, for any differentiable f:R→R such that $\mathbb{E}f(X)^2<∞$ the Gaussian Poincare inequality states that $$Var(f(X))≤\mathbb{E}[f′(X)^2]$$. I'd like to know what is the equivalent bound for normal random variables with variance $\sigma^2 \neq 1$, if there is such an equivalent. I tried reading up on the Gaussian Poincare inequality and where the fact that $\sigma^2 = 1$ comes in, and couldn't find my way around it. I'd love some help. Thanks!

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  • $\begingroup$ At least you always can always set $X=Z/\sigma$, when $Z \sim N(0,\sigma^2)$. $\endgroup$
    – fes
    Jul 17, 2020 at 16:26
  • $\begingroup$ Thanks! I think that might do it! $\endgroup$
    – Hadasdas
    Jul 19, 2020 at 6:54
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    $\begingroup$ I hope my math is correct: we set $X=Z/\sigma$ when $Z~N(0,\sigma^2)$ and define $g(X) = f(\sigma X)$ which means $g(X) = f(Z)$. For the derivative we have that: $g'(X) = \left(f(\sigma X)\right)' = \sigma f'(\sigma X) = \sigma f'(Z)$ Therefore: $$Var(f(Z)) = Var(f(\sigma X) = Var(g(X)) \leq \mathbb{E}[g'(X)^2] = \mathbb{E}[\sigma^2 f'(Z)^2] = \sigma^2\mathbb{E}[f'(Z)^2] $$ $\endgroup$
    – Hadasdas
    Jul 19, 2020 at 8:27
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    $\begingroup$ Looks correct to me. $\endgroup$
    – fes
    Jul 19, 2020 at 9:57

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More generally, for differentiable $f: \mathbb{R}^d \to \mathbb{R}$, and $d$-dimensional Gaussian vector $X \sim N(0, \Sigma$), we have $$ \text{Var}(f(X)) \le \mathbb{E} \langle \Sigma \nabla f(X), \nabla f(X) \rangle $$

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