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Question:

Let X be a standard Gaussian r.v.

Use Stein's characterization $Ef'(X) = E(Xf(X))$

to prove the Poincare inequality $E|f(X)-Ef(X)|^2 \leq E|f'(X)|^2$.

This looks like Markov's inequality could be used to prove the inequality.

However, I'm not quite sure how to piece it together.

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You want to prove that $E|f(X)-Ef(X)|^2 \leq E|f'(X)|^2$. This inequality doesn't change if you add a constant to $f(x)$, so without loss of generality you can assume that $Ef(X)= 0$. So, all we need is that $E|f(X)|^2 \leq E|f'(X)|^2$.

To show that we will use Stein's method and define a function $g$ such that $g'(x) - x g(x) = f(x)$. This implies that $f'(x) = g''(x) - x g'(x) - g(x)$.

We will use the following two equalities derived by Stein's Lemma: $$E[ f'(X)g(X) ] + E[ f(X)g'(X) ] = E(f(X)g(X))' = E [ X f(X) g(X)] \,\,\,\,\,\, \textrm{(1)}$$ also $$E[ g'(X) g'(X) ] + E[ g''(X) g(X) ] = E(g'(X)g(X))' = E [ X g'(X) g(X)]\,\,\,\,\,\, \textrm{(2)}$$

So, we have that $E[ f(X)^2 ] = E[ f(X) f(X) ] = E[ (g'(X) - X g(X)) f(X) ] = -E[ f'(X)g(X) ]$, where the last equality follows by $(1)$.

Moreover, we also have that: $$E[ f(X)^2 ] = -E[ f'(X)g(X) ] = -E[ (g ''(X) - X g'(X) - g(X) ) g(X) ] = E[ g'(X)^2 + g(X)^2 ]$$ where we applied $(2)$ for the last equality. This implies that $E[ f(X)^2 ] \le E[ g(X)^2 ]$.

Finally, by the Cauchy-Swartz we have that: $$E[ f(X)^2 ] = -E[ f'(X)g(X) ] \le \sqrt{ E[ f'(X)^2 ] E[ g(X)^2 ] } \le \sqrt{ E[ f'(X)^2 ] E[ f(X)^2 ] }$$

This completes the proof of the Poincare Inequality.

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  • $\begingroup$ It might be helpful to say how you know that such a function $g$ exists. Can you write it down explicitly, or are you invoking existence of solutions to ODEs? $\endgroup$ – Nate Eldredge Mar 12 '15 at 15:05
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    $\begingroup$ Stein has proved the existence of such a solution under mild conditions, e.g. absolute continuity. See lemma 1 in stat.duke.edu/courses/Fall10/sta205/lec/topics/dstn.pdf for a more precise statement. I assumed that such mild conditions hold in this case as It is impossible to prove the statement without any conditions since expectations might not even be defined. $\endgroup$ – thelionkingrafiki Mar 12 '15 at 15:15

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