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I'm currently reading Roman Vershynin's High-Dimensional Probability. For Proposition $2.1.2$, I wonder how the lower bound is obtained. I understand that the lower bound is correct, but I don't know where the term $-3x^{-4}$ comes from. Thanks.

Proposition 2.1.2

Analyst_311419
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1 Answers1

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It just removes a term to help the proof. Namely, for any $x>0$, $$ \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \geq (1-3x^{-4})\cdot \frac{1}{\sqrt{2\pi}} e^{-x^2/2} $$ since you just remove something positive from the LHS; and then you can use the "rabbit-out-of-the-hat" identity he gives to give an exact expression for the integral of the RHS.

Clement C.
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  • Why it's $3x^{-4}$ but not some other term? – Analyst_311419 Jan 13 '22 at 08:18
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    That term is just chosen to get something whose integral can be explicitly computed. Maybe try to go backwards: start with the RHS you want to obtain, differentiate it, you'll get that term. – Clement C. Jan 13 '22 at 08:35