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I tried taking exponentials and using Markov's Inequality, but this only gave me an upper bound of $\sqrt{2\pi}$. I'm not sure how to begin to approach this question - can anyone give a hint?

acernine
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  • Did you try directly, $\frac{1-\Phi(x)}{\phi(x)} -\frac{1}{x} =h(x)$? – Alex May 26 '20 at 15:35
  • @Alex I tried finding $h'(x)$, but I still got an expression containing $P(X>x)$ so I wasn't sure how to proceed from there. – acernine May 26 '20 at 15:40
  • I haven't tried, but derivative of $\Phi$ is $\phi$, also at 0 this right inequality holds – Alex May 26 '20 at 16:00
  • @StubbornAtom That question does give the upper bound, but not the lower bound (as far as I can see), so it answers half of the question. Thanks for bringing it to my attention. – acernine May 26 '20 at 16:00
  • No it completely answers your question. Look at the answer by Prof. Sarwate. – StubbornAtom May 26 '20 at 16:01
  • @StubbornAtom You're right, I just looked and their answer does give exactly what I need. Thanks. – acernine May 26 '20 at 16:21

1 Answers1

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You can use these known "double inequality" for the Gaussian (Bounds and Approximations in "Q-function", Wikipedia)

Applying this double inequality to your exercise you immediately get

$\frac{x}{x^2+1}<\frac{\mathbb{P}[X>x]}{\phi(x)}<\frac{1}{x}$

To prove what you need it is enough to prove that, for $x>0$,

$\frac{1}{x}-\frac{1}{x^3}<\frac{x}{x^2+1}$

tommik
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