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I have been reading Amir Dembo's book, and at the very beginning, I found this result that came across and unfortunately, I cannot derive it by myself. So, I'm looking for some help.

It happens that for a sequence of IID standard normal random variables $X_i$, for $i=1,...,n$. We obtain the empirical mean as:

$\hat{S}_n = \frac{1}{n} \sum_{i=1}^n X_i$.

Then, the claim starts by noting that:

$P ( |\hat{S}_n | \geq \delta ) = 1 - \frac{1}{\sqrt{2\pi}} \int_A e^{-x^2/2} dx$;

Therefore:

$\frac{1}{n} \log P ( |\hat{S}_n | \geq \delta ) \to_{n\to \infty} -\frac{\delta^2}{2}$.

The above result is the one I cannot obtain. I've tried taking the logarithm of the $P ( |\hat{S}_n | \geq \delta )$ but no luck. I mean, I end up with the logarithm of the integral of $e^{x^2/2}$ which is equivalent to logarithm of a summation, so no way to move forward.

Does anybody know what the trick is? Thanks!!

This is on page 2 of Dembo's book in Large deviations.

kentropy
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1 Answers1

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You need a calculus fact about the tail behavior of the Gaussian distribution: $P(Z>t) \sim \phi(t)/t$ as $t\to\infty$ (see this old SE answer or a formula such as the one in the Wikipedia article).

Your $S_n\sim N(0,1/n)$ so $P(|S_n|\ge \delta)=P(|Z|\ge \sqrt n \delta)$ for a standard normal $Z\sim N(0,1)$. So use $t=\sqrt n \delta$ in the formulas in the paragraph above. (Note that $\log \phi(t) \approx -t^2/2$ for large $t$, where $\phi(t)=\exp(-t^2/2)/\sqrt{2\pi}$ is the standard normal density funciton, and that $P(|Z|>t)=2P(Z>t)$, at least when $t>0$.)

kimchi lover
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