Bounding the normal tail

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Let Z \sim \mathcal{N}(0,1). For large t, we can bound the normal tail as follows:

\dfrac{\phi(t)}{t}\left( 1 - \dfrac{1}{t^2}\right) \leq \mathbb{P}(Z > t) \leq \dfrac{\phi(t)}{t}.

This means that for large t, \mathbb{P}(Z > t) = 1 - \Phi(t) \approx \dfrac{\phi(t)}{t}.

The proof of the second inequality is fairly straightforward:

\begin{aligned} \mathbb{P}(Z > t) &= \int_t^\infty \frac{e^{-x^2/2}}{\sqrt{2\pi}} dx \leq \int_t^\infty \frac{x}{t} \frac{e^{-x^2/2}}{\sqrt{2\pi}} dx \\ &= \left[-\frac{1}{t} \frac{e^{-x^2/2}}{\sqrt{2\pi}} \right]_t^\infty = \frac{\phi(t)}{t}. \end{aligned}

There might be an easier proof of the first inequality, but the one I found goes through a slightly tighter bound: \mathbb{P}(Z > t) \geq \dfrac{t}{1+t^2} \phi(t). We use a trick similar to that above to prove this inequality:

\begin{aligned} \mathbb{P}(Z > t) &\geq \int_t^\infty \frac{t^2}{x^2} \frac{e^{-x^2/2}}{\sqrt{2\pi}} dx \\ &= \left[ -\frac{t^2}{x} \frac{e^{-x^2/2}}{\sqrt{2\pi}} \right]_t^\infty + \int_t^\infty \frac{t^2}{x} (-x) \frac{e^{-x^2/2}}{\sqrt{2\pi}} dx \\ &= t\phi(t) - t^2 \mathbb{P}(Z > t), \\ (1 + t^2) \mathbb{P}(Z > t) &\geq t\phi(t), \\ \mathbb{P}(Z > t) &\geq \dfrac{t}{1+t^2}\phi(t). \end{aligned}

Since \dfrac{t}{1+t^2} \geq \dfrac{t^2 - 1}{t^3} (cross-multiply and check), the first inequality above follows.

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  1. Pingback: Approximating large normal quantiles | Statistical Odds & Ends

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