Exercise 2.2: Mills Ratio
Let \(\phi(z) = \frac{1}{\sqrt{2 \pi}} e^{- z^2 / 2}\) be the density of the standard normal \(Z \sim \gauss (0, 1)\) variate.
(a)
The result follows from \(\phi' (z) = - z \phi(z)\).
(b)
Since \(\P (Z \geq z) = \int_z^\infty \phi (x) \, \sd x \overset{\text{(a)}}{=} - \int_z^\infty \frac{\phi'(x)}{x} \, \sd x\) the result can be obtained by repeated application of integration by parts: \begin{align} \P (Z \geq z) & = \left[ - \frac{\phi(x)}{x} \right]_z^\infty - \int_z^\infty \frac{\phi(x)}{x^2} \, \sd x \newline & = \frac{\phi(z)}{z} + \int_z^\infty \frac{\phi’(x)}{x^3} \, \sd x \newline & \overset{\text{(i)}}{=} \phi(z) \left(\frac{1}{z} - \frac{1}{z^3} \right) + 3 \int_z^\infty \frac{\phi(x)}{x^4} \, \sd x \newline & = \phi(z) \left(\frac{1}{z} - \frac{1}{z^3} \right) - 3 \left\lbrace \left[ \frac{\phi(x)}{x^5} \right]_z^\infty + 5 \int_z^\infty \frac{\phi(x)}{x^6} \, \sd x \right\rbrace \newline & \overset{\text{(ii)}}{=} \phi(z) \left(\frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5} \right) - 15 \int_z^\infty \frac{\phi(x)}{x^6} \, \sd x \, . \end{align} Since \(\phi \geq 0\), the lower bound follows from (i) and the upper bound from (ii).
Notes
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The quantity \(\P (Z \geq z) \, / \, \phi (z)\) is called the Mills ratio. The above result shows that for large \(z\), the ratio is close to \(1 / z\).
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The result derived in (b) can be made sharper by continuing with the repeated integration by parts. This can then be used to obtain increasingly sharp bounds on the tail probabilities of the standard normal distribution, which can be easier to work with than the exact expression in terms of the Gaussian CDF.