Showing that $\lim_{x \rightarrow \infty} - \frac{\log (2 \Phi(\alpha \sqrt{x}))}{x} = \frac{\alpha^2}{2}$.

Question

How can I show that $$ \lim_{x \rightarrow \infty} - \frac{\log (2 \Phi(\alpha \sqrt{x}))}{x} = \frac{\alpha^2}{2}, $$ where $\Phi$ denotes the cdf of the standard normal distribution and $\alpha < 0$?

I tried to use $$ \Phi(y) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \exp(-\tfrac{y^2}{2}) \Big( y + \frac{y^3}{3} + \frac{y^5}{3 \cdot 5} + \ldots \Big). $$ Then $$ \log(2\Phi(y)) = -\frac{y^2}{2} + \log\bigg(\frac{2}{\sqrt{2\pi}} \Big( y + \frac{y^3}{3} + \frac{y^5}{3 \cdot 5} + \ldots \Big)\bigg) $$ and thus $$ - \frac{\log (2 \Phi(\alpha \sqrt{x}))}{x} = \frac{\alpha^2}{2} - \frac{\log(\frac{2}{\sqrt{2\pi}} ( (\alpha \sqrt{x}) + \frac{(\alpha \sqrt{x})^3}{3} + \frac{(\alpha \sqrt{x})^5}{3 \cdot 5} + \ldots ))}{x}. $$ Now it remains to show that $$ \lim_{x \rightarrow \infty} - \frac{\log(\frac{2}{\sqrt{2\pi}} ( (\alpha \sqrt{x}) + \frac{(\alpha \sqrt{x})^3}{3} + \frac{(\alpha \sqrt{x})^5}{3 \cdot 5} + \ldots ))}{x} = 0. $$

Answer 1

By L'Hopital's rule we have

$$-\lim_{x \to \infty} \frac{\log(2 \Phi(\alpha \sqrt{x}))}{x} = -\lim_{x \to \infty} \frac{2 \phi(\alpha\sqrt{x})}{2 \Phi(\alpha \sqrt{x})}\frac{\alpha}{2 \sqrt{x}}$$

Since $\phi'(x) = -x\phi(x)$, a second application of L'Hopital's rule gives

$$-\lim_{x \to \infty} \frac{\log(2 \Phi(\alpha \sqrt{x}))}{x} = \lim_{x \to \infty} \frac{-(\alpha^2/2) \phi(\alpha\sqrt{x})(-\alpha/2)}{\sqrt{x} \phi(\alpha\sqrt{x})(\alpha/(2 \sqrt{x})) + \Phi(\alpha \sqrt{x})/(2 \sqrt{x})}\\ = \lim_{x \to \infty} \frac{\alpha^2}{2}\frac{ \phi(\alpha\sqrt{x})(\alpha/2)}{ \phi(\alpha\sqrt{x})(\alpha/2) + \Phi(\alpha \sqrt{x})/(2 \sqrt{x})}\\ = \lim_{x \to \infty} \frac{\alpha^2}{2}\frac{ 1}{ 1 + \Phi(\alpha \sqrt{x})/(\alpha \sqrt{x} \phi(\alpha \sqrt{x}))} \\ = \frac{\alpha^2}{2},$$

where the final limit follows from $\Phi(z)/(z \phi(z)) \to 0$ as $z \to -\infty$, also by L'Hopital's rule.

Answer 2

If you use l'Hôpital's rule, the limit is equal to $$\lim_{x \to \infty} - \frac{\log (2 \Phi(\alpha \sqrt{x}))}{x} = - \frac{\alpha}{2} \lim_{x \to \infty} \frac{\phi(\alpha \sqrt{x})}{\sqrt{x} \ \Phi(\alpha\sqrt{x})} = \frac{\alpha^2}{2}.$$

The last step is due to the Mills Ratio $\frac{\phi(y)}{y (1 - \Phi(y))} \to 1$ as $y \to \infty$.