$-\log\Phi(x) \sim \frac{x^2}{2}$ as $x \to -\infty$ for $\Phi(x) = (2\pi)^{-\frac 12}\int_{-\infty}^x \mathrm{e}^{-t^2/2}\,\mathrm{d}t$

Question

For $$\Phi(x) = (2\pi)^{-\frac 12}\int_{-\infty}^x \mathrm{e}^{-t^2/2}\,\mathrm{d}t,$$ it is claimed in the proof of Lemma 8.12 of this book that we have the asymptotic relation $-\log\Phi(x) \sim \frac{x^2}{2}$ as $x \to -\infty$. However, I do not see any way of deriving it from the so-called Gordon's inequality, which says that $$1 - \Phi(t) \sim \frac{1}{t\sqrt{2\pi}}\mathrm{e}^{-t^2/2}$$ for $t \gg 1$ being sufficiently large. May I know how can we obtain the advertised asymptotic relation as $x \to -\infty$?

Answer 1

Note that $-\ln\left( \Phi(x)\right) \sim \frac{x^2}{2}$ as $x \to -\infty$ is equivalent with $\ln \left( 1 - \Phi(t) \right) \sim -\frac{t^2}{2}$ as $t \to +\infty$, due to relation $\Phi(x) + \Phi(-x) = 1$ (and setting $t = -x$).

But since $1-\Phi(t) \sim \frac{1}{t\sqrt{2\pi}}\exp(-\frac{t^2}{2})$, then by definition $t\sqrt{2\pi}(1-\Phi(t))\exp(\frac{t^2}{2}) \to 1$. Taking $\ln(\cdot)$ both sides, we see that $\ln(t\sqrt{2\pi}) + \ln(1-\Phi(t)) + \frac{t^2}{2} \to 0 $. After dividing LHS by $\frac{t^2}{2}$ we don't change the limit, hence $$ \frac{2\ln(t\sqrt{2\pi})}{t^2} + \frac{2\ln(1-\Phi(t))}{t^2} +1 \xrightarrow[t \to \infty]{} 0, $$ or in other words $\frac{2}{t^2}\ln(1-\Phi(t)) \to -1$ as $t \to \infty$. This is exactly as saying $\ln(1-\Phi(t)) \sim -\frac{t^2}{2}$.

Answer 2

I try to turn the way around: for $x\to +\infty$, $\log ( (2\pi)^{-\frac 12}\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t )\sim -x^2/2$. Since this is asymptotic, we omit the $(2\pi)^{-\frac 12}$ and only focus on the integral.

Notice that $$\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t\ge \int_{x}^{x+1/x} \mathrm{e}^{-t^2/2}\,\mathrm{d}t\ge\int_{x}^{x+1/x} \mathrm{e}^{-(x+1/x)^2/2}\,\mathrm{d}t=\frac 1x\mathrm{e}^{-\frac{x^2+2+1/x^2}{2}}.$$

So when taking logarithms, we have $\log(\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t)\ge-\frac{x^2}{2}+O(\log x)$.

Furthermore, notice that for $t\ge x$, we have $t^2\ge x^2+2x(t-x)$. So we have, for $x>0$,

$$\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t\le \int_{x}^{\infty} \mathrm{e}^{\frac{-x^2-2x(t-x)}{2}}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\int_{x}^{\infty} \mathrm{e}^{-x(t-x)}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\int_{0}^{\infty} \mathrm{e}^{-xt}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\frac{1}{x}.$$

So when taking logarithms, we also have $\log(\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t)\le-\frac{x^2}{2}+O(\log x)$.

Therefore $\log \Phi(x)\sim -\frac{x^2}{2}$. What's more, we can prove that $\log \Phi(x)= -\frac{x^2}{2}-\log x+O(1)$.

Answer 3

Let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Notice that $$\begin{align} -\frac{d}{dx}\Big(\Big(\frac1x-\frac1{x^3}\Big)\phi(x)\Big)&=(1-3x^{-4})\phi(x)\\ &\leq \phi(x)\leq (1+x^{-2})\phi(x)=-\frac{d}{dx}\Big(\frac{\phi(x)}{x}\Big) \end{align}$$ For $a<0$, after tntegrating over $(-\infty,a]$, we obtain $$\Big(\frac1{a^2}-\frac1a\Big)\phi(a)\leq \Phi(a)\leq-\frac{1}{a}\phi(a)$$ and so $$\lim_{a\rightarrow-\infty}\frac{-a\Phi(a)}{\phi(a)}=1$$ which (using L'Hospital rule for example) gives $$\lim_{a\rightarrow-\infty}\frac{-\log(\Phi(a))}{a^2/2}=1$$