Limit of normal CDF

Question

I'm trying to prove that $\Phi\left((1-\epsilon)\sqrt{2\log(n)}\right)^n \to 0$ as $n \to \infty$ (where $\Phi$ is the standard normal CDF) for any $\epsilon > 0$. I'm having problems. I've tried to relate it to the complimentary error function using the fact that $\Phi(x) = 0.5\operatorname{erfc}(-\frac{x}{\sqrt{2}})$, but this hasn't helped me.

Answer 1

Edit: The question has changed after I posted. Please refer to angryavian's answer.

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$$ \def\dif{\mathrm{d}} $$

By the definition of CDF, $\Phi(x) := 1 - \int_x^{+\infty} \text{PDF}\,\dif x$, where the second term (complementary CDF, CCDF) approaches to zero as $x \to +\infty$. Therefore, the limit $\Phi(\sqrt{2 \log n})^n$ is in the form of $1^{+\infty}$, and the key is to estimate how fast the CCDF approaches to zero:

• If CCDF approaches to zero faster than $1/n$, then the limit is zero one.
• If CCDF is equivalent to $a/n$, then the limit is the finite number $\lim_{n \to +\infty} (1 - a/n)^n = e^{-a}$.
• If CCDF approaches to zero faster than $1/n$, then the limit goes to infinity.

Fortunately, there is a well-known asymptotic approximation of the standard normal CCDF:

$$ \int_x^{+\infty} \frac{e^{-x^2/2}}{\sqrt{2\pi}} \,\dif x \sim \frac{e^{-x^2/2}}{\sqrt{2\pi}} \frac{1}{x}. $$

Proof. Apply the L'Hôpital's rule.

$$ \begin{aligned} \frac{\dif}{\dif x} \text{LHS} &= -\frac{e^{-x^2/2}}{\sqrt{2\pi}}. \\ \frac{\dif}{\dif x} \text{RHS} &= \frac{e^{-x^2/2}}{\sqrt{2\pi}} \left(\frac{-x}{x} -\frac{1}{x^2}\right) = -\frac{e^{-x^2/2}}{\sqrt{2\pi}} \left(1 + \frac{1}{x^2}\right). \end{aligned} $$

The ratio of them approaches to $1$ as $x \to +\infty$. $\blacksquare$

Now substitute $x = \sqrt{2 \log n}$ into the formula, we have

$$ 1 - \Phi(\sqrt{2 \log n}) \sim \frac{e^{-(\sqrt{2 \log n})^2/2}}{\sqrt{2\pi}} \frac{1}{\sqrt{2 \log n}} = \frac{1/n}{\sqrt{2\pi}} \frac{1}{\sqrt{2 \log n}}. $$

It's obvious that it approaches to zero faster than $1/n$.

Answer 2

From the Mills ratio bound $\frac{1-\Phi(u)}{\phi(u)} \ge \frac{1}{u} - \frac{1}{u^3}$ (e.g. here) we have $$\Phi((1-\epsilon)\sqrt{2 \log n}) \le 1 - \underbrace{\frac{1}{n^{(1-\epsilon)^2}\sqrt{2\pi}} \left(\frac{1}{(1-\epsilon)\sqrt{2 \log n}} - \frac{1}{((1-\epsilon)\sqrt{2 \log n})^{3}}\right)}_{=: f_\epsilon(n)}.$$ Raising to the $n$th power and taking the logarithm yields $$n \log \Phi((1-\epsilon)\sqrt{2 \log n}) \le n \log (1 - f_\epsilon(n)) = - n f_\epsilon(n)(1-h(f_\epsilon(n))),$$ where we use the fact that $\log(1-x)=-x(1-h(x))$ where $h$ satisfies $h(x) \to 0$ as $x \to 0$. We know $f_\epsilon(n) \to 0$ as $n \to \infty$, so $1-h(f_\epsilon(n)) \to 1$. Further, $nf_\epsilon(n) = c_\epsilon n^{1-(1-\epsilon)^2}/\sqrt{\log n} \to \infty$. Thus, $$n \log \Phi((1-\epsilon)\sqrt{2 \log n}) \to -\infty$$ and $$\Phi((1-\epsilon)\sqrt{2 \log n})^n \to 0.$$

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The work below was for an earlier version of OP's question, and shows that $\Phi(\sqrt{2 \log n})^n \to 1$ rather than $\to 0$. That is, the $1-\epsilon$ in the statement of the problem is important.

The following is a rephrasing of Y.D.X.'s solution, but points out that the the limit of $\Phi(\sqrt{2 \log n})^n$ should be $1$ rather than $0$.

From the Mills ratio bound $\frac{1-\Phi(u)}{\phi(u)} \le \frac{1}{u}$ (see e.g., here for a simple proof) we have $$\Phi(\sqrt{2 \log n}) \ge 1 - \frac{1}{\sqrt{2\pi}} \frac{1}{n \sqrt{2\log n}}.$$

Let $c = 2\sqrt{\pi}$. Raising both sides to the $n$th power and taking a logarithm yields

$$n \log \Phi(\sqrt{2 \log n}) \ge n \log\left(1 - \frac{1}{c n \sqrt{\log n}}\right) = -n \cdot \frac{1}{c n\sqrt{\log n}} \left(1- h\left(\frac{1}{cn\sqrt{\log n}}\right)\right)$$ where the last step is from Taylor's theorem which implies $\log(1-x) = -x(1-h(x))$ where $h(x) \to 0$ as $x \to 0$. Taking $n \to \infty$, the right-hand side becomes zero. Thus $n \log \Phi(\sqrt{2 \log n}) \to 0$ and $\Phi(\sqrt{2 \log n})^n \to 1$.