Rate of convergence for the sign of sample mean

Question

What is the rate of convergence for $\textrm{sign}(\frac{1}{n}\sum_{i=1}^{n}X_{i})$? ( $X_{i}$ are independent and identically distributed as $F$ satisfying the conditions for central limit theorem)

Answer 1

If $\mathbb{E}[X_i]=\mu<0$, with $\text{var}(X_i)=\sigma^2$, then \begin{align}\mathbb P(\bar X_n \ge 0)&=\mathbb P(\bar X_n \ge \mu-\sqrt{n}\mu\sigma/\sqrt{n}\sigma)\\ &=\mathbb P\left(\bar X_n \ge \mu-\frac{\sqrt{n}\mu}{\sigma}\underbrace{\frac{\sigma}{\sqrt{n}}}_{\text{sd}(\bar X_n)}\right)\\ &=\mathbb P\left(\frac{\sqrt{n}}{\sigma}\{\bar X_n - \mu\} \ge -\frac{\sqrt{n}\mu}{\sigma}\right)\\ &\le -\frac{\sigma}{\sqrt{n}\mu}\dfrac{\exp\{-[\sqrt{n}\mu/\sigma]^2/2\}}{\sqrt{2\pi}}\qquad\text{for the Normal cdf, with $\mu<0$}\\ &=-\frac{\sigma}{\sqrt{n}\mu}\dfrac{\exp\{-n\mu^2/2\sigma^2\}}{\sqrt{2\pi}}\end{align} So the rate of $\text{sign}(\bar X_n)$ not going to the right value is of order $$\exp\{-n\mu^2/2\sigma^2\}/\sqrt{n}$$ Note that the bound $$\mathbb P\left(\frac{\sqrt{n}}{\sigma}\{\bar X_n - \mu\} \ge -\frac{\sqrt{n}\mu}{\sigma}\right)\le -\frac{\sigma}{\sqrt{n}\mu}\dfrac{\exp\{-[\sqrt{n}\mu/\sigma]^2/2\}}{\sqrt{2\pi}}$$ can be obtained as $$\mathbb P(X\ge x) =\int_x^\infty\phi(u)\,\text{d}u <\int_x^\infty\frac ux\phi(u)\,\text{d}u =\int_{\frac{x^2}{2}}^\infty\frac{e^{-v}}{x\sqrt{2\pi}}\,\text{d}v=-\biggl.\frac{e^{-v}}{x\sqrt{2\pi}}\biggr|_{\frac{x^2}{2}}^\infty=\frac{\phi(x)}{x}$$ when $X\sim\mathcal N(0,1)$.