Statistical problems involving confidence intervals for a population mean can be framed in terms of the following

weighting function:$$w(\alpha, n) \equiv \frac{t_{n-1,\alpha/2}}{\sqrt{n}} \quad \quad \quad \quad \text{for } 0<\alpha<1 \text{ and } n > 1.$$

For example, the standard classical $1-\alpha$ level confidence interval for the mean of an infinite super-population can be written as:

$$\text{CI}(1-\alpha) = \Bigg[ \bar{x}_n \pm w(\alpha, n) \cdot s_n \Bigg].$$

It is trivial to establish the limits $\lim_{\alpha \downarrow 0} w(\alpha, n) = \infty$ and $\lim_{\alpha \uparrow 1} w(\alpha, n) = 0$ using the quantile function of the T-distribution. In the context of confidence intervals, this tells us that the interval shrinks to a single point as we decrease the confidence level, and increases to the whole real line as we increase the confidence level. Another intuitive property that should hold is that the interval shrinks to a single point as we get more and more data, which means that:

$$\lim_{n \rightarrow \infty} w(\alpha, n) = 0.$$

Question:Please provide a proof for this latter property of the weighting function.

More information:For any mathematical readers who are unfamiliar with the critical points of the T-distribution, the value $t_{n-1, \alpha/2}$ is a function of $n$ defined by the implicit equation:$$\frac{\alpha}{2} = \frac{1}{\sqrt{(n-1) \pi}} \cdot \frac{\Gamma(\tfrac{n}{2})}{\Gamma(\tfrac{n-1}{2})} \int \limits_{t_{n-1, \alpha/2}}^\infty \Big( 1+ \frac{r^2}{n-1} \Big)^{-n/2} dr.$$

**Answer**

# Proof with Chebyshev’s inequality

Here is a proof using Chebyshev’s inequality $Pr(|T|\geq k\sigma) \leq \frac{1}{k^2}$.

If we fill in $\sigma_{t_\nu} = \frac{\nu}{\nu-2}$ and set $1/k^2=\alpha = Pr\left(|T|\geq t_{\nu,\alpha/2}\right)$ then we have a limit

$$Pr\left(|T|\geq \frac{\nu}{\nu-2}\frac{1}{\sqrt{\alpha}}\right) \leq Pr\left(|T|\geq t_{\nu,\alpha/2}\right) $$

thus $t_{\nu,\alpha/2}$ will be bounded above by

$$t_{\nu,\alpha/2} \leq \frac{\nu}{\nu-2}\frac{1}{\sqrt{\alpha}}$$

adding the obvious lower bound and devide by $\sqrt{\nu+1}$

$$0 \leq \frac{t_{n-1,\alpha/2}}{\sqrt{\nu+1}} \leq \frac{\nu}{\sqrt{\nu+1}\left(\nu-2\right)}\frac{1}{\sqrt{\alpha}} $$

which squeezes $t_{n-1,\alpha/2} / \sqrt{n}$ to zero for $n \to \infty$

**Attribution***Source : Link , Question Author : Ben , Answer Author : Community*