Prove that $t_{n-1, \alpha/2} / \sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$

Question

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.

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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 1

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 $n \to \infty$

Answer 2

I'm sure there is an easier way to do this, but the result is immediate from the following:

Proposition: Let $F$ be a continuous distribution function and $F_n$ a sequence of distribution functions such that $F_n \to F$ weakly (i.e., in distribution). Then $F_n(x) \to F(x)$ uniformly in $x$.

Proof: Using continuity and monotonicity, for any natural number $m$ we can select $x_0, x_1, \ldots, x_m$ such that $F(x_j) = j/m$ (taking $x_0 = -\infty$ and $x_m = \infty$). By weak convergence and the fact that $F$ is continuous, $F_n(x_j) \to F(x_j)$. For any $y$, find an interval $[x_{j-1}, x_{j}]$ containing $y$ and note that $|F_n(y) - F(y)| \le \sup_j |F_n(x_j) - F(x_j)| + |F(x_j) - F(x_{j-1})| \to \frac{1}{m}$. Hence $\varlimsup_{n \to \infty} \sup_y |F_n(y) - F(y)| \le \frac 1 m$ and because $m$ was arbitrary we get $\sup_y |F_n(y) - F(y)| \to 0$.

Next, it is a well-known application of Slutsky's theorem that the $t_{n-1}$ converges in distribution to a standard normal distribution. The previous result implies that $F_n(t_{n-1,\alpha}) - F(t_{n-1,\alpha}) \to 0$, i.e., $F(t_{n-1,\alpha}) \to \alpha$. Applying the normal quantile function to both sides, we get $t_{n-1,\alpha} \to z_{\alpha}$.

Hence $t_{n-1,\alpha} \to z_{\alpha}$ implying $\frac{t_{n-1,\alpha}}{g(n)} \to 0$ for any $g(n) \to \infty$ (in particular, $g(n) = \sqrt n$).

Answer 3

Geometrical proof

Geometrical view

Consider the observed sample as a point in n-dimensional Euclidean space and the estimation of the mean as the projection of an observation $x_1,x_2,...,x_n$ onto the model line $x_1=x_2= ... = x_n = \bar{x}$.

The t-score can be expressed as ratio of two distances in this space

• the distance between the projected point and the population mean $$\sqrt{n}(\bar{x}-\mu)$$
• the distance between this point and the observation $$\sqrt{\sum_{i=1}^n(\hat{x}-x_i)^2}$$

This is related to the tangent of the angle between the observation and the line on which it is projected.

$$\frac{t}{\sqrt{n-1} } =\frac{\sqrt{n}(\bar{x}-\mu)}{\sqrt{\sum_{i=1}^n(\hat{x}-x_i)^2}} = \frac{1}{\tan{\theta}}$$

Equivalence t-distribution and angle distribution

In this geometrical view the probability of the t-score being higher than some value is equivalent to the probability of the angle being less than some value:

$$Pr(|T|>t_{n-1,\alpha/2}) = 2 Pr(\theta \leq \theta_{\nu,\alpha}) = \alpha$$

Or

$$\frac{t_{n-1,\alpha/2}}{\sqrt{n-1}} = \frac{1}{\tan \theta_{\nu,\alpha}}$$

You could say that the t-score relates to the angle of the observation with the line of the theoretic model. For points outside the confidence interval (then $\mu$ is further away from $\bar{x}$ and the angle will be smaller) the angle will be below some limit $\theta_{\nu,\alpha}$. This limit will change with more observations. If the limit of this angle $\theta_{\nu,\alpha}$ goes to 90 degrees for large $n$ (the cone shape getting more flat, ie less pointy and long) then this means that the size of the confidence interval becomes smaller and approaches zero.

Angle distribution as relative area of the cap of an n-sphere

Due to symmetry of the joint probability distribution of independent normal distributed variables, every direction is equally probable and the probability for the angle to be within a certain region is equal to the relative area of the cap of an n-sphere.

The relative area of this n-cap is found by integrating the area of a n-frustum:

$$\begin{array}{rcl} 2 Pr(\theta \leq \theta_c)& =& 2 \int_{\frac{1}{\sqrt{1+\tan(\theta_c)^2}}}^1 \frac{(1-x^2)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dx \\ &=& \int_{\frac{1}{{1+\tan(\theta_c)^2}}}^1 \frac{t^{-0.5}(1-t)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt \\ &=& I_{\frac{1}{{1+\tan(\theta_c)^2}}}\left(\frac{1}{2},\frac{n-1}{2} \right) \end{array}$$

where $I_x(\cdot,\cdot)$ is the upper regularized incomplete beta function.

Limit of angle

If $\theta_{n,\alpha}$ goes to 90 degrees for $n \to \infty$ then $t_{n-1,\alpha/2}/\sqrt{n}$ goes to zero.

Or an inverse statement: for any angle smaller than 90 degrees the relative area of that angle on a n-sphere, decreases to zero when $n$ goes to infinity.

Intuitively this means that all area of a n-sphere concentrates to the equator as the dimension $n$ increases to infinity.

Quantitatively we can show this by using the expression

$$\int_{a}^1 \frac{t^{-0.5}(1-t)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt < \int_{a}^1 \frac{(1-a)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt = \frac{(1-a)^{\frac{n-1}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} = L(n)$$

and consider the difference between $L(n+2)$ and $L(n)$.

At some point the decrease in the denominator $$\frac{B(\frac{1}{2},x+1)}{B(\frac{1}{2},x)} = \frac{x}{x+\frac{1}{2}}$$ will be taken over by the decrease in the numerator $$\frac{(1-a)^{\frac{n+1}{2}}}{(1-a)^{\frac{n-1}{2}}} = 1-a$$ and the function $L(n)$ decreases to zero for $n$ to infinity.

Answer 4

We have

\begin{align} \frac{\alpha}{2} &= \int \limits_{t_{n-1, \alpha/2}}^\infty \lim_{n \to \infty}\frac{1}{\sqrt{(n-1) \pi}} \cdot \frac{\Gamma(\tfrac{n}{2})}{\Gamma(\tfrac{n-1}{2})} \Big( 1+ \frac{r^2}{n-1} \Big)^{-n/2} dr \\[10pt] &= \int \limits_{t_{n-1, \alpha/2}}^\infty \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}r^2} dr \\[10pt] &= 1-\Phi(t_{n-1, \alpha/2}) \\[10pt] &\approx 1-\left[\frac{1}{2}+\varphi(t_{n-1, \alpha/2}) \left(t_{n-1, \alpha/2}+\frac{(t_{n-1, \alpha/2})^3}{3}+\frac{(t_{n-1, \alpha/2})^5}{15} + \dots \right) \right] \end{align}

which implies that the second term in the boxed brackets can be at most $\frac{1}{2}$ since the maximum $\alpha$ can be is $1$. Note that $\varphi(x)$ is the pdf of normal distribution. This approximation is also based on this.

So $$ 0 < \alpha \approx 1+2\varphi(t_{n-1, \alpha/2}) \left(t_{n-1, \alpha/2}+\frac{(t_{n-1, \alpha/2})^3}{3}+\frac{(t_{n-1, \alpha/2})^5}{15} + \dots \right) <1 $$