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:

$$\frac{\alpha}{2} = \frac{1}{\sqrt{(n-1) \pi}} \cdot \frac{\Gamma(\tfrac{n}{2})}{\Gamma(\tfrac{n-1}{2})} \int \limits_{t_{n-1, \alpha/2}}^\infty \Big( 1+ \frac{r^2}{n-1} \Big)^{-n/2} dr.$$

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    $\begingroup$ Isn’t this trivial due to the fact that students t converges to a normal in total variation? The numerator goes to the constant $z_{\alpha/2}$. $\endgroup$
    – guy
    Commented Jul 23, 2018 at 14:34
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    $\begingroup$ @guy: What constitutes "trivial" depends on the level of the person you're asking. The reason I ask this question is that this is a problem I have posed to a couple of grad-students in my teaching, and I want to see what kinds of proofs people on this site come up with (so I can compare this with the kinds of answers given by students). Grad-students usually have a rough idea of how to approach the proof, but struggle with the details. Most try to show $t_{n-1, \alpha/2} \rightarrow z_{\alpha/2} < \infty$ and go from there, but others try to solve it by establishing a bound on the integral. $\endgroup$
    – Ben
    Commented Jul 24, 2018 at 2:23
  • $\begingroup$ (I am also hoping to see the most elegant proof, as a good way to present the result.) $\endgroup$
    – Ben
    Commented Jul 24, 2018 at 2:25
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    $\begingroup$ I'd be interested in seeing how elegant someone can get this. Personal opinion, but I would argue that using the densities is not elegant because it is bringing in additional structure that is not needed. The proof in my answer has two important lessons. First, the proposition is a useful bit of probability theory and can give some idea how DFs behave. Second, the proof technique (which can be restated in terms of covering numbers) is very useful for proving many things. $\endgroup$
    – guy
    Commented Jul 24, 2018 at 3:49

4 Answers 4


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 $n \to \infty$

  • $\begingroup$ I usually avoid "nice answer" comments but this one is really elegant! (+1 obviously!) $\endgroup$
    – usεr11852
    Commented Sep 1, 2018 at 21:44
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    $\begingroup$ @usεr11852 you might like my geometrical proof where the distribution of the t-statistic is related to the distribution of an angle in n-space. $\endgroup$ Commented Sep 2, 2018 at 15:39

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, $g(n) = \sqrt n$).


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.

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

Geometrical sketch

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:

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


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

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 vs t

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.

  • $\begingroup$ I [heart] this answer ---wish I could give it more than +1. $\endgroup$
    – Ben
    Commented Nov 4, 2020 at 16:58
  • $\begingroup$ I think I wrote you an email a while back asking if you'd like to write a joint paper on this (since your geometric analysis is beautiful) but didn't hear back. Interested? $\endgroup$
    – Ben
    Commented Nov 4, 2020 at 17:00

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

  • $\begingroup$ Welcome to the site, @guerhuerh. Please consider registering your account & becoming a regular member here (there is information in the My Account section of our help center). We'd love to have you. If you'd like, you can take our tour, which has information for new users. $\endgroup$ Commented Jul 23, 2018 at 14:29
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    $\begingroup$ This is an interesting approach, but it is unclear to me how you justify the first step (where you have added the limit operation inside the integral). This gives an implicit equation for $t_{n-1, \alpha/2}$ that is false. $\endgroup$
    – Ben
    Commented Jul 24, 2018 at 2:27
  • $\begingroup$ @Ben: Wouldn't $\lim_{n \to \infty} \frac{\alpha}{2} = \frac{\alpha}{2}$? $\endgroup$ Commented Sep 14, 2018 at 1:15
  • $\begingroup$ @PEV: Yes, but then the right-hand-side should have the limit outside the integral, not inside it. Since the limits in the integral depend on $n$ the right-hand-side is presently a function of $n$, so it is not equal to $\alpha/2$. $\endgroup$
    – Ben
    Commented Sep 14, 2018 at 1:18
  • $\begingroup$ @PEV: I think the method could be modified by separating the integral into two parts, one which replaces the lower limit with $z_{\alpha/2}$ and one that gives the remainder. The former would then converge to what is already shown, and we would then have to show that the latter converges to zero. Like I said, I think this is an interesting approach. It might have merit as a method, and might be able to be altered to be correct, but it is presently incorrect. $\endgroup$
    – Ben
    Commented Sep 14, 2018 at 1:21

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