This question is related to another recent question on CV.SE relating to the weighting function that commonly appears in confidence intervals for a population mean:.

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

This weighting appears in the standard classical $1-\alpha$ level confidence interval for the mean of an infinite super-population, which can be written as $\text{CI}(1-\alpha) = [ \bar{x}_n \pm w(\alpha, n) \cdot s_n ]$. Normally we treat the input $n>1$ as an integer, but it is possible to extend the treatment to consider it a real value (since the T-distribution is well-defined for non-integer degrees-of-freedom). In such a case a natural question is what happens to the weighting when we take $n \downarrow 1$. Intuitively, it is reasonable to conjecture that the interval extends to the whole real line in this case$^\dagger$, so that:

$$\lim_{n \downarrow 1} w(\alpha, n) = \infty \quad \quad \quad \quad \quad \quad \text{for all } 0<\alpha<1.$$

Question: Please find the limiting value of the weighting function in this case, and provide an adequate proof of the limit (which may or may not be the above conjectured value).

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:

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

$^\dagger$ A caveat on this statement must be noted, which is that $s_n$ is not generally considered to be well-defined for $n<2$. The assertion that the confidence interval would extend to the whole real line requires some extended meaning for $s_n$ for $1<n<2$. Really what we are asking about is if the weighting becomes infinite.

  • 1
    $\begingroup$ Asymptotically $$w\left(\alpha, 1 + \frac{1}{y}\right) \sim \frac{\alpha ^ {-y}}{ \sqrt{y}}$$ as $y\to \infty.$ $\endgroup$
    – whuber
    Commented Jul 27, 2018 at 14:32
  • 1
    $\begingroup$ @user21478: You can ask a question on this site with 1 reputation. $\endgroup$
    – Ben
    Commented Jul 30, 2018 at 0:24
  • $\begingroup$ @user21478, click the gray ASK QUESTION link at the top of the page & ask there. $\endgroup$ Commented Jul 30, 2018 at 0:51

1 Answer 1


Noting that the division by $\sqrt{n}$ does not change the result (because it converges to $1$) and writing $\nu=n-1$ and $2\gamma = 1-\alpha,$ the problem is to analyze the behavior of the function $x_\gamma(\nu)$ defined implicitly by

$$\gamma = \frac{\Gamma(\nu/2+1/2)}{\Gamma(1/2)\Gamma(\nu/2)}\int_0^{x_\gamma(\nu)}\left(1 + \frac{r^2}{\nu}\right)^{-1/2 - \nu/2} \frac{dr}{\sqrt{\nu}}.$$

The change of variable $x \sqrt{\nu} = r$ yields

$$\gamma = \frac{\Gamma(\nu/2+1/2)}{\Gamma(1/2)\Gamma(\nu/2)}\int_0^{x_\gamma(\nu)/\sqrt{\nu}}\left(1 + x^2\right)^{-1/2 - \nu/2} dx.$$

Because for all $\nu \gt 0$ the integrand is bounded above by $(1+x^2)^{-1/2},$ a lower bound for $x_\gamma(\nu)$ is given by the solution $t$ to the equation

$$\eqalign{ \gamma &= \frac{\Gamma(\nu/2+1/2)}{\Gamma(1/2)\Gamma(\nu/2)}\int_0^{t/\sqrt{\nu}}\left(1 + x^2\right)^{-1/2} dx \\ &= \frac{\Gamma(\nu/2+1/2)}{\Gamma(1/2)\Gamma(\nu/2)} \operatorname{Asinh}\left(\frac{t}{\sqrt{\nu}}\right), }$$

which is readily solved to produce

$$x_\gamma(\nu) \ge t = \sinh\left(\frac{\gamma\,\Gamma(1/2)\Gamma(\nu/2)}{\Gamma(\nu/2+1/2)}\right)\sqrt{\nu}.$$

The divergence of $x_\gamma(\nu)$ as $\nu\to 0$ (from above) is immediate, because $\Gamma(\nu/2+1/2)\to\Gamma(1/2),$ $\Gamma(\nu/2) \approx 2/\nu$ for small positive $\nu,$ and $\operatorname{Asinh}(C) \approx \exp(C)/2$ for large positive $C.$

(The rate of divergence is substantially greater than suggested by this lower bound, because it is fairly crude.)


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