Find $\lim_{n \downarrow 1} t_{n-1, \alpha/2} / \sqrt{n}$ and prove the limit

Question

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.

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.)