Prove that $t_{n-1, \alpha/2}$ is strictly decreasing in $n$

This question is related to a couple of other questions (here and here) relating to some material that I teach in sampling theory. Consider the critical points of the T-distribution defined as follows. The value $$t_{n-1, \alpha/2}$$ is a function of $$n$$ defined by the implicit equation:

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

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). I'm trying to come up with a (relatively simple) proof that this critical point value is decreasing in $$n$$, but I seem to be making quite a mess. Differentiation using Leibniz rule gets part way, but it quickly becomes a bit of a mess. Consequently, I would like to solicit proofs of this result --- this simpler the better.

Question: How do you prove that this critical point is strictly decreasing in $$n$$?

• You will need to analyze the error term, so I can't help thinking some of the analyses at stats.stackexchange.com/questions/449097 (to which you contributed!) would be helpful.
– whuber
Mar 20, 2021 at 13:43

This looks like a good opportunity to discuss important relationships among the Student $$t$$ distributions. The analysis needed to demonstrate them is elementary, requiring only the basic concepts of differential Calculus, and with the right strategy can be reduced to a simple algebraic calculation.

There is a classic set of plots comparing the density functions (right panel of the figure) and the distribution functions (left panel) of Student $$t$$ variables as their parameter $$\nu$$ is varied:

Although these cannot show the full extents of the graphs, which go from $$-\infty$$ to $$\infty,$$ the curves do suggest that when $$\nu^\prime \gt \nu \gt 0,$$

1. All these densities are symmetric around $$0;$$

2. The density for $$\nu$$ is lower near $$0$$ than the density for $$\nu^\prime;$$

3. The density for $$\nu$$ is higher asymptotically than the density for $$\nu^\prime$$ (that is, the distribution with small parameter $$\nu$$ has heavier tails); and

4. There is just one positive number (depending on $$\nu$$ and $$\nu^\prime$$) where the density functions for $$\nu^\prime$$ and $$\nu$$ cross.

All but the last claim are straightforward (and obvious) to prove, so let's get to the heart of the matter: a study of how two Student $$t$$ densities relate to one another for positive arguments $$x.$$ (Claim (1) justifies the focus on positive values.) The worry is that two different Student $$t$$ densities might wiggle around each other several times as their argument $$x$$ grows large, alternating between which has the larger density. Intuitively that shouldn't be the case, but how to prove it?

The following analysis is motivated by a focus on simplifying away the obstacles. What would these obstacles be? Consider the expression for the distribution function,

$$F(t;\nu) = \int_{-\infty}^t f(x,\nu)\,\mathrm{d}x = \frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\,\Gamma\left(\frac{\nu}{2}\right)} \int_{-\infty}^t \left(1 + \frac{x^2}{\nu}\right)^{-(\nu+1)/2}\,\mathrm{d}x.$$

This is required to define the critical point $$t_\nu(\alpha/2)$$ as given in the question.

From left to right we are confronted in turn with the apparent need to analyze (1) a ratio of Gamma functions, (2) an integral, (3) a fractional power, and (4) a reciprocal quadratic function. The strategy of comparing two distributions, starting with the obvious equalities $$F(0,\nu)=1/2=F(0,\nu^\prime)$$ and $$\lim_{t\to \infty} F(t,\nu) = 1 = \lim_{t\to\infty}F(t,\nu^\prime),$$ can eliminate the first obstacle. Comparing the density functions avoids dealing directly with the integral. To deal with the powers, let's compare the densities by taking the logarithm of their ratios. The log is positive when the numerator exceeds the denominator and negative otherwise. Although this introduces the logarithm as a new complication, by differentiating it we are reduced to a manageable rational function.

To this end, for $$x\ge 0$$ define

$$h(x,\nu^\prime,\nu) = \log\left(\frac{f(x,\nu^\prime)}{f(x,\nu)}\right) = \log(f(x,\nu^\prime)) - \log(f(x,\nu)).$$

See the left panel of the next figure for a plot of $$h.$$ This is a typical shape of the graph, no matter what $$\nu^\prime \gt \nu$$ might be.

Claim (2) is that $$h(0,\nu^\prime,\nu)\gt 0$$ while claim (3) is that $$h(x,\nu^\prime,\nu)\lt 0$$ for all sufficiently large $$x.$$ Our concern is what happens at intermediate values $$x$$ where $$h$$ drops from its maximum down to negative values.

The right panel plots the derivative of $$h.$$ I will prove that the derivative has exactly one zero at $$x=1.$$ Because the simplification strategy is a good one, this is an easy calculation based on computing the logarithmic derivative of $$f(x,\nu):$$

$$\frac{\mathrm{d}\log f(x,\nu)}{\mathrm{d}x} = -\frac{(\nu+1)x}{\nu + x^2}.$$

Consequently

$$\frac{\mathrm{d}h(x,\nu^\prime,\nu)}{\mathrm{d}x} = \frac{(\nu+1)x}{\nu + x^2} - \frac{(\nu^\prime+1)x}{\nu^\prime + x^2} = \frac{(\nu^\prime-\nu)}{(\nu+x^2)(\nu^\prime+x^2)}\,x\,(1-x^2).$$

Since $$\nu,$$ $$\nu^\prime,$$ and $$x^2$$ are all positive, this is a continuous function for all $$x\ge 0.$$ It can cross zero, then, only where $$x(1-x^2)=0.$$ The only solution for $$0\lt x \lt \infty$$ is $$x=1,$$ as claimed.

Let's unroll the implications.

• $$h$$ increases from $$x=0$$ (where it is positive) to $$x=1$$ and thereafter decreases, eventually becoming negative (which is a restatement of Claim (3)).

• Therefore Claim (4) holds: any two different Student $$t$$ densities cross at exactly one positive number (potentially depending on those two densities, of course).

• Consequently the distribution function $$t \to F(t,\nu^\prime)$$ rises more steeply from its value of $$1/2$$ at $$t=0$$ compared to $$t\to F(t,\nu)$$ and can never cross that graph: the two graphs eventually converge as $$t\to \infty,$$ where they both squeeze up to a height of $$1.$$

• Thus, for any $$0\lt p \lt 1,$$ the middle portion $$p$$ of the probability distribution with $$\nu^\prime$$ degrees of freedom is strictly contained within the middle portion $$p$$ of the distribution with $$\nu\lt \nu^\prime$$ degrees of freedom.

The last conclusion is equivalent to saying $$t_\nu((1-p)/2)$$ is a strictly decreasing function of $$\nu$$, QED.

• how do you have the time to provide these beautiful in depth explorations? Asking for a friend Mar 22, 2021 at 16:27
• @bdeonovic I get insomnia sometimes. Most of my longer replies were developed in the middle of the night as an effort to get back to sleep! That's one reason I choose solutions that avoid long calculations--I just can't keep track of everything with my eyes closed ;-).
– whuber
Mar 22, 2021 at 16:34
• Glad I could entertain you during your insomnia @whuber. I note your demonstration that $h$ is stritly quasi-concave in $x$, and so the T-densities cross at only one point. You then say that the implication is that the distribution raises more steeply and does not cross. Could you please elaborate on demonstrating that step?
– Ben
Mar 25, 2021 at 2:19
• @Ben Could you clarify which implication you're asking about?
– whuber
Mar 25, 2021 at 2:53
• Let me just have another look at it first --- I'm sure I'll process it and then find I don't need anything further.
– Ben
Mar 25, 2021 at 4:57