# is it the case that increasing degrees of freedom always makes every tail of a t-distribution smaller?

As per the title. Say I have X a random variable that is a 0-centered t-student.

Can I affirm that P(X>a) decreases when I increase the degrees of freedom of X?

Looking at the image in the wikipedia case makes me think this is the case (https://en.wikipedia.org/wiki/Student's_t-distribution) but I am not sure.

Also, I have been told that assuming less degrees of freedom is "conservative", which also points in that direction.

If this is indeed the case, a proof would also be appreciated

• The answer is yes when $a$ is positive. I suspect a fairly short proof might be afforded by representing the Student t as a variance mixture of Gaussians.
– whuber
Jul 27 '20 at 22:15
• I have a vague recollection that it's possible to show that for given $x$ in the tail of the t density $K(\nu)\left(1 + \frac{x^2}{\nu}\right)^{-(\nu+1)/2}$ $= K(\nu)\left(1+\frac{x^2}{\nu}\right)^{-1/2}\left(1+\frac{x^2}{\nu}\right)^{-\nu/2},$ where the last factor converges to $e^{-.5x^2},$ is decreasing in $\nu.$ Jul 27 '20 at 23:33

As degrees of freedom $$\nu$$ increase, the tails of Student's t distribution contain less probability, with the normal distribution being the limiting case.

• As $$\nu = n-1$$ increases, quantile 0.975 $$q$$ decrease to the normal value 1.96. For example, a t confidence interval $$\bar X \pm qS/\sqrt{n}$$ gets closer to the z confidence interval $$\bar X \pm 1.96 \sigma/\sqrt{},$$ for known population standard deviation $$\sigma.$$

• For the standard normal distribution, the probability $$p = P(-1.96 < Z < 1.96) = 0.95.$$ As $$\nu$$ increases, $$p = P(-1.96, < T < 1.96)$$ increases to the normal value.

Many elementary textbooks say that, for $$\nu = 30,$$ the t distribution is sufficiently close to normal for some practical purposes. But $$\mathsf{T}(\nu=30)$$ is hardly the same as $$\mathsf{Norm}(0,1).$$

Here are graphs of $$q$$ and $$p$$ for $$\nu = 1, 2, \dots, 200.$$ The R code used to make the plots is shown below.

df = 1:200
q = qt(.975, df)
pu = pt(1.96, df);  pl = pt(-1.96, df);  p = pu-pl
par(mfrow=c(1,2))
plot(df,q, type="l", ylim=c(1.96,2.25), xaxs="i", main="Quantile 0.975")
abline(h = 1.96, col="blue")
plot(df,p, type="l", ylim=c(.925,.95), xaxs="i", main="P(-1.96< T < 1.96)")
abline(h = diff(pnorm(c(-1.96,1.96))), col="blue")
par(mfrow=c(1,1))

• I think the $n=30$ criterion comes from 1) testing at 5% level and 2) the need to limit the size of the tables at the end of the book ... Jan 31 '21 at 6:08