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

  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Jul 27 at 22:15
  • $\begingroup$ 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 $\nu.$ $\endgroup$ – BruceET Jul 27 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.

enter image description here

df = 1:200
q = qt(.975, df)
pu = pt(1.96, df);  pl = pt(-1.96, df);  p = pu-pl 
 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")
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.