5
$\begingroup$

In the scope of a project, I need to find the sum of two independent $t$-distributions.

I know that in the general case, the sum of two $t$-distributed random variables is not $t$-distributed. However, I would like to know what might be the special cases where the sum of two $t$-distributions is a $t$ distribution.

It would also be nice if someone can give me an example of two such $t$-distributions.

$\endgroup$
2
  • 6
    $\begingroup$ It is well known, and pointed out in several threads here, that the sum of two iid Cauchy (Student $t(1)$) variables again has a Cauchy distribution. See stats.stackexchange.com/a/36037/919 for a nice explanation. I believe that is the only case of this type, as one can show by examining products of Bessel $K$ functions (which are the characteristic functions in the Student t family). Examining the behavior of these functions at infinity might be a simple way to demonstrate this result. $\endgroup$
    – whuber
    May 19, 2022 at 21:24
  • 2
    $\begingroup$ In the scope of a project, I need to find the sum of two independent t-distributions. You could try and tell more about this in a new question. $\endgroup$ May 20, 2022 at 15:06

2 Answers 2

9
$\begingroup$

Characteristic functions are handy for studying sums of independent variables, because the c.f. of the sum is the product of the c.f.s of the addends and c.f.s determine the distribution.

The c.f. of a Student t variable with parameter ("degrees of freedom") $\nu \ge 1$ is, up to a constant of proportionality $C(\nu)$ we don't need to consider, given by

$$\phi_\nu(t) = C(\nu) K_{\nu/2}(|t|\sqrt{\nu})\,|t|^{\nu/2}\tag{*}$$

where $K$ is the Bessel Function of the Second Kind. I promise we won't need to learn much about this function in the following analysis.

Interpreting the question a little more broadly to accommodate a possible change of scale in the result, it asks for all solutions $\nu,\kappa,\mu,\sigma$ to the equation

$$\phi_\nu\,\phi_\kappa= \phi_\mu\circ \lambda_\sigma$$

where all the subscripts are Student t parameters (necessarily positive) and the new scale $\sigma$ is positive; $\lambda_\sigma (t) = \sigma t$ is the change of scale. According to $(*),$ this equation is equivalent to

$$K_{\nu/2}(|t|\sqrt{\nu})\, K_{\kappa/2}(|t|\sqrt{\kappa}) |t|^{(\nu + \kappa)/2}\ \propto\ K_{\mu/2}(|t|\sigma \sqrt{\mu}) |t|^{\mu/2}.\tag{**}$$

As $|t|$ grows large, $K_\alpha(|t|)$ is asymptotically proportional to

$$K_\alpha(|t|) \ \sim\ C(\alpha)e^{-|t|}|t|^{-1/2}\left(1 + \frac{4\alpha^2-1}{8|t|} + O\left(|t|^{-2}\right)\right).$$

Comparing the leading terms of $(**)$ shows us that

$$e^{-(\sqrt\nu + \sqrt\kappa)|t|}|t|^{(\nu + \kappa)/2-1}\ \propto\ e^{-\sigma\sqrt\mu|t|}|t|^{(\mu-1)/2},$$

equivalent to $\nu+\kappa-1=\mu$ and $\sqrt{\nu}+\sqrt{\kappa} = \sigma\sqrt{\mu}.$ Consequently, one or both of the original random variables must have a Student $t(1)$ distribution. Moreover, taking $\kappa=1$ in that case, whence $\mu=\nu,$ we find $\sigma\sqrt{\nu} = \sqrt{\nu}+1,$ entailing

$$\sigma = 1 + \frac{1}{\sqrt{\nu}}.$$

Comparing the $|t|^{-1}$ terms in the asymptotic expansion shows

$$\frac{4\nu^2-1}{8|t|} =\frac{4\nu^2-1}{8\sigma|t|} + O\left(|t|^{-2}\right)$$

which is possible only when $1 = \sigma = 1 + 1/\sqrt{\nu}$ or $4\nu^2-1=0:$ the only solution is $\nu=1.$

The sum of two independent Student t variables has a Student t distribution (up to scale) only when both variables have one degree of freedom; and in that case, the resulting distribution has one degree of freedom and a scale factor of $2.$

As a check, here is a QQ plot of $|X+Y|$ for a million independent realizations of a Student $t(2)$ variable $Y$ and Student $t(1)$ variable $X$ (shown on a log-log scale due to the heavy tail).

Figure

(To avoid plotting all million values, only the 0.5%, 1%, ..., 99.5% quantiles of the data are shown.)

The agreement with a $\sigma$ - scaled Student $t(2)$ distribution is good in the middle, but grows increasingly worse out in the tail: adding the heavy-tailed $X$ to $Y$ has given the sum a heavier tail, even after scaling. It is this deviation at the upper right that the analysis of the $O(|t|^{-1})$ term in the c.f. revealed.

$\endgroup$
3
  • $\begingroup$ So if do the sum of two distribution where each one of them has a degree of freedom of 1. I Will get a t distribution with a degree of freedom 2? $\endgroup$
    – sam
    May 20, 2022 at 14:04
  • $\begingroup$ No: this answer shows you obtain a scaled t distribution with $1+1-1=1$ d.f. I should have made that more explicit in my summary, so I will edit it. $\endgroup$
    – whuber
    May 20, 2022 at 14:07
  • $\begingroup$ Thank you it is clearer now. $\endgroup$
    – sam
    May 20, 2022 at 14:10
1
$\begingroup$

Here is a much less formal approach that (mainly) relies on the fact that moments of order $\nu$ or larger do not exist for Student's t-distributions with $\nu$ degrees of freedom.

If $X_\nu$ and $Y_\nu$ have independent Student's t-distributions with $\nu$ degrees of freedom where the moments of order $\nu$ or higher don't exist, then for the sum ($X_\nu +Y_\nu$) will have the same moments that exist and the same moments that don't exist. That implies that if the sum were to have a Student's t-distribution, then that distribution would also need to have exactly $\nu$ degrees of freedom.

But for $\nu >2$ the variance is $\nu/(\nu-2)$ for a Student's t-distribution and the sum would have variance $2\nu/(\nu-2)$. But those two variances can't be the same for $\nu>2$. So for $\nu>2$, there is no sum of two independent Student's t-distributions degrees of freedom $\nu$ that results in a Student's t-distribution.

That leaves only $\nu=1$ and $\nu=2$ to check.

$\endgroup$
1
  • $\begingroup$ Provided we consider non-integral moments, too, this is a nice approach (+1), but it's incomplete: it leaves all real values $\nu$ in the interval $(0,2)$ to check. This is not just a theoretical quibble: Student t distributions with non-integral degrees of freedom are used in several common tests and are often used to approximate other distributions. $\endgroup$
    – whuber
    Jun 17, 2022 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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