# What is the distribution of the difference of two-t-distributions

... and why ?

Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:

• $E(X_1-X_2)=\mu_1-\mu_2$
• $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$

Now let's say $X_1$, $X_2$ are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?

This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.

-
this is related question which might be of interest. – mpiktas May 16 '11 at 14:46
Google the Satterthwaite t-test, the CABF t-test (Cochran's approximation to the Behrens-Fisher), and the Behrens-Fisher problem. – whuber May 16 '11 at 15:51
For the special case where the degrees of freedom is 1 (the Cauchy distribution) the answer to the original question is 1. The sum (or difference) of two independent Cauchy distributed random variables is Cauchy with scale parameter $2$, but then again, the Cauchy distribution does not even have a mean value. – NRH May 16 '11 at 17:20
@whuber thank you, apparently I didn't know that – steffen May 17 '11 at 7:24

The sum of two independent t-distributed random variables is not t-distributed. Hence you cannot talk about degrees of freedom of this distribution, since the resulting distribution does not have any degrees of freedom in a sense that t-distribution has.

-
 @mpiktas: Dumb question. If the t-distribution with n-1 df can be derived from the sum of n indepent normal distributions (see wikipedia) and given df high enough so that the t-distribution approximates the normal distribution, doesn't derive from that that the sum of t-distributions is again a t-distribution ? – steffen May 16 '11 at 14:15 @mpiktas: What about the test-statistic of the t-test, which seems to be derived from the difference of two t-distributions ? – steffen May 16 '11 at 14:17 @steffen, no. It will be approximately normal, since you will add two approximately normal distributed normal variables. t-distribution with high df is approximately normal, but approximately normal is not necessarily t-distribution with high df. – mpiktas May 16 '11 at 14:19 @steffen, t-test statistic is derived from the difference of two normals not two t-distributions. Note that definition of t distribution is a fraction of normal and square root of chi-square. – mpiktas May 16 '11 at 14:21 @mpiktas thank you for the response ! I often make the mistake to mix "precisely equivalent" with "close enough for practical purpose" which was the case here. I am going to reformulate this question in a few minutes since it is, well, rather stupid :). – steffen May 16 '11 at 14:31
show 1 more comment