# 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.

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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