... 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:
- $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$
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.