suggests that the sum of two t-distributions is never t distributed.   With t distribution I mean the (non-standardized) t distribution with location and scale parameter.  Now, let $Y_1,Y_2$ be independently t distributed with same dof $\nu$, location $\mu$ and scale $\sigma$.  Then the sum is given by $X = Y_1+Y_2 = (\mu+\sqrt{\nu/V}\sigma Z_1) + (\mu+\sqrt{\nu/V}\sigma Z_2)$, where $V$ is $\chi^2$ distributed with $\nu$ degrees of freedom and $Z_1,Z_2\stackrel{iid}{\sim}N(0,1)$. Since the sum of two independent $N(0,1)$ is $N(0,2)$, it follows that $X = 2\mu+\sqrt{\nu/V}\sqrt{2}{\sigma}Z$, where $Z\sim N(0,1)$. Thus $X$ is t distributed with unchanged dof, twice the location and new scale $\sqrt{2}\sigma$.   So am I missing something?
... 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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