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This question already has an answer here:

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

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marked as duplicate by whuber Apr 29 '16 at 14:56

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    $\begingroup$ I actually interpret the response there as saying it's not the case in general, not that it can never be the case. But in any case your argument is looks wrong. You used the same $V$ in $Y_1$ and $Y_2$. You should have different $V$s. $\endgroup$ – Glen_b Aug 17 '13 at 10:12
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    $\begingroup$ Thanks! You're right. In fact using the same $V$ in $Y_1$ and $Y_2$ results in a bivariate t distribution for $Y_1$ and $Y_2$! And linear combinations are then t distributed, because the t distribution is an elliptical distribution. Sometimes you miss the forest for the trees... $\endgroup$ – user29242 Aug 17 '13 at 10:17
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I think you are wrong, because the Student-t includes the Gaussian and the Cauchy distribution. It is well known that sum of Cauchy is Cauchy and sum of normal is normal - so there are at least two contradictions!

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    $\begingroup$ This isn't a contradiction at all. You founf two cases where it is true! unfortunately, it is not true in any other case. $\endgroup$ – kjetil b halvorsen Apr 29 '16 at 8:01

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