How can I compare 2 means that are Laplace distributed? I want to compare 2 sample means for 1-minute-stock returns. I assume they are Laplace distributed (already checked) and I split the returns into 2 groups. How can I check whether they are significantly different? 
I think I cannot treat them like a Normal distribution, because even though they are more than 300 values, the QQ-plot shows that there is a huge difference to a Normal distribution
 A: Assuming both Laplace distributions have the same variance, 
a) the likelihood ratio test would involve a test statistic like:
$\mathcal{L}=\frac{\prod_{i=1}^n \frac{1}{2\hat{\tau}} \exp(-\frac{|x_i-\hat{\mu}|}{\hat{\tau}})}{\prod_{i=1}^{n_1}\frac{1}{2\hat{\tau}_1} \exp(-\frac{|x_i-\hat{\mu}_1|}{\hat{\tau}_1})\cdot \prod_{i=n_1+1}^{n} \frac{1}{2\hat{\tau}_2} \exp(-\frac{|x_i-\hat{\mu}_2|}{\hat{\tau}_2})}$
Taking logs, cancelling/simplifying and multiplying by $-2$.
$\,-2\mathcal{l} = 2(n\log(\hat{\tau})-n_1\log(\hat{\tau}_1)-n_2\log(\hat{\tau}_2))\;\qquad$ (where $\mathcal{l}=\log(\mathcal{L})$)
where $\hat{\tau}=m$, the mean absolute deviation from the median in the combined sample and $\hat{\tau}_i=m_i$, the mean absolute deviation from the median in sample $i$.
According to Wilks' theorem this is asymptotically distributed as $\chi^2_1$ under the null, so for a 5% test you'd reject if that exceeded $3.84\,$.
Simulation experiments suggest that the test is anticonservative at small sample sizes (the probability of rejecting is somewhat higher than nominal), but by about n=100, it seems to be at least reasonable (you get on the order of 5.3% - 5.4% rejection rate under the null for a nominal 5% test, for example; for $n_1,n_2>300$ it seems to be nearer to 5.25%).
b) We'd also expect that $\frac{\tilde{\mu}_1- \tilde{\mu}_2}{\sqrt{v}}$ would be a good test statistic (where $\tilde{\mu}$ represents the sample median and $v=2\hat{\tau}^2(\frac{1}{n_1}+\frac{1}{n_2})$); if I haven't made an error in there, in large samples like yours it would be approximately normally distributed under the null, with mean 0 and variance 1, where $\hat{\tau}^2$ could be based on the square of the mean absolute deviation from the mean in the combined sample, $m^2$, though I expect it would in practice tend to work better basing it on a sample-weighted average of the two sample $m^2_i$'s
$^\dagger$ .
$\dagger$ (Edit: simulation suggests the normal approximation is fine but the variance calculation is not correct above; I can see what the problem is now but I still have to fix it. The permutation version of this test (see item (c)) should still be fine).
c) Another alternative would be to perform a permutation test based on either of the above statistics. (One of the answers here gives an outline of how to implement the permutation test for a difference in medians.)
d) You could always do a Wilcoxon/Mann-Whitney test; it will be considerably more efficient than trying to use a t-test at the Laplace.
e) Better than (d) for Laplace data would be Mood's median test; while often recommended against in books, when dealing with Laplace data it will show good power. I expect it would have similar power to the permutation version of the asymptotic test of difference in medians (one of the tests mentioned in (c)). 
The question here gives an R implementation that uses a Fisher test, but that code can be adapted to use a chi-square test instead (which I'd suggest in even moderate samples); alternatively there's example code for it (not as a function) here. 
The median test is discussed in Wikipedia here, though not in much depth (the linked German translation has a little more information). Some books on nonparametrics discuss it. 
