How to determine what type of distribution a data set follows? I computed a histogram of a data set
 
and I am trying to figure out what type of distribution this is. 
Does anyone have any idea? 
It looks like a symmetric decaying exponential but I don't know how to represent this as an equation.
 A: There are an infinite number of possibilities -- it certainly looks symmetric and with a spike at the center, but it's impossible to say much about whether the decay is really exponential (a plot of log-counts might give a clearer idea -- it should look close to linear until the count gets small).
(What if it decays as some power function for example? How could you tell? Or what if it's exponential near the spike but more like a power function in the tail? Or any number of other things that might look more or less exponential? What if it's not quite symmetric? )
If it were symmetric exponential either side, that's normally called a Laplace or double exponential distribution.
$f(x;\mu,\tau) = \frac{1}{2\tau} \exp \left( -\frac{|x-\mu|}{\tau} \right) $ for $x,\mu\in \mathbb{R}$ and $\tau>0$

So that looks somewhat similar to what you have. But what about this?

The big problem is even if you investigate it a bunch of ways and it looks consistent with symmetric exponential decay each time, that doesn't mean it actually is. It might just be fairly close to that distribution.
[To be honest I think your distribution is actually more peaked and heavy tailed than a double exponential. If you look at the log-histogram counts or log-histogram proportions I expect that instead of looking like noise about a pair of straight lines joined at the peak, it will curve "in" toward the middle, and be more flared out in the tail.]
The same applies more generally -- you can possibly identify distributions that are plausible but you can't know they're correct. As the sample size grows to be large, you are generally able to rule out almost any simple closed-form distribution (real data is unlikely to actually conform perfectly to any of them, and eventually you can tell).
