First notice that median minimizes the L1 norm (see [here][1] or [here][2] for learning more on L1 and L2)

$$ \DeclareMathOperator*{\argmin}{arg\,min}
\text{median}(x) = \argmin_s \sum_i |x_i - s|^1  $$

while mean minimizes L2

$$ \text{mean}(x) = \argmin_s \sum_i |x_i - s|^2  $$

now, recall that Normal distributions' $\mu$ parameter can be estimated using [sample mean][3], while the [MLE estimator][4] for Laplace distribution $\mu$ parameter is median. So using Normal distribution is equivalent to L2 norm optimization and using Laplace distribution, to using L1 optimization.

---

Hurley, W. J. (2009) [An Inductive Approach to Calculate the MLE for the Double Exponential Distribution][5]. *Journal of Modern Applied Statistical Methods: 8*(2), Article 25.

Kotz, S., Kozubowski, T., Podgorski, K. (2001). [*The Laplace Distribution and Generalizations: A Revisit with Applications to Communications, Economics, Engineering, and Finance.*][6] Boston: Birkhäuser.


  [1]: http://www.johnmyleswhite.com/notebook/2013/03/22/modes-medians-and-means-an-unifying-perspective/
  [2]: https://rorasa.wordpress.com/2012/05/13/l0-norm-l1-norm-l2-norm-l-infinity-norm/
  [3]: https://en.wikipedia.org/wiki/Normal_distribution#Sample_mean
  [4]: https://en.wikipedia.org/wiki/Laplace_distribution#Parameter_estimation
  [5]: http://digitalcommons.wayne.edu/jmasm/vol8/iss2/25
  [6]: https://books.google.at/books?id=cb8B07hwULUC&lpg=PA22&dq=laplace%20distribution%20exponential%20characteristic%20function&hl=fr&pg=PA66#v=onepage&q=median&f=false