# Sufficient statistic $\sum_{j=1}^{n} |x_{j}|$ for laplace distribution

Let be $$X_{1},\ldots , X_{n}$$ random variables independent and identically distributed with density function: $$f_{\theta}(x)=\dfrac{1}{2}e^{-|x-\mu|}, \quad x,\mu \in \mathbb{R}$$

Find the joint density function $$X=(X_{1},\ldots X_{n})$$ and show that $$\sum_{j=1}^{n} |x_{j}|$$ is a sufficient statistic for $$\mu$$.

I have that, $$f_{\theta}(x_{1},\ldots,x_{n})=\dfrac{1}{2^{n}}e^{-\sum_{i=1}^{n}|x_{i}-\mu|}$$ but I do not know how to use the factorization theorem to show that $$\sum_{j=1}^{n} |x_{j}|$$ is a sufficient statistic for $$\mu$$.

Also, How can I express this density function as an exponential family? I have problems with the expression $$e^{-\sum_{i=1}^{n}|x_{i}-\mu|}$$.

• Welcome to CV. If this question relates to a class exercise, please see stats.stackexchange.com/tags/self-study/info and add the tag to modify the question accordingly. Commented Nov 22, 2021 at 16:54
• Prior to searching for a sufficient statistic, (a) examine whether or not this is an exponential family and (b) check the Pitman-Koopman theorem. Commented Nov 22, 2021 at 16:55
• Alternative: plot the likelihood and check whether or not the resulting curve can be set by a single value like $\sum_i |x_i|$. Commented Nov 22, 2021 at 17:15
• Where does this assertion come from? You cannot "show" something that is not true. That's why I always advocate that the OP needs to provide the source of problem when asking a proof question. Commented Dec 19, 2023 at 3:23

and show that $$\sum_{j=1}^{n} |x_{j}|$$ is a sufficient statistic for $$\mu$$.

It is difficult to show this because it is not a sufficient statistic. The Laplace distribution has no sufficient statistic except for the entire sample.

In an earlier version of this post I tried to argue that the sample median was a sufficient statistic instead. But that is a fruitless exercise, since the Laplace family is not in the exponential family, and only distribution families of the exponential family type can have a sufficient statistic that doesn't grow with increasing sample size (related to the Pitman Koopman Darmois theorem discussed here: Undergraduate-level proofs of the Pitman–Koopman–Darmois theorem).

I do not know how to use the factorization theorem

You have to write the likelihood as a product of two functions. One function is allowed to have parameters in it, and one is not. The one with the parameter in it will have the sufficient statistic in it.

Also, How can I express this density function as an exponential family? I have problems with the expression

Once you have a definition of an example family (sometimes they are written in different ways), you just need to write your likelihood like that. It will simply be notation matching.

• @Taylor I gave the downvote because I found the answer particularly not helpful. For example the OP said they were having problem identifying density as exponential family and your suggestion is "just need to write your likelihood like that. It will simply be notation matching." This is not helpful. Especially because in the OP's case the Laplace distribution is NOT an exponential family in the $\mu$ parameter Commented Nov 22, 2021 at 21:51
• The correctness of the content is not only reason to vote "Voting up a question or answer signals to the rest of the community that a post is interesting, well-researched, and useful, while voting down a post signals the opposite: that the post contains wrong information, is poorly researched, or fails to communicate information." stats.stackexchange.com/help/why-vote Commented Nov 22, 2021 at 21:57
• @bdeonovic remember that you are not the intended audience. it is written to help students new to mathematical statistics. regarding your example, i believe it can be helpful, for some, to be reassured that the nature of problems of identifying whether a density belongings to an exponential family is not "deep." That will assuage concerns and divert attention towards what you are concerned with. Commented Nov 22, 2021 at 22:06
• I have to agree with @bdeonovic in this case: the answer in its present form suggests a course of investigation that is doomed to be confusing and fruitless.
– whuber
Commented Nov 22, 2021 at 23:12
• @Taylor I agree that answers are directed at the OP. I meant that your answer was particularly not useful for the OP. Lets hope that this wasn't some kind of homework problem. As a student I would be very confused if my professor gave me such a problem! Commented Nov 23, 2021 at 14:57