How does one find the sample median of for a group of iid random variables with Laplace distribution?

Question

I know a median is $$\frac{1}{2} = \int_{-\infty}^{\mu} f(x)dx$$ I understand how to solve this for simple distributions. However, I am learning how to do it for iid samples, which I haven't done before. In this current case, I have iid $X_1,...,X_n$ where they have a distribution $$\frac{1}{2b}e^{-|x_i-\theta|/b}$$ Note, assume $n = 2m+1$, $m\in \Bbb N$.

I calculated the MLE for this set of samples and want to now show the median is equal to $\hat{\theta}$

---

My Question:

How do I find the median of iid samples with a a Laplace distribution?

---

I know for an abstract discrete set of samples like $X_1,...,X_n$ I can order them like $X_{(1)},...,X_{(n)}$ from smallest to largest. I know my median should be some value like $X_{(m+1)}$. But I don't really understand how the Laplace distribution fits into that. To find the median for a Laplace distribution, I would think I could just use that integral formula. But that doesn't make sense if I am considering a discrete set of samples. In short, I think I am confusing finding the median of a distribution with finding the median for a set of iid samples with the same distribution. And I can't figure out what to do.

Answer 1

For a Laplacian distribution $\frac 12\exp(-|x-\theta|)$ with unknown mean (and median) $\theta$, the likelihood function of $n = 2m+1$ samples is $$L(\theta; x_1, x_2, \ldots, x_n) = \left(\frac 12\right)^n \prod_{i=1}^n \exp(-|x_i-\theta|) = \left(\frac 12\right)^n \exp\left(-\sum_{i=1}^n |x_i-\theta|\right).$$ The maximum value of $L$, regarded as a function of $\theta$, occurs at that point where $\sum_{i=1}^n |x_i-\theta|$ has minimum value. But, this function is a continuous piecewise linear function with slope $2i-n$ for $\theta \in (x_{(i)}, x_{(i+1)})$ where $x_{(i)}$ is the $i$-th order statistic of the data. Thus, the slope is negative for $\theta < x_{(m+1)}$ (where $x_{(m+1)}$ is the median of the $n=2m+1$ sample values) and positive for $\theta > x_{(m+1)}$. That is, the MLE estimator for $\theta$ which is the mean as well as for the median is $x_{(m+1)}$.