# consistency of mle of double exponential distribution

Let $y_i\sim DE(\mu, \sigma), i=1,2,...,n, \ i.i.d$.

Where $DE$ represents the double exponential distribution. The the MLE of $\sigma$ is:

$\hat\sigma$ = $\frac{1}{n} \sum_{i=1}^{n}|y_i-med(y_i)|$, where $med$ refers to the median of the $y_i$s.

How can I show that the MLE is consistent?

I have learned some methods to prove consistency of random variable, but I don't know how to deal with the absolute value and median.

• Note that the title doesn't match the body. You're not computing the MLE of the distribution, but rather the MLE of the scale parameter, $\sigma$. What are the methods you know? – Glen_b Sep 28 '17 at 13:59

To close this one, here is a way to prove consistency constructively, without invoking the general properties of the MLE that make it a consistent estimator.

Because the sample is i.i.d., it is ergodic-stationary. This means that sample moments and quantiles tend to their theoretical true counterparts. So

$$\hat \sigma =\frac{1}{n} \sum_{i=1}^{n}|y_i-med(y_i)| \to_p E|Y-\mu|$$

This is the probability limit of the estimator, and we have to show that $\sigma = E|Y-\mu|$ to prove consistency.

We have that $Y-\mu \equiv Z \sim DE(0,\sigma)$. So

$$E|Y-\mu|=E|Z| = \int_{-\infty}^{\infty}|z| \frac 1{2\sigma}\exp\{-|z|/\sigma\}dz$$

$$=2\int_{0}^{\infty}(z/\sigma) \frac 1{2}\exp\{-z/\sigma\}dz$$

Simplify $2$ and multiply and divide by $\sigma$

$$E|Y-\mu|=E|Z| = \sigma \int_{0}^{\infty}(z/\sigma) \exp\{-z/\sigma\}d(z/\sigma)$$

But the integrand is now the expected value of a standard exponential random variable, so the integral equals unity. Therefore we obtain

$$\text{plim } \hat \sigma = E|Y-\mu|= \sigma$$