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Let $X_1, \dots, X_n$ be i.i.d. from the $DE(\theta,b)$ distribution (DE is the double exponential). (a) Show that the MLE $\theta$ is the sample median. (b) Prove that the MLE $\theta$ is asymptotically efficient.

For part (a): $$L(\theta) = \frac{1}{(2b)^n}e^{-\sum_{i=1}^{n}\frac{|X_i-\theta|}{b}}$$

Let $g(\theta) = \sum_{i=1}{n}|X_i-\theta|$ and $L(\theta)$ is the largest when $g(\theta)$ is the smallest.

Since $g(\theta)$ is continuous function of $\theta$, so $$g'(\theta) = \sum_{i=1}^{n}[-1\times 1\{X_i>\theta\}+1\{X_i<\theta\}]= \begin{cases} \text{positive if } \theta > \operatorname{median}(X_i) \\\text{negtive if } \theta < \operatorname{median}(X_i) \end{cases}$$

So, MLE $\hat{\theta}$ is the sample median.

For part (b):

$lnf_\theta(x_1) = -ln2b-\frac{|x_1-\theta|}{b}$

$\frac{\partial}{\partial\theta}lnf_\theta(x_1)=\frac{1}{b}$

$I(\theta) = E[\frac{\partial}{\partial\theta}lnf_\theta(x_1)]^2=\int_{-\infty}^{\infty}\frac{1}{b^2}\frac{1}{2b}e^{-\frac{|x_1-\theta|}{b}}dx=\frac{1}{2b^3}\int_{0}^{\infty}be^{-y}dy=\frac{1}{2b^2}$

where $y=\frac{|x_1-\theta|}{b}$, $\frac{dy}{dx}=\frac{1}{b}dx$, $y \in(0,\infty)$

so, $\mathcal{v(\theta_0)}=\frac{1}{I(\theta_0)}=2b^2$

But, in class, the professor showed us that We have proved that for the sample median $M_n$, $\sqrt{n}(M_n-\theta) \xrightarrow{L} N(0, \frac{1}{4(f_\theta(\theta))^2})$ and $f_\theta(\theta)) = \frac{1}{2b}e^{-\frac{|\theta-\theta|}{b}}=\frac{1}{2b}$

So, $\sqrt{n}(M_n-\theta) \xrightarrow{L} N(0, \frac{1}{4(f_\theta(\theta))^2})=N(0, b^2)$

I suppose to get $I(\theta)=\frac{1}{b^2}$ right? Could you help me out?

anonyx2
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    Why is it divergent? – user10354138 Apr 01 '21 at 17:33
  • I used a calculator, the result tells me that it's divergent. – anonyx2 Apr 01 '21 at 17:39
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    The integral converges as long as $b > 0$. – A rural reader Apr 01 '21 at 17:44
  • Shouldn't you be looking at $I(\theta)$ instead of $I(b)$? – StubbornAtom Apr 01 '21 at 18:23
  • Thanks! I will try that way. – anonyx2 Apr 01 '21 at 18:41
  • If $b$ is a known quantity in this exercise, then $\frac{\partial}{\partial\theta}\ln f_\theta(x)=\frac{1}{b}$ should directly imply $E\left[\frac{\partial}{\partial\theta}\ln f_\theta(X)\right]^2=\frac1{b^2}$. But technically, for $x\ne \theta$, I think you should have $\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=\frac1b$ if $x>\theta$ and $-\frac1b$ if $x<\theta$. Notice that the derivative does not exist at $x=\theta$. This doesn't change the final answer though. – StubbornAtom Apr 01 '21 at 20:05

0 Answers0