1
$\begingroup$

Let $X_1,X_2,...,X_n$ be a random sample from a population with probability density function $$f(x\mid\theta)=\frac{\theta}{2}e^{-\theta|x|} \,,\,-\infty < x < \infty\,,\,\theta>0$$

Then a UMVUE of $\theta$ is ?

Can someone tell me if i did everything right or not in the following steps. I dont have answer written for this problem in my solution manual.

1.Its an even function that's why pdf changes as $$f(x\mid\theta)=\theta e^{-\theta x} ,0 < x < \infty,\theta>0$$ 2.Using Rao blackwell theorem i found MVUE $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$

Now MVUE is $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$ or $\dfrac{n-1}{\sum_{i=1}^{n} |X_i|}$ ? Please give your thoughts and tell me where did i follow wrong track?

$\endgroup$
3
  • $\begingroup$ $\theta > 0$ given. $\endgroup$
    – Daman
    Jan 22, 2018 at 16:37
  • $\begingroup$ So much for my reading comprehension! I'll delete my pointless comment, if I'm still able. $\endgroup$
    – jbowman
    Jan 22, 2018 at 16:39
  • $\begingroup$ Its totally fine. $\endgroup$
    – Daman
    Jan 22, 2018 at 16:41

1 Answer 1

2
$\begingroup$

Yes, $f(x\mid\theta)$ is an even function of $x$, but how can the pdf 'change' completely in your calculations? You carried out calculations for an exponential density whereas you are given a Laplace distribution.

Indeed, the sample is drawn from a Laplace distribution with scale parameter $1/\theta$ and location parameter zero. We can conclude that the joint density $f_{\theta}$ belongs to the one-parameter exponential family. Using this, we can find a complete sufficient statistic for $\theta$.

Due to independence, joint density of $\mathbf X=(X_1,X_2,\cdots,X_n)$ is

\begin{align} f_{\theta}(\mathbf x)&=\prod_{i=1}^n\frac{\theta}{2}e^{-\theta\,|x_i|} \\&=\left(\frac{\theta}{2}\right)^ne^{-\theta\sum_{i=1}^n|x_i|} \\&=\exp\left[-\theta\sum_{i=1}^n|x_i|+n\ln\left(\frac{\theta}{2}\right)\right]\quad,\,\mathbf x\in\mathbb R^n\,,\,\theta>0 \end{align}

Clearly, a complete sufficient statistic for the family of distributions $\{f_{\theta}:\theta>0\}$ is $$T(\mathbf X)=\sum_{i=1}^n|X_i|$$

By Lehmann-Scheffe theorem, an unbiased estimator of $\theta$ based on $T$ will be the UMVUE of $\theta$.

It is a simple exercise to show that $|X_i|\sim\mathcal{Exp}(\theta)$ independently for each $i$, where $\theta$ denotes rate of the distribution. As such, we have $T\sim\mathcal{Ga}(\theta,n)$ with density $$g(t)=\frac{\theta^n e^{-\theta t}t^{n-1}}{\Gamma(n)}\mathbf1_{t>0}$$

We find that

\begin{align} E\left(\frac{1}{T}\right)&=\int_0^{\infty}\frac{1}{t}\frac{\theta^n e^{-\theta t}t^{n-1}}{\Gamma(n)}\,dt \\&=\frac{\theta^n}{\Gamma(n)}\frac{\Gamma(n-1)}{\theta^{n-1}} \\&=\frac{\theta}{n-1} \end{align}

So the UMVUE of $\theta$ is $$\frac{n-1}{T}=\frac{n-1}{\sum_{i=1}^n|X_i|}$$

$\endgroup$
3
  • $\begingroup$ Can you tell me when you calculated$ E\bigg(\dfrac{1}{T}\bigg)$ how come you substituted $\dfrac{1}{t}$ our $T=\dfrac{1}{\sum |X_i|}$. Just explain to me Expectation part. $\endgroup$
    – Daman
    Jan 10, 2019 at 16:29
  • 1
    $\begingroup$ @Damn1o1 $T$ is the sum of $|X_i|$'s, and the distribution of $T$ is known. So, $E\,[h(T)]=\int h(x)g(x)\,dx$ for any function $h$ where $g$ is the pdf of $T$. Courtesy this theorem. $\endgroup$ Jan 10, 2019 at 17:52
  • 1
    $\begingroup$ So I understood it this year last year it went above my head. Thanks man! $\endgroup$
    – Daman
    Jan 11, 2019 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.