3
$\begingroup$

Let $X_1,\ldots,X_n$ be iid with pdf

$$f(x\mid\theta) = \exp(\theta -x) I(x)_{(\theta, \infty)}$$

It is asked to find an unbiased estimator of $\theta$ that is a function of a sufficient statistic for $\theta$.

By factorization theorem, we show that $X_{(1)}$ is a sufficient statistical for $\theta.$ And, since

$$E(X) = \theta +1$$

the estimator $ \bar{X} -1$ is unbiased. So, by the Rao Blackell theorem,

$$W=E(\bar{X}-1\mid X_{(1)})$$ is an unbiased estimator that is function of the sufficient statistical. But it seems very complicated to evaluate the distribution of $\bar{X}-1\mid X_{(1)}$. How can I find this distribution? Is there a better unbiased estimador that I can use, in this case?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ Why go roundabout? Obtaining the distribution of $X_{(1)}$ and its expected value is not difficult. Have you tried that, instead of going through the sample mean? $\endgroup$ – Alecos Papadopoulos Aug 28 '14 at 22:36
  • $\begingroup$ No, I didn't. Do you think we can find a unbiesed estimator that is function of the minimum? It would let the work of calculatingthe conditional distribution a little more easy.Another attempt: We could use, for example, $X_1 -1$ instead of $\bar{X}$! But I dont see a good way of finding the distribution $\endgroup$ – Giiovanna Aug 29 '14 at 0:08
  • 1
    $\begingroup$ Giovanna, it is really easy to find the density function of the minimum in this case, and then calculate the expected value of the minimum, which will immediately give you an unbiased estimator based on the minimum. First find the CDF of $X$, then apply the standard formula for the density of the minimum. $\endgroup$ – Alecos Papadopoulos Aug 29 '14 at 0:36
  • $\begingroup$ The minimum pdf is given by $ n \exp^{n-1}(\theta - x)$. In this case, the expectation is not a "beautiful" function of theta so I can construct an estimator using it. Can you explain it a little more? $\endgroup$ – Giiovanna Aug 29 '14 at 21:32
  • $\begingroup$ The CDF of $X$ is $F_X(x) = 1- \exp\{\theta-x\}$. The CDF of the minimum is $1- [1-F_X(x_{(1)})]^n$. So the pdf is $ne^{n(\theta -x_{(1)})}I(x)_{(\theta, \infty)}$. The integral $\int_{\theta}^{\infty} x_{(1)}ne^{n(\theta -x_{(1)})}dx_{(1)}$ presents no difficulties. This is $E(X_{(1)})$. You will find that it is a linear function of $\theta$ and so correcting for the bias is obvious. $\endgroup$ – Alecos Papadopoulos Aug 29 '14 at 21:50
6
$\begingroup$

We have $$F_X(x) = \int_{\theta}^{x}e^{\theta -t} dt = -e^{\theta}e^{-t}\Big|^{x}_{\theta} = 1 - e^{\theta -x} $$

Since $F_{X_{(1)}}(x_{(1)}) = 1 -[1-F_X(x_{(1)})]^{n}$, the density function of the minimum order statistic is

$$f_{X_{(1)}}(x_{(1)}) = nf_X(x_{(1)})[1-F_X(x_{(1)})]^{n-1}I(x)_{(\theta, \infty)} = ne^{\theta -x_{(1)}}[e^{\theta -x_{(1)}}]^{n-1}I(x)_{(\theta, \infty)}$$

$$\Rightarrow f_{X_{(1)}}(x_{(1)}) =ne^{n(\theta -x_{(1)})}I(x)_{(\theta, \infty)}$$

Then

$$E[X_{(1)}] = \int_\theta^{\infty}x_{(1)}ne^{n(\theta -x_{(1)})}dx_{(1)} =\theta+\frac 1n$$

and so

$$ \hat \theta = X_{(1)} -\frac 1n$$

is an unbiased estimator based on the sufficient statistic.

$\endgroup$
2
$\begingroup$

The fact is that Alecos' answer is the easiest way to handle the problem, but the problem can be solved via Rao-Blackwell as well. Start with the joint density $$f(x_1,..., x_n | \theta) = e^{-\sum x_i + n\theta}\prod{I_{\theta < x_i}(x_i)}. $$ We know that $X_{[1]}$ is a sufficient statistic, so this factors as $$f(x_1,...,x_n)=e^{-\sum x_i +nx_{[1]}}\times e^{-n(x_{[1]}-\theta)}\cdot I_{\theta < x_1}(x_{[1]}).$$ Now apply the Rao-Blackwell Theorem to this form of the density, and correct for the bias. You could also incorporate the shift ($\bar{X} - 1$) explicitly in the joint density, in which case the bias correction is incorporated directly.

$\endgroup$
  • $\begingroup$ To apply Rao Blackwell, I need an unbiased estimator (for example, $X_1 -1$) and I need to find the distribution of the sufficient statistics given this estimator But it not seems so easy to find the distribution of $X_{(1)}|X_1-1$ $\endgroup$ – Giiovanna Aug 29 '14 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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