Rao-Blackwell Theorem – How to Obtain an Estimator via Rao-Blackwell Theorem

Question

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!

Answer 1

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.