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Let $X_1,..,X_n$ be a random sample of $X\sim\text{Exp}(\lambda)$ with $f(x;\lambda)=\frac{1}{\lambda}e^{-\frac{1}{\lambda}x}I_{[0,\infty]}(x)$

i) Find a unbiased estimator of $\lambda$ based only on $X_{(1)}=\min(X_i)$

ii) Apply the Rao-Blackwell theorem to find an estimator better than the one you found in i)

For i) I find that $\hat{\lambda}=nX_{(1)}$ is unbiased estimator for $\lambda$. Now for the part ii) that is the problem.

I know that $f(x;\lambda)\in$ the exponential family and $T=\sum X_i$ is a complete and sufficient statistic. Then since $\phi_T=E[nX_{(1)}|\sum X_i]$ produces a unbiased estimator, then it needs to be UMVUE.

So I calculated the Cramer-Rao Lower Bound and find $\text{var}_\lambda\geq \frac{\lambda^2}{n}$, finally I just take $\phi_T=\frac{\sum X_i}{n}$

Now my doubts are:

  1. Is my reasoning correct?
  2. Do I always need to calculate the conditional distribution?
  3. Is there any simple way to find this conditional?
  4. How could I find the conditional in this case?
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    $\begingroup$ Since your reasoning is based on Cramer-Rao and not Rao-Blackwell, you do not answer the question. My opinion is that yes indeed you need to compute the conditional expectation. Maybe remarking that $$\sum_i X_i=\sum_i X_{(i)}$$could help. $\endgroup$
    – Xi'an
    May 19, 2015 at 14:49
  • $\begingroup$ @Xi'an But I used the Cramer-Rao lower bound just for confirm my idea, if I were asked to just find a better estimator could I do that? $\endgroup$
    – user72621
    May 19, 2015 at 15:05

1 Answer 1

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To find the conditional expectation $\operatorname E[nX_{(1)}\mid T]$ as part of the Rao-Blackwellization, we can use the independence between $\frac{X_{(1)}}{T}$ and $T$.

This independence can be argued by Basu's theorem since $T=\sum\limits_{i=1}^n X_i$ is a complete sufficient statistic and $\frac{X_{(1)}}{T}=\frac{X_{(1)}/\lambda}{T/\lambda}$ is an ancillary statistic, i.e. its distribution is free of $\lambda$.

So, $$\operatorname E\left[X_{(1)}\mid T\right]=\operatorname E\left[\frac{X_{(1)}}{T}\cdot T\mid T\right]=T\operatorname E\left[\frac{X_{(1)}}{T}\mid T\right]=T\operatorname E\left[\frac{X_{(1)}}{T}\right] \,,\text{ a.e. }\tag{1}$$

Again,

$$\operatorname E\left[X_{(1)}\right]=\operatorname E\left[\frac{X_{(1)}}{T}\cdot T\right]=\operatorname E\left[\frac{X_{(1)}}{T}\right] \operatorname E\left[T\right]\,,$$

so that (since $\operatorname E\left[T\right]\ne 0\,$)

$$\operatorname E\left[\frac{X_{(1)}}{T}\right]=\frac{\operatorname E\left[X_{(1)}\right]}{\operatorname E\left[T\right]} \tag{2}$$

$(1)$ and $(2)$ together imply

$$\operatorname E\left[nX_{(1)}\mid T\right]=nT\cdot\frac{\operatorname E\left[X_{(1)}\right]}{\operatorname E\left[T\right]}=nT\cdot \frac{\lambda/n}{n\lambda}=\frac{T}n \,\,,\text{ a.e. }$$

The answer is of course not surprising since $T/n$ is UMVUE of $\lambda$ by Lehmann-Scheffé (or by the fact that $T/n$ is unbiased for $\lambda$ and its variance attains the Cramér-Rao bound for $\lambda$), and we know that UMVUE is unique whenever it exists.

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