# Rao-Blackwell exponential distribution

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?
• 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. May 19, 2015 at 14:49
• @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?
– user72621
May 19, 2015 at 15:05

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.