# Finding UMVUE for uniform distribution $U(\alpha, \beta)$

Let $$X = (X_1, X_2, \ldots, X_n)$$ be a sample from uniform distribution $$U(\alpha, \beta): \alpha, \beta \in \mathbb{R}, \alpha < \beta$$. I have to find UMVUE for the parameters $$\alpha, \beta$$.

Using factorization theorem I showed that $$T(X) = (\min\{ X_1, X_2, \ldots, X_2\}, \max\{ X_1, X_2, \ldots, X_n \})$$ is a sufficient statistic.

I think that I should use Lehmann–Scheffé theorem. Should I solve this problem by fixing $$\beta$$ and then computing the UMVUE for $$\alpha$$? Then vive versa?

I tried to find an unbiased estimator for $$\alpha$$ first. Thinking of $$\beta$$ as fixed I found that $$\mathbb{E}(2X_1 - \beta) = \alpha.$$ Thus using Lehmann–Scheffé theorem UMVUE would be $$\mathbb{E}(2X_1 - \beta|T(X)).$$ How can I find conditional expected value when $$T(X)$$ is a vector?

On the other hand I fixed $$\beta$$ in order to find my unbiased estimator so should I calculate $$\mathbb{E}(2X_1 - \beta|X_{(1)})$$?

I am a bit confused. I would appreciate any hints or tips.

$$T=(X_{(1)},X_{(n)})$$ is not only sufficient but a complete sufficient statistic which is needed here, $$X_{(k)}$$ being the $$k$$th $$\,(1\le k\le n)$$ order statistic.

Since the $$X_i$$'s are i.i.d $$U(a,b)$$ variables, $$Y_i=\frac{X_i-a}{b-a}$$ are i.i.d $$U(0,1)$$ variables, $$\, 1\le i\le n$$.

Now it is well-known that $$Y_{(1)}\sim \text{Beta}(1,n)$$ and $$Y_{(n)}\sim\text{Beta}(n,1)$$, implying $$E(Y_{(1)})=\frac{1}{n+1}$$ and $$E(Y_{(n)})=\frac{n}{n+1}$$. So all you have to do is solve for $$a$$ and $$b$$ from the equations

$$E(X_{(1)})=\frac{b-a}{n+1}+a\\ E(X_{(n)})=\frac{(b-a)n}{n+1}+a$$

You would get $$a$$ and $$b$$ as unbiased estimators of some function of $$T$$, and those will be the corresponding UMVUEs by Lehmann-Scheffe theorem.

• Does that mean that my UMVUE will be a vector too? Commented Jan 13, 2019 at 20:19
• @Hendrra No, the UMVUEs are linear combinations of $X_{(1)}$ and $X_{(n)}$. Commented Jan 13, 2019 at 20:21
• @Hendrra If you have found an unbiased estimator of your parameter based on a complete sufficient statistic, then that is bound to be the UMVUE and it equals the conditional expectation ( the latter is sometimes difficult to compute). This follows from Lehmann-Scheffe. Commented Jan 13, 2019 at 20:46
• @Hendrra The linear combination part is irrelevant (it's only here that the UMVUEs are linear combinations, not in general). Commented Jan 13, 2019 at 21:01
• Commented Feb 2, 2023 at 18:23