Obtaining the UMVUE for $U(-\theta, \theta)$ for $\frac{\theta}{1+\theta}$

Question

Let $X_1, X_2, .., X_n$ be $n$ random variables from $U(-\theta, \theta)$. We need to obtain the UMVUE of $\frac{\theta}{1+\theta}$.

I already derived that $Y = Max|X_i|$ is a complete and sufficient statistic (CSS). The pdf of this statistic is given by:

$f(y) = \frac{ny^{n-1}}{\theta^n}, \text{for }, 0<y<\theta$.

I need to obtain a a statistic such that it will be a function of CSS. I have observed one thing that:

$E(\frac{1}{y}) = \frac{n}{(n-1)\theta} => E(\frac{n-1}{n}\frac{1}{y}) = \frac{1}{\theta}$

It further implies that: $E(\frac{n-1}{n}\frac{1}{y} + 1) = \frac{1}{\theta} + 1 = \frac{1+\theta}{\theta}$

Which is the reciprocal of what i want. I want an unbiased estimate of the function of CSS. I get a lot of these type of problems where I have a CSS but I am struggling to get the Unbiased estimate of the desired parameter (or function of parameter). Can someone please suggest me what would be the UMVUE here? Also, Are there some of the standard procedures that can be followed while searching for Unbiased estimator?

Answer 1

As you know any function of a CSS is a UMVUE of its expected value. We wish to find a function of the CSS that has expected value desired. So $$\begin{split}E(g(Y))&=\frac{\theta}{1+\theta}\\ \int _0^\theta g(y)\frac{ny^{n-1}}{\theta^n}dy&=\frac \theta {1+\theta}\\ n\int _0^\theta g(y)y^{n-1}dy&=\frac{\theta^{n+1}}{1+\theta}\\ n\left[g(\theta)\theta^{n-1}\cdot 1-0\right]&=\frac{\theta^n(n\theta+n+1)}{(\theta+1)^2}\text{ take the derivative of both sides wrt $\theta$}\\ g(\theta)&=\frac{\theta(n\theta+n+1)}{n(1+\theta)^2}\end{split}$$

Therefore, replacing the $\theta$ with Y... you can conclude that $g(Y)=\frac{Y(nY+n+1)}{n(1+Y)^2}$ has a certain expected value, and hence is a UMVUE of a certain something...