Finding the MVUE of a two-sided interval of a normal

[rayge]

Our task is to determine if $P(-c \le X \le c)$ has a minimum variance unbiased estimator for a sample from a distribution that is $N(\theta,1)$. The one-sided interval $P(X \le c) = \Phi(x - \theta)$ is unique, so constructing an MVUE is just a matter of applying Rao-Blackwell and Lehmann-Scheffe.

However for our case, $P(-c \le X \le c)$ is the same for $\theta$ and $-\theta$. So it seems like the MVUE isn't unique. I'm wondering if you can make a decision rule like choosing one unbiased estimator when $\theta \ge 0$ and the other when $\theta < 0$, but now instead of two non-unique unbiased estimators, we have three. Any thoughts? Is a MVUE just not possible?

 

[Stephen Tashi]

Our task is to determine if $P(-c \le X \le c)$ has a minimum variance unbiased estimator for a sample from a distribution that is $N(\theta,1)$.

I assume this means that $c$ is given and $\theta$ is unknown. So you can't employ a rule that depends on knowing the sign of $\theta$.

 

[rayge]

What if we construct two MVUE's, one for $P(X \le c)$, and one for $P(X \le -c)$, and then subtract one from the other? It still seems like we have the same problem, where the MVUE is not one-to-one...

 

[Stephen Tashi]

However for our case, $P(-c \le X \le c)$ is the same for $\theta$ and $-\theta$.

There is ambiguity if you estimate $P(-c \le X \e c)$ first and try to estimate $\theta$ from that estimate. However the problem you stated doesn't insist we estimate $\theta$ in that manner. Wouldn't the simplest try be to estimate $\theta$ from the sample mean and then estimate $P(-c \le X \le c)$ from that estimate?