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

In summary, the task at hand is to determine if P(-c \le X \le c) has a minimum variance unbiased estimator for a sample from a normal distribution with mean \theta and variance 1. The one-sided interval P(X \le c) = \Phi(x - \theta) is unique, so constructing an MVUE can be done using Rao-Blackwell and Lehmann-Scheffe. However, for our case, P(-c \le X \le c) is the same for \theta and -\theta, making the MVUE not unique. It is suggested to use two MVUEs, one for P(X \le c) and one for P(X \le -c), and then subtract
  • #1
rayge
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Our task is to determine if [itex]P(-c \le X \le c)[/itex] has a minimum variance unbiased estimator for a sample from a distribution that is [itex]N(\theta,1)[/itex]. The one-sided interval [itex]P(X \le c) = \Phi(x - \theta)[/itex] is unique, so constructing an MVUE is just a matter of applying Rao-Blackwell and Lehmann-Scheffe.

However for our case, [itex]P(-c \le X \le c)[/itex] is the same for [itex]\theta[/itex] and [itex]-\theta[/itex]. 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 [itex]\theta \ge 0[/itex] and the other when [itex]\theta < 0[/itex], but now instead of two non-unique unbiased estimators, we have three. Any thoughts? Is a MVUE just not possible?
 
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  • #2
rayge said:
Our task is to determine if [itex]P(-c \le X \le c)[/itex] has a minimum variance unbiased estimator for a sample from a distribution that is [itex]N(\theta,1)[/itex].

I assume this means that [itex] c [/itex] is given and [itex] \theta [/itex] is unknown. So you can't employ a rule that depends on knowing the sign of [itex] \theta [/itex].
 
  • #3
What if we construct two MVUE's, one for [itex]P(X \le c)[/itex], and one for [itex]P(X \le -c)[/itex], and then subtract one from the other? It still seems like we have the same problem, where the MVUE is not one-to-one...
 
  • #4
rayge said:
However for our case, [itex]P(-c \le X \le c)[/itex] is the same for [itex]\theta[/itex] and [itex]-\theta[/itex].

There is ambiguity if you estimate [itex] P(-c \le X \e c) [/itex] first and try to estimate [itex] \theta [/itex] from that estimate. However the problem you stated doesn't insist we estimate [itex]\theta [/itex] in that manner. Wouldn't the simplest try be to estimate [itex] \theta [/itex] from the sample mean and then estimate [itex] P(-c \le X \le c) [/itex] from that estimate?
 
  • #5


I can provide a response to the above content by saying that the concept of MVUE (Minimum Variance Unbiased Estimator) is a statistical tool used to estimate a population parameter with minimum possible variance and without any bias. In the case of a two-sided interval of a normal distribution, the task is to determine if there exists an MVUE for the parameter \theta.

In the given scenario, we have a normal distribution with a mean of \theta and a variance of 1. The one-sided interval P(X \le c) = \Phi(x - \theta) is unique, and we can apply Rao-Blackwell and Lehmann-Scheffe to construct an MVUE.

However, when we consider the two-sided interval P(-c \le X \le c), we see that it is the same for \theta and -\theta. This leads us to believe that the MVUE may not be unique in this case. It is possible to have multiple unbiased estimators for the same parameter, and this can make it challenging to determine the MVUE.

One approach to dealing with this issue could be to use a decision rule, as mentioned in the content. This approach involves choosing one unbiased estimator when \theta \ge 0 and another when \theta < 0. However, this would result in three non-unique unbiased estimators, which may not be ideal.

In conclusion, it may be challenging to determine an MVUE for the parameter \theta in this scenario. It is possible that an MVUE does not exist in this case, or it may require further exploration and analysis to find a suitable solution. As a scientist, it is important to carefully consider all possible options and make informed decisions based on statistical principles and methods.
 

1. What is the MVUE of a two-sided interval of a normal distribution?

The MVUE, or minimum variance unbiased estimator, is a statistical method used to estimate the parameters of a normal distribution. In this case, the MVUE of a two-sided interval of a normal distribution is the most efficient and unbiased estimate of the mean.

2. How is the MVUE calculated for a two-sided interval of a normal distribution?

The MVUE is calculated by taking the average of the sample mean and the sample variance. This estimate is then adjusted to account for the sample size and any known parameters of the distribution.

3. What are the advantages of using the MVUE for a two-sided interval of a normal distribution?

The MVUE is advantageous because it is unbiased, meaning that it provides an estimate that is not systematically higher or lower than the true value of the parameter. Additionally, the MVUE has the lowest possible variance, making it the most efficient estimate.

4. Can the MVUE be used for any sample size?

Yes, the MVUE can be used for any sample size. However, as the sample size increases, the MVUE becomes closer to the true value of the parameter and the variance decreases.

5. How is the MVUE affected by the shape of the normal distribution?

The MVUE is not affected by the shape of the normal distribution as long as the distribution is symmetric. However, if the distribution is skewed or has outliers, the MVUE may not be the most efficient estimate.

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