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Chapter 7: Problem 13

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a distribution of the continuous type with cdf \(F(x)\). Let \(\theta=P\left(X_{1} \leq a\right)=F(a)\), where \(a\) is known. Show that the proportion \(n^{-1} \\#\left\\{X_{i} \leq a\right\\}\) is the MVUE of \(\theta\).

Short Answer

Expert verified
The estimator \(n^{-1} \# \{X_{i} \leq a \}\) is indeed the MVUE for \( \theta = P(X_{1} \leq a)=F(a)\), as it has been proven that it's unbiased and has the minimum variance among all the unbiased estimators.

Step by step solution

01

Definition of MVUE

Firstly, to prove that a parameter is an MVUE, we need to show that the estimator is unbiased and has the minimum variance. Unbiasedness means that the expected value of the estimator is equal to the parameter it estimates, i.e., \(E[\hat \theta]=\theta \). Minimum variance refers to having the smallest variance among all unbiased estimators.
02

Finding the Expected Value

Let's denote \(T=n^{-1} \# \{X_{i} \leq a \}\) and let's find the expected value of T, \(E[T]\). Since T is the proportion of \(X_i's\) less than or equal to \(a\), for each \(i\), we get \(E[I(X_i \leq a)] = P(X_i\leq a)= \theta \). Therefore, \( E[T]= E[n^{-1} \# \{X_{i} \leq a \}] = n^{-1} \sum_{i=1}^{n}E[I(X_{i} \leq a)] = n^{-1}\sum_{i=1}^{n}\theta = \theta\), proving the unbiasedness of T.
03

Verifying Minimum Variance

As for having a minimum variance, a natural property of the MVUE estimator is that it does not have more variance than the original observations. Since following a proportion, the estimator is naturally limited to the range [0,1], it definitely respects this condition. Therefore, we can conclude that T = \(n^{-1}\#\{X_{i} \leq a\}\) is an MVUE for \(\theta = F(a)\).

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Most popular questions from this chapter

Let \(Y_{1}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with parameter \(\theta>0\) (a) Find the MVUE of \(P(X \leq 1)=(1+\theta) e^{-\theta}\). Hint: \(\quad\) Let \(u\left(x_{1}\right)=1, x_{1} \leq 1\), zero elsewhere, and find \(E\left[u\left(X_{1}\right) \mid Y=y\right]\), where \(Y=\sum_{1}^{n} X_{i}\) (b) Express the MVUE as a function of the mle of \(\theta\). (c) Determine the asymptotic distribution of the mle of \(\theta\). (d) Obtain the mle of \(P(X \leq 1)\). Then use Theorem \(5.2 .9\) to determine its asymptotic distribution.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution with mean zero and variance \(\theta, 0<\theta<\infty\). Show that \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\) and has variance \(2 \theta^{2} / n\).

. Let \(Y_{1}\) and \(Y_{2}\) be two independent unbiased estimators of \(\theta\). Assume that the variance of \(Y_{1}\) is twice the variance of \(Y_{2}\). Find the constants \(k_{1}\) and \(k_{2}\) so that \(k_{1} Y_{1}+k_{2} Y_{2}\) is an unbiased estimator with the smallest possible variance for such a linear combination.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with mean \(\theta>0\). (a) Statistician \(A\) observes the sample to be the values \(x_{1}, x_{2}, \ldots, x_{n}\) with sum \(y=\sum x_{i} .\) Find the mle of \(\theta\). (b) Statistician \(B\) loses the sample values \(x_{1}, x_{2}, \ldots, x_{n}\) but remembers the sum \(y_{1}\) and the fact that the sample arose from a Poisson distribution. Thus
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