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Welche der folgenden Aussagen (a) bis (c) sind richtig: (a) Der Anteil \(\theta\) wird bei einer einfachen Stichprobe durch die relative Häufigkeit \(\widehat{\theta}\) in der Stichprobe geschätzt. Bei dieser Schätzung ist der MSE umso größer, je näher \(\theta\) an \(0.5\) liegt. (b) \(\bar{X}\) ist stets ein effizienter Schätzer für \(\mathrm{E}(X)\). (c) Eine nichtideale Münze zeigt,,"Kopf" mit Wahrscheinlichkeit \(\theta\). Sie werfen die Münze ein einziges Mal und schätzen $$ \widehat{\theta}= \begin{cases}1, & \text { falls die Münze , Kopf" zeigt. } \\\ 0, & \text { falls die Münze ,Zahl" zeigt. }\end{cases} $$ Dann ist diese Schätzung erwartungstreu.

Short Answer

Expert verified
(a) In a simple random sample, the mean squared error of the relative frequency estimator for the parameter theta is larger for values of theta closer to 0.5. (b) The sample mean is always an efficient estimator for the expected value of X. (c) The estimator for the parameter theta, based on a single coin toss, is an unbiased estimator for a non-ideal coin that shows "Head" with probability theta. Answer: (a) and (c) are correct, while statement (b) is not correct without additional conditions or context.

Step by step solution

01

Statement (a) - Analyzing the MSE of a simple random sample

To determine if statement (a) is correct, we need to examine the relationship between the mean squared error (MSE) and the parameter theta in a simple random sample. Mean squared error can be expressed as: $$ \text{MSE}(\widehat{\theta}) = \mathrm{Bias}(\widehat{\theta})^2 + \mathrm{Var}(\widehat{\theta}) $$ When the estimator is unbiased, the bias term is zero, and the MSE is equal to the variance of the estimator. For a simple random sample, the unbiased estimator of the parameter \(\theta\) is the relative frequency \(\widehat{\theta}\) in the sample. In the case of a Bernoulli distribution, the variance of the estimator can be expressed as: $$ \mathrm{Var}(\widehat{\theta}) = \frac{\theta(1-\theta)}{n} $$ where n is the sample size. The variance is maximized when \(\theta = 0.5\). Therefore, the MSE is larger for values of \(\theta\) closer to 0.5. Statement (a) is correct.
02

Statement (b) - Efficiency of the sample mean

Statement (b) states that the sample mean \(\bar{X}\) is always an efficient estimator for the expected value of X. An efficient estimator is one that has the smallest variance among all unbiased estimators. The sample mean is known to be an unbiased estimator for the expected value of X and, under certain conditions, can be considered efficient according to the Cramér-Rao lower bound. However, the statement says "always," which is not entirely accurate, as there may be cases where other estimators might be more efficient. Statement (b) is not correct without additional conditions or context.
03

Statement (c) - Unbiasedness of a single coin toss estimator

In statement (c), we are given a non-ideal coin that shows "Head" with probability \(\theta\). We want to determine if the proposed estimator based on a single coin toss, represented by \(\widehat{\theta}\), is unbiased. An estimator is unbiased if its expected value is equal to the true value of the parameter being estimated: $$ \mathrm{E}(\widehat{\theta}) = \theta $$ Let us calculate the expected value of the proposed estimator: $$ \mathrm{E}(\widehat{\theta}) = 1 \cdot \mathrm{P}(\text{"Head"}) + 0 \cdot \mathrm{P}(\text{"Tail"}) = \theta $$ As the expected value of the proposed estimator equals the true value of the parameter, statement (c) is correct, and the estimator is unbiased. In conclusion, statements (a) and (c) are correct, while statement (b) is not correct without additional conditions or context.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Squared Error (MSE)
The Mean Squared Error (MSE) is a concept often encountered in the field of statistical estimation. It's a measure used to determine how close an estimator is to the true parameter. Mathematically, MSE is expressed as the sum of the variance and the square of the bias of the estimator: \[ \text{MSE}(\widehat{\theta}) = \mathrm{Bias}(\widehat{\theta})^2 + \mathrm{Var}(\widehat{\theta}) \]
  • Bias: This is the difference between the expected value of the estimator and the true value of the parameter. If the estimator is unbiased, this term is zero.
  • Variance: This indicates how much the estimator's estimation might differ from the expected value. Essentially, it measures the estimator's consistency.
For a Bernoulli distribution, as one example, the variance is given as \( \frac{\theta(1-\theta)}{n} \), where \( n \) is the sample size. The MSE reaches its peak when \( \theta = 0.5 \), meaning that the estimation tends to be less accurate around this value.
Unbiased Estimator
An unbiased estimator plays a crucial role in statistics because it provides the expected value equal to the parameter it estimates. In simpler terms, if you repeatedly collect samples and calculate the estimator, on average, it should equal the true parameter of the population. This is concisely presented as: \[ \mathrm{E}(\widehat{\theta}) = \theta \] An example of this concept is noted in the context of estimating the probability of a coin showing "Head." If we suggest that \( \widehat{\theta} = 1 \) if "Head" shows and \( \widehat{\theta} = 0 \) if "Tail" shows on a single toss, this is an unbiased estimator of the true probability \( \theta \). The main benefit of an unbiased estimator is that, on average, it doesn't systematically overestimate or underestimate the parameter. This makes unbiased estimators preferable in many statistical applications.
Efficiency of Estimators
The efficiency of an estimator refers to its degree of reliability, especially its variance. An efficient estimator is the one that possesses the smallest possible variance amongst all unbiased estimators for a given parameter, making it a prime choice for analysis. The concept of efficiency can often be understood in the context of the Cramér-Rao lower bound, which provides a benchmark for the variance of unbiased estimators. If an estimator reaches this lower bound, it is deemed an efficient estimator. However, it is essential to realize that certain statements regarding efficiency may require additional contexts. For example, the exercise mentioned that the sample mean \( \bar{X} \) is not always the most efficient estimator in all situations. The efficiency can depend on other conditions or distributions specific to the dataset or parameter settings.In summary, while efficient estimators are sought for precise estimation, it's crucial to assess their efficiency within appropriate contexts and validate them against lower variance requirements.

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Most popular questions from this chapter

Es seien \(X_{1}, \ldots, X_{n} \quad \mathrm{im}\) Intervall \([0, \theta]\) i.i.d.gleichverteilt. (a) Bestimmen Sie den ML-Schätzer für \(\theta\) und daraus einen erwartungstreuen Schätzer für \(\theta\). (b) Hat der ML-Schätzer oder der erwartungstreue Schätzer den kleineren MSE? (c) Bestimmen Sie ein Konfidenzintervall für \(\theta\) zum Niveau \(1-\alpha\).

Welche der folgenden Aussagen sind richtig? (a) Die Likelihood-Funktion hat stets genau ein Maximum. (b) Für die Likelihood-Funktion \(L(\theta \mid x)\) gilt stets \(0 \leq\) \(L(\theta \mid x) \leq 1 .\) (c) Die Likelihood-Funktion \(L(\theta \mid x)\) kann erst nach Vorlage der Stichprobe berechnet werden.

Bei der Suche nach medizinisch wirksamen Substanzen werden 1000 von Wissenschaftlern gesammelte Pflanzen auf ihre Wirksamkeit getestet. Dabei bedeute \(\mu=0\) Wirkungslosigkeit und \(\mu \neq 0\) potenzielle Wirksamkeit. Das Testniveau sei \(\alpha=10 \%\). Falls alle Pflanzen in Wirklichkeit wirkungslos sind, wie gro \(ß\) ist mit hoher Wahrscheinlichkeit der Anteil der Pflanzen, denen fälschlicherweise Wirksamkeit unterstellt wird: (a) unbekannt. (b) genau \(10 \%\) (c) zwischen 8 und \(12 \%\). Der größte Schaden für das Unternehmen besteht darin, wenn wirksame Pflanzen übersehen werden. Wie können Sie diese Problem durch geeignete Wahl der Hypothesen, des Niveaus und des Stichprobenumfangs lösen?

Der Ausschussanteil in einer laufenden Produktion sei \(\theta\). Es werden unabhängig voneinander zwei einfache Stichproben vom Umfang \(n_{1}\) bzw. \(n_{2}\) gezogen. Dabei seien \(x_{1}\) bzw. \(x_{2}\) schlechte Stücke getroffen worden. \(\theta\) wird jeweils geschätzt durch \(\widehat{\theta}_{(i)}=\frac{x_{i}}{n_{i}}\). Wie lassen sich beide Schätzer kombinieren?

Es sei \(X\) binomialverteilt: \(X \sim B_{n}(\theta)\). Was sind die ML-Schätzer von \(\mathrm{E}(X)\) und \(\operatorname{Var}(X)\) und wie gro \(\beta\) ist der Bias von \(\widehat{\mu}\) und von \(\widehat{\sigma^{2}}\). Warum geht der Bias von \(\widehat{\sigma^{2}}\) nicht mit wachsendem \(n\) gegen 0 ?

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