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If \(\mathrm{A}^{\prime}\) is a binomial random variable, show that (a) \(P=X / n\) is an unbiased estimator of \(p ;\) (b) \(P^{\prime}=\frac{X+\sqrt{n} / 2}{n+w}\) is a biased estimator of \(p\).

Short Answer

Expert verified
The estimator \(P = X/n\) is unbiased because its expectation is equal to \(p\). However, the estimator \(P^{\prime} = (X + \sqrt{n}/2) / (n + \sqrt{n})\) is biased as its expectation deviates from \(p\) for \(n ≠ 0\).

Step by step solution

01

Prove that \(P = X / n\) is an unbiased estimator of \(p\)

We know that for a binomial distribution, \[E(X) = np\] where \(E\) is the expectation operator. We can find the expectation of \(P\) by substituting \(X\) by \(Pn\):\[\begin{align*}E(P) &= E(X/n) \ &= E(n \cdot p / n) \ &= E(p) \ &= p\end{align*}\] Therefore, \(P\) is an unbiased estimator of \(p\).
02

Prove that \(P^{\prime} = (X + \sqrt{n}/2) / (n+\sqrt{n})\) is a biased estimator of \(p\)

Applying the expectation operator to \(P^{\prime}\):\[\begin{align*}E(P') &= E \left( \frac{X + \sqrt{n}/2}{n + \sqrt{n}} \right) \ &= E \left( \frac{np + \sqrt{n}/2}{n + \sqrt{n}} \right) \ &= p \cdot \frac{n + \sqrt{n}/2}{n + \sqrt{n}} \ &= p \cdot \frac{1 + \sqrt{1/n}/2}{1 + \sqrt{1/n}} \end{align*}\]Notice that the above expression is clearly different from \(p\) for \(n ≠ 0\). Therefore, \(P'\) is a biased estimator of \(p\)

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Key Concepts

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

Unbiased Estimator
In probability theory and statistics, an estimator is a rule or formula that allows us to make deductions about unknown parameters based on observed data. An unbiased estimator is a special kind of estimator with a unique property: its expected value is equal to the parameter it estimates. Using the example of \(P = X / n\), we're trying to estimate the probability \(p\) of success in a binomial distribution. The estimation procedure is unbiased because the expected value of \(P\), denoted as \(E(P)\), equals \(p\). For the binomial distribution, the expectation of the random variable \(X\) is given by:
  • \(E(X) = np\)
To find \(E(P)\), we evaluate it as follows:
  • \(E(P) = E(X/n) = (1/n) \cdot E(X) = (1/n) \cdot np = p\)
This shows that the average or expected value of \(P\) is indeed \(p\). Hence, \(P\) is an unbiased estimator of \(p\). By relying on unbiased estimators, we ensure that on average, our estimates are accurate.
Biased Estimator
When an estimator's expected value does not equal the parameter it tries to estimate, it is called a biased estimator. The presence of bias means that there is an inherent systematic error in the estimation process. Take, for instance, \(P' = \frac{X + \sqrt{n}/2}{n + \sqrt{n}}\), which is a supposed estimate of \(p\). To determine if it is biased, we compute its expected value:
  • \(E(P') = E \left( \frac{np + \sqrt{n}/2}{n + \sqrt{n}} \right)\)
  • \(= p \cdot \frac{n + \sqrt{n}/2}{n + \sqrt{n}}\)
Simplifying, this does not equal \(p\). The calculation shows that \(E(P')\) results in a value different from \(p\) due to the additional \(\sqrt{n}\) term in both the numerator and the denominator. Thus, \(P'\) is a biased estimator.Bias in an estimator can be intentional for simpler calculations or to reduce variance, but it often requires adjustments to correct the error.
Binomial Distribution
The binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of independent experiments, each with the same probability of success. This makes it particularly useful for scenarios where you have two possible outcomes, such as a coin flip. Parameters of the binomial distribution include:
  • \(n\): The total number of trials
  • \(p\): The probability of success on an individual trial
If \(X\) is a binomial random variable, it represents the number of successes in \(n\) trials with probability \(p\). The outcomes are often modelled in situations like survey responses or natural phenomena. A critical property of the binomial distribution is its expectation and variance, which are given by:
  • Expected value: \(E(X) = np\)
  • Variance: \( ext{Var}(X) = np(1-p)\)
The binomial distribution is foundational in understanding and applying probabilistic models and is crucial in constructing unbiased and biased estimators alike.

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Most popular questions from this chapter

A clinical trial is conducted to determine if a certain type of inoculation has an effect on the incidence of a certain disease. A sample of 1000 rats was kept in a controlled environment for a period of 1 year and 500 of the rats were given the inoculation. Of the group not given the drug, there were 120 incidences of the disease, while 98 of the inoculated group contracted it. If we call \(p_{1}\) the probability of incidence of the disease in uninoculated rats and \(p_{2}\) the probability of incidence after receiving the drug, compute a \(90 \%\) confidence interval for \(p_{1}-p_{2}\).

A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a \(99 \%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b) What can we assert with \(99 \%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23.500 kilometers per year?

A consumer group is interested in comparing operating costs for two different types of automobile engines. The group is able to find 15 owners whose cars have engine type \(A\) and 15 who have engine type B. All 30 owners bought their cars at roughly the same time and all have kept good records for a certain 12 month period. In addition, owners were found that drove roughly the same mileage. The cost statistics are \(y_{A}=\$ 87.00 / 1,000\) miles, \(y_{B}=\$ 75.00 / 1,000\) miles, \(s_{A}=\$ 5.99,\) and \(S B-\$ 4.85 .\) Compute a \(95 \%\) confidence interval to estimate \(\mu_{A}-\mu_{B_{1}}\) the difference in the mean operating costs. Assume normality and equal variance.

An automotive company is considering two types of batteries for its automobile. Sample information on the life of the battery is being used. Twenty batteries of type \(A\) and twenty batteries of type \(B\) are being used. The summary statistics are \(X A=32.91\), \(x_{B}=30.47, s_{A}=1.57,\) and \(s_{B}-1.74 .\) Assume the data on each battery are normally distributed and assume \(\sigma_{A}=\sigma_{B}\) (a) Find a \(95 \%\) confidence interval on \(\mu_{A}-\mu_{B}\). (b) Draw some conclusion from (a) that provides some insight into whether \(A\) or \(B\) should be adopted.

An alternative form of estimation is accomplished through the method of moments. The method involves equating the population mean and variance to the corresponding sample mean \(\overline{\mathcal{T}}\) and sample variance \(\mathrm{s}^{2}\) and solving for the parameter, the result being the moment estimators. In the case of a single parameter, only the means are used. Give an argument that in the case of the Poisson distribution the maximum likelihood estimator and moment estimators are the same.

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