91Ó°ÊÓ

Suppose the true average growth \(\mu\) of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type, the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(E\left(X_{i}\right)=\mu\), \(V\left(X_{i}\right)=\sigma^{2}\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu, V\left(Y_{i}\right)=4 \sigma^{2}\right]\). a. Show that for any \(\delta\) between 0 and 1 , the estimator \(\hat{\mu}=\) \(\delta \bar{X}+(1-\delta) \bar{Y}\) is unbiased for \(\mu\). b. For fixed \(m\) and \(n\), compute \(V(\hat{\mu})\), and then find the value of \(\delta\) that minimizes \(V(\hat{\mu})\). [Hint: Differentiate \(V(\hat{\mu})\) with respect to \(\delta\).]

Step by step solution

01

Understanding the Problem

We need to show that an estimator \( \hat{\mu} = \delta \bar{X} + (1-\delta) \bar{Y} \) is unbiased for \( \mu \) using the given observations of two plant types. Then, we need to find the variance \( V(\hat{\mu}) \) and the value of \( \delta \) that minimizes this variance.

02

Calculate the Expectation of \( \hat{\mu} \)

The expectation of an estimator \( \hat{\mu} = \delta \bar{X} + (1-\delta) \bar{Y} \) is given by \( E(\hat{\mu}) = \delta E(\bar{X}) + (1-\delta) E(\bar{Y}) \). Since \( E(X_i) = \mu \) and \( E(Y_i) = \mu \), we have \( E(\bar{X}) = \mu \) and \( E(\bar{Y}) = \mu \). Thus, \( E(\hat{\mu}) = \delta \mu + (1-\delta) \mu = \mu \). This shows \( \hat{\mu} \) is an unbiased estimator.

03

Calculate the Variance of \( \hat{\mu} \)

The variance of \( \hat{\mu} \) is \( V(\hat{\mu}) = V(\delta \bar{X} + (1-\delta) \bar{Y}) = \delta^2 V(\bar{X}) + (1-\delta)^2 V(\bar{Y}) \). We use \( V(\bar{X}) = \frac{\sigma^2}{m} \) and \( V(\bar{Y}) = \frac{4\sigma^2}{n} \). Substitute these into the variance formula: \( V(\hat{\mu}) = \delta^2 \frac{\sigma^2}{m} + (1-\delta)^2 \frac{4\sigma^2}{n} \).

04

Differentiate \( V(\hat{\mu}) \) with respect to \( \delta \)

To minimize \( V(\hat{\mu}) = \delta^2 \frac{\sigma^2}{m} + (1-\delta)^2 \frac{4\sigma^2}{n} \), differentiate with respect to \( \delta \). We have \( \frac{d}{d\delta} V(\hat{\mu}) = 2\delta \frac{\sigma^2}{m} - 2(1-\delta) \frac{4\sigma^2}{n} \). Set this derivative to zero: \( 2\delta \frac{\sigma^2}{m} = 2(1-\delta) \frac{4\sigma^2}{n} \).

05

Solve for \( \delta \)

Simplify \( \frac{\delta}{m} = \frac{4(1-\delta)}{n} \). Clear fractions: \( n\delta = 4m - 4m\delta \). Rearrange to \( (n + 4m)\delta = 4m \). Thus, \( \delta = \frac{4m}{n + 4m} \).

06

Conclusion

The minimum variance of \( \hat{\mu} \) is obtained when \( \delta = \frac{4m}{n + 4m} \). Therefore, \( V(\hat{\mu}) \) is minimized for this \( \delta \).

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Key Concepts

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

Estimator

An estimator is a statistical rule or formula that gives us an idea of an unknown parameter within a population based on sample data. In this example, we have the estimator \( \hat{\mu} = \delta \bar{X} + (1-\delta) \bar{Y} \). This estimator is attempting to estimate the true average growth, \( \mu \), of two types of plants.

To understand how the estimator works, we calculate its expected value, which is the average result it would produce over many samples. For the estimator \( \hat{\mu} \) to be unbiased, its expected value must equal the true parameter value, \( \mu \). The calculation here showed that \( E(\hat{\mu}) = \mu \), confirming that it's indeed unbiased. This is because the average of both observations, \( \bar{X} \) and \( \bar{Y} \), are themselves unbiased estimators of \( \mu \).

• Estimation is crucial in statistics as it allows predictions based on sample data.
• An unbiased estimator provides results centered on the true parameter on average.

Variance Minimization

To understand variance minimization of an estimator like \( \hat{\mu} \), consider that variance measures the spread or variability of an estimator. Lower variance leads to more precise estimates. For optimal estimation, you want an estimator that not only is unbiased but also has the smallest possible variance. This ensures that repeated estimates give consistently close results.

In this exercise, variance minimization was achieved by finding the right value of \( \delta \) that reduces the variance \( V(\hat{\mu}) \). We determined the variance expression: \[ V(\hat{\mu}) = \delta^2 \frac{\sigma^2}{m} + (1-\delta)^2 \frac{4\sigma^2}{n} \] By setting its derivative with respect to \( \delta \) to zero, we found the value \( \delta = \frac{4m}{n + 4m} \), which minimizes the variance.

• Variance minimization improves the consistency of estimator predictions.
• Calculating variances and derivatives allows one to optimize estimation.

Bias

Bias in the context of statistics refers to the systematic errors that cause estimations to deviate from the true parameter value. In other words, if an estimator consistently overestimates or underestimates the parameter across different samples, it is a biased estimator.

In this context, ensuring that \( \hat{\mu} \) is unbiased is crucial. It's calculated to confirm that the expectation of our estimator is precisely the true parameter \( \mu \). Hence, the estimator \( \hat{\mu} \) offers a reliable way to assess the central tendency of plant growth, as we're confident that any observations are a reflection of actual average behavior and not a result of systematic bias.

• An unbiased estimator produces estimates that are on target, statistically.
• Bias reduction is essential for improving the accuracy of statistical predictions.