Question

Consider the situation of the last exercise, but suppose we have the following two independent random samples: (1) \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample with the common pdf \(f_{X}(x)=\theta^{-1} e^{-x / \theta}\), for \(x>0\), zero elsewhere, and \((2) Y_{1}, Y_{2}, \ldots, Y_{n}\) is a random sample with common pdf \(f_{Y}(y)=\theta e^{-\theta y}\), for \(y \geq 0\), zero elsewhere. The last exercise suggests that, for some constant \(c, Z=c \bar{X} / \bar{Y}\) might be an unbiased estimator of \(\theta^{2}\). Find this constant \(c\) and the variance of \(Z\). Hint: Show that \(\bar{X} /\left(\theta^{2} \bar{Y}\right)\) has an \(F\) -distribution.

Short answer

The constant \(c\) was found to be 1. The variance of \(Z\) is \(Var(Z) = c^2 Var(\bar{X} / \bar{Y})\). However, exact value of the variance depends on the degrees of freedom in F-distribution.

Step-by-step solution

Define the terms

Define the random variables \(X\) and \(Y\) and their corresponding Probability Density Functions (PDF). \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample with pdf \(f_{X}(x) = \theta^{-1} e^{-x / \theta}\), \(x>0\), zero elsewhere, and \(Y_{1}, Y_{2}, \ldots, Y_{n}\) is a random sample with common pdf \(f_{Y}(y) = \theta e^{-\theta y}\), for \(y \geq 0\), zero elsewhere.

Find \(E[\bar{X}/\bar{Y}]\)

Calculate \(E[\bar{X}/\bar{Y}]\) as \(E[\bar{X}]\) divided by \(E[\bar{Y}]\). By using the property of expectation \(E[g(X)] = \int g(x)f(x)dx\), we can find the expectation of \(X\) and \(Y\) respectively. \(E[X] = \int_{0}^{\infty} x f_{X}(x) dx = \theta\) and \(E[Y] = \int_{0}^{\infty} y f_{Y}(y) dy = 1/\theta\). So \(E[\bar{X}/\bar{Y}] = E[\bar{X}] / E[\bar{Y}] = \theta / (1/\theta) = \theta^2\).

Find the constant \(c\)

Use the result from Step 2 to find constant \(c\). Given that \(Z=c \bar{X} / \bar{Y}\) is an unbiased estimator of \(\theta^{2}\), so \(E[Z] = \theta^2\). We have, \(E[Z] = c E[\bar{X}/\bar{Y}] = \theta^2\). Hence, \(c = 1\).

Find the variance of \(Z\)

Use the result from step 3 and the formula of the variance of a constant times a random variable to find the variance of \(Z\). This results in \(Var(Z) = c^2 Var(\bar{X} / \bar{Y})\). However, because \(\bar{X} /\left(\theta^{2} \bar{Y}\right)\) has an \(F\) -distribution, it follows that \(Var(\bar{X} /\left(\theta^{2} \bar{Y}\right))\) can be looked up from the table of \(F\) -distribution.

Key concepts

Probability Density Functions

Understanding the concept of probability density functions (PDFs) is central to grasping the nature of continuous random variables in statistical analysis. A PDF describes the likelihood of a random variable taking on a certain value. With continuous random variables, the PDF gives the probability that the variable falls within a certain range, since the probability at any single point is zero due to the infiniteness of real numbers.

In the given exercise, we are presented with two different PDFs. The PDF for the random sample \(X_1, X_2, ..., X_n\) is given by \(f_X(x) = \theta^{-1} e^{-x/\theta}\), which applies to values of \(x > 0\) and is zero elsewhere. This means the likelihood of observing a specific \(x\) decreases exponentially as \(x\) increases, influenced by the parameter \(\theta\).

The second PDF for the random sample \(Y_1, Y_2, ..., Y_n\) is \(f_Y(y) = \theta e^{-\theta y}\) for \(y \geq 0\), and zero elsewhere. This distribution has a different shape, with the likelihood of observing a specific \(y\) value increasing as the parameter \(\theta\) increases, and decreasing exponentially as \(y\) increases.

These PDFs tell us a lot about the behavior of our random variables and inform us how to compute various statistics such as the mean (\(E[X]\) or \(E[Y]\)) by integrating the PDF over all possible values of the variable.

Unbiased Estimator

An unbiased estimator is a statistical term referring to an estimation method for which the expected value of the estimate equals the parameter being estimated. In simpler terms, an unbiased estimator gives us, on average, the true value of the population parameter.

In the problem we're examining, we're tasked with determining whether the statistic \(Z = c \bar{X} / \bar{Y}\) is an unbiased estimator of \(\theta^2\). To qualify as unbiased, the expected value of \(Z\), denoted as \(E[Z]\), must be equal to \(\theta^2\). We discovered that this is indeed true when the constant \(c\) is set to 1.

Steps to Confirm Unbiasedness

• Calculate Expectations: We first find the expected values of \(\bar{X}\) and \(\bar{Y}\), which are \(\theta\) and \(1/\theta\) respectively.
• Division of Expectations: By dividing these expected values, we get the ratio \(\theta^2\) since \(E[\bar{X}/\bar{Y}] = E[\bar{X}] / E[\bar{Y}]\).
• Determine Constant \(c\): With \(E[Z] = c E[\bar{X}/\bar{Y}]\) and knowing that \(E[Z]\) should be \(\theta^2\), we find that \(c\) must be 1 to satisfy the condition for \(Z\) to be unbiased.

By this method, we're able to affirm the property of unbiasedness for \(Z\) when it's calculated with \(c = 1\). The variance of \(Z\), which is another consideration, must also be calculated to fully understand the distribution of our estimator.

F-distribution

The F-distribution is a probability distribution that arises frequently when dealing with ratios of variances and is central to the analysis of variance (ANOVA), among other applications. It is an asymmetrical distribution that is bound to the positive side of the x-axis and is defined by two different degrees of freedom.

In reference to the exercise, we are interested in showing that the statistic \(\bar{X} / (\theta^2 \bar{Y})\) follows an F-distribution. This is significant because if it follows this distribution, we can utilize F-distribution tables to determine probabilities and variances associated with our statistic Z.

The F-distribution has the unique property that, if the null hypothesis is true, the ratio of the two variances will closely follow an F-distribution. This allows statisticians to make inferences about the ratio being significantly different from what would be expected under the null hypothesis. From knowing that \(\bar{X} / (\theta^2 \bar{Y})\) has an F-distribution, we can infer the variance of our statistic \(Z\) that, when \(c=1\), gives us an unbiased estimator of \(\theta^2\).