Question

Given \(f(x ; \theta)=1 / \theta, 00\), formally compute the reciprocal of $$n E\left\\{\left[\frac{\partial \ln f(X: \theta)}{\partial \theta}\right]^{2}\right\\}$$ Compare this with the variance of \((n+1) Y_{n} / n\), where \(Y_{n}\) is the largest observation of a random sample of size \(n\) from this distribution. Comment.

Short answer

The reciprocal of \(nE\left\{\left[\frac{\partial \ln f(X; \theta)}{\partial \theta}\right]^2\right\}\) is \(2\theta^2 / n\) and the variance of \((n+1) Y_{n} / n\) is \((n+1)^2 * \frac{\theta^2}{4n^2} = \frac{(n+1)^2}{4n}\). Comparing these two, shows that they are not equal indicating that there is information loss when using \(Y_{n}\) as a statistic in estimation of \(\theta\).

Step-by-step solution

Compute the derivative of the natural logarithm of f(x; \(\theta\)).

First, differentiate the natural logarithm of \(f(x ; \theta)\) with respect to \(\theta\), \(\frac{\partial}{\partial\theta} \ln f(x ; \theta)= -\frac{1}{\theta}\).

Square the derivative and find its expectation

Now, square the derivative: \(\left[-\frac{1}{\theta}\right]^2 = \frac{1}{\theta^2}\). Then find its expectation by integrating \(f(x;\theta) * [\frac{1}{\theta^2}] = \frac{1}{\theta^3}\) over the interval [0,\(\theta\)]. This yields \(E\left\{\left[\frac{d \ln f(X: \theta)}{d \theta}\right]^2\right} = \frac{1}{2\theta^2}\). So, the reciprocal of \(n * \frac{1}{2\theta^2}\) is equal to \(2\theta^2 / n\).

Find variance of \((n+1) Y_{n} / n\)

Let us derive the variance of \((n+1) Y_{n} / n\). Here, since \(Y_n\) follows a uniform distribution [0, \(\theta\)], it has variance of \(\frac{\theta ^2}{4n^2}\). Thus, the variance of \((n+1) Y_{n} / n\) is \((n+1)^2 * \frac{\theta^2}{4n^2} = \frac{(n+1)^2}{4n}\)

Comparison and Comment

Comparing the results from Step 2 and 3, we realize that \(\frac{2\theta^2}{n}\) is not equal to \(\frac{(n+1)^2}{4n}\). This states that the reciprocal of the expectation of the squared derivative of \(\ln f(X: \theta)\) is not equal to the variance of \((n+1) Y_{n} / n\). This indicates that there is some amount of information loss when using \(Y_{n}\) as a statistic in estimation of \(\theta\) when samples are drawn from this uniform distribution.

Key concepts

Expectation of a function of a random variable

When studying probability and statistics, understanding the expectation of a function of a random variable is crucial. The expectation, or expected value, is the sum of the probability-weighted outcomes of a random variable. Mathematically, if we have a random variable, X, and a function g(X), then the expectation of g(X), denoted as E[g(X)], is calculated by integrating or summing the product of g(X) with the probability density function (pdf) or probability mass function (pmf) of X.

For instance, if X is continuous with a pdf f(x), E[g(X)] is found by \[\begin{equation} \int_{-\b{x}}^{+\b{x}} g(x)f(x)dx \end{equation}\]\
. This computation is central to numerous statistical methods because it helps to predict the average outcome of a process described by the function g(X) when subjected to the variability of the random variable X.

In the context of the exercise, the expectation of the squared derivative of the natural logarithm of f(X; \(\b{\theta}\)), E\left\{\left[\frac{\partial ln f(X: \(\b{\theta}\))}{\partial\(\b{\theta}\)}\right]^{2}\right\}, is found by multiplying this squared derivative by the pdf and integrating over the range of X. This concept is pivotal in estimating the efficiency and accuracy of statistical estimators.

Variance of a statistic

Variance measures the spread or dispersion of a set of data points or the values of a random variable around its mean. When we talk about the 'variance of a statistic', we're referring to how much that statistical measure is expected to vary from one sample of data to another. A statistic here is any function of the data, such as the mean, median, or in our scenario, \((n+1) Y_{n} / n\), where \(Y_{n}\) is the maximum of a uniform random sample.

Calculating variance involves squaring the differences between each data point and the mean, then taking the average of these squared differences. This can be expressed in a formula: \[\begin{equation}\text{Var}(X) = E[(X - E[X])^2]\end{equation}\]\br>, where E[X] is the expectation of the random variable X. In the context of a uniform distribution, the formula is slightly different, but the underlying concept of measuring spread remains the same.

Understanding the variance of a statistic is essential in data analysis. It helps determine the reliability of an estimate: a smaller variance indicates that the estimate is more consistently close to the mean, implying better precision.

Derivatives in statistics

In statistics, derivatives provide a way to assess the sensitivity of one variable in relation to another. They're particularly useful in maximum likelihood estimation, where we need to know how the likelihood function changes with respect to the parameters of a model.

Mathematically, this involves taking the derivative of the likelihood function with respect to these parameters. For instance, in the given exercise, differentiating the natural logarithm of the probability density function with respect to \(\theta\) yields the rate of change of the log-likelihood as \(\theta\) varies. This derivative is a cornerstone in deriving estimators and understanding their properties, such as bias and efficiency.

Furthermore, the second derivative, or the curvature of the likelihood function, is related to the confidence intervals and standard errors of the estimators. Derivatives hence play a significant role in statistical inference and hypothesis testing.

Uniform distribution properties

The uniform distribution is one of the simplest probability distributions in statistics, where all intervals of the same length within the distribution's range are equally probable. There are two types: continuous and discrete. In continuous uniform distribution, the probability density function (pdf) is constant between two bounds, a and b, which is the case in our exercise.

Some important properties of the continuous uniform distribution are:

• The mean is \(\frac{a+b}{2}\).
• The variance is \(\frac{(b-a)^2}{12}\), showcasing how the data points are distributed around the mean.
• It is symmetrical, so the mean and median are equal.

These properties are essential in understanding how uniform distribution behaves and assists in calculating probabilities, expected values, variances, and other statistical measures.

In our exercise, knowing that \(Y_{n}\) follows a uniform distribution over the interval [0, \(\theta\)], helps in deriving its variance and informs us about its behavior as a statistic within the given context.