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Prove that for the family of uniform distribution on the interval \([0, \theta]\), \(\max \left(X_{1}, X_{2}, \ldots, X_{n}\right)\) is the MLE for \(\theta\).

Short Answer

Expert verified
For a family of uniform distributions on the interval [0, \( \theta \)](, )\max \left(X_{1}, X_{2}, \ldots, X_{n} \right) is indeed the Maximum Likelihood Estimator for \( \theta \).

Step by step solution

01

Define the Likelihood Function

The likelihood function \( L(\theta; x) \) for a uniform distribution is the product of the density functions of individual observations (independent samples). In the case of a uniform distribution on the interval [0, \( \theta\)], the density function for each observation is \(1/\theta\). As we have 'n' observations, the likelihood function is \[L(\theta; x) = (1/\theta)^n\] if \(0 ≤ x_i ≤ \theta\) for all \( i = 1, 2, ..., n\), and 0 otherwise.
02

Find the Maximum Likelihood Estimator

In order to maximize likelihood function we derive the natural logarithm of the likelihood function and equate it to zero. But in this specific case this won't be feasible because likelihood function is decreasing with \( \theta \). Instead we notice that if \( \theta \) is less than any observation then likelihood function becomes zero. The maximum \( \theta \) that includes all observations, without making the likelihood function zero is \( \max \left( X_{1}, X_{2}, ..., X_{n} \right) \) Hence the maximum observed value is the MLE.

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Key Concepts

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

Uniform Distribution
Uniform distribution is a type of probability distribution where all outcomes are equally likely. When we say a variable is uniformly distributed in the interval \( [0, \theta] \), it means that the probability of the variable falling within any subinterval of \( [0, \theta] \) is the same. This property makes the uniform distribution quite unique among probability distributions.

In the context of the exercise, the uniform distribution gives us a simple probability density function (PDF), which is constant (\(1/\theta\)) on its interval and zero everywhere else. This characteristic facilitates the process of deriving the maximum likelihood estimation (MLE) because the likelihood function depends on the PDF of the data.
Likelihood Function
The likelihood function is a fundamental concept in statistical inference. It is defined as a function of the parameter(s) of a statistical model, given specific observed data. Essentially, the likelihood function measures how likely it is to obtain the observed data as a function of the parameters.

In our exercise, the likelihood function \(L(\theta; x)\) is obtained by taking the product of the density functions for each data point, assuming that they are independent. The likelihood for the uniform distribution is particularly straightforward, as it is simply the product of \(1/\theta\) for each observation \(x_i\) within the interval \(0 \leq x_i \leq \theta\). Outside this interval, the likelihood is zero since it's impossible for our data to arise from this distribution.
Statistical Inference
Statistical inference involves making conclusions about a population based on data sampled from it. The process typically involves establishing a model for the data, then using observed data to infer the values of parameters of the model. Inferences can be made through various methods, such as hypothesis testing, confidence intervals, and, pertinent to our exercise, estimation.

Estimation itself can take different forms, and maximum likelihood estimation is a powerful method within the inferential toolkit. It seeks the values of the parameters that make the observed data most likely under the statistical model in question. The outcomes of statistical inference allow us to transition from individual sample observations to general statements about a larger population.
Parameter Estimation
Parameter estimation is the process of using data to determine the values of the parameters of a statistical model. The maximum likelihood estimation (MLE) method, which is the focus of our exercise, finds the parameter values that maximize the likelihood function. When we estimate parameters, we're attempting to find the best values that describe the underlying population from which the sample is drawn.

Interestingly, MLE doesn't always require complicated calculus to find a solution. As shown in our exercise, logical reasoning can also lead us to the best estimate for a parameter. In the case of the uniform distribution, the highest value of \(x \) that does not invalidate the likelihood function (i.e., does not turn it into zero) is the MLE for \(\theta\). This would be the maximum value in the sample, supporting the intuitive notion that in a uniform distribution, the upper bound parameter \(\theta\) should be at least as large as the biggest observation.

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Most popular questions from this chapter

Consider repeated observation on a \(m\) -dimensional random variable with mean \(E\left(X_{i}\right)=\mu, i=1,2, \ldots, m, \quad \operatorname{Var}\left(X_{i}\right)=\sigma^{2}, i=1,2, \ldots, m\) and \(\operatorname{Cov}\left(X_{i}, X_{j}\right)=\rho \sigma^{2}, i \neq j .\) Let the \(i\) th observation be \(\left(x_{1 i}, \ldots, x_{m i}\right)\) \(i=1,2, \ldots, n\). Define $$ \begin{array}{c} \bar{X}_{i}=\frac{1}{m} \sum_{j=1}^{m} X_{j i} \\ W_{i}=\sum_{j=1}^{m}\left(X_{j i}-\bar{X}_{i}\right)^{2}, \\ B=m \sum_{i=1}^{n}\left(\bar{X}_{i}-\bar{X}\right)^{2}, \\ W=W_{1}+\cdots+W_{n} . \end{array} $$ where \(B\) is sum of squares between and \(W\) is sum of squares within samples. 1\. Prove (i) \(\left.W \sim(1-\rho) \sigma^{2} \chi^{(} m n-n\right)\) and (ii) \(B \sim(1+(m-1) \rho) \sigma^{2} \chi^{2}(n-1)\). 2\. Suppose \(\frac{(1-\rho) B}{(1+(m-1) \rho) W} \sim F_{(n-1),(m n-n)} .\) Prove that when \(\rho=0, \frac{W}{W+B}\) follows beta distribution with parameters \(\frac{m n-n}{2}\) and \(\frac{n-1}{2}\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from uniform distribution on an interval \((0, \theta)\). Show that \(\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\) is consistent estimator of \(\theta e^{-1}\).

A random variable \(X\) has PDF $$ f(x ; \theta)=\frac{1}{2} e^{-|x-\theta|}, \quad-\infty

If \(X\) and \(Y\) are independent exponential random variable with parameter \(\lambda\), then show that \(\frac{X}{X+Y}\) follows uniform distribution on \((0,1)\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from Poisson distribution with parameter \(\lambda\). Show that \(\alpha \bar{X}+(1-\alpha) s^{2}, 0 \leq \alpha \leq 1\), is a class of unbiased estimators for \lambda. Find the UMVUE for \(\lambda\). Also, find an unbiased estimator for \(e^{-\lambda}\).

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