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Show that the family of uniform distributions on the intervals \([{\bf{0}},{\bf{\theta }}]\) for \({\bf{\theta }} > {\bf{0}}\) is not an exponential family as defined in Exercise 23. Hint: Look at the support of each uniform distribution.

Short Answer

Expert verified

A uniform distribution does not belong to an exponential family.

Step by step solution

01

Given information

It is given that a random variable X follows uniform distribution on the interval\([0,\theta ]\).

Therefore, the pdf is,

\(f\left( x \right) = \frac{1}{\theta },0 < x < \theta \)

02

Define an exponential distribution

A class of families belonging to exponential distribution is defined as follows:

The parameter space is\(\left\{ {{P_\theta }:\theta \in \Theta } \right\},\Theta \in {R^k}\), the real valued functions exists if:

\({p_\theta }\left( x \right) = \exp \left[ {\sum\limits_i^k {{n_i}\left( \theta \right){T_i}\left( x \right) - B\left( \theta \right)} } \right]h\left( x \right)\) , where x denotes the sample space.

03

Checking the uniform distribution’s family

Since the uniform distribution has an indicator function, where\(0 < x < \theta \). This cannot be expressed inside the exponential product part that is, \(\exp \left[ {\sum\limits_i^k {{n_i}\left( \theta \right){T_i}\left( x \right)} } \right]\).

Therefore, since the support of x is the indicator function, which cannot be expressed as the exponential family, therefore it has been proved.

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Most popular questions from this chapter

The Pareto distribution with parameters\({{\bf{x}}_{\bf{0}}}\)andα\(\left( {{{\bf{x}}_{\bf{0}}}{\bf{ > 0}}\;{\bf{and}}\;{\bf{\alpha > 0}}} \right)\)is defined in Exercise 16 of Sec. 5.7.Show that the family of Pareto distributions is a conjugate family of prior distributions for samples from a uniformdistribution on the interval (0, θ), where the value of the endpointθis unknown.

Assume that \({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\) form a random sample from a distribution that belongs to an exponential family of distributions as defined in Exercise 23 of Sec. 7.3. Prove that\({\bf{T = }}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{\bf{d}}\left( {{{\bf{X}}_{\bf{i}}}} \right)} \) is a sufficient statistic for θ.

Question: Consider the conditions of Exercise 2 again. Suppose that the prior distribution of θ is as given in Exercise 2, and suppose again that 20 items are selected at random from the shipment.

a. For what number of defective items in the sample will the mean squared error of the Bayes estimate be a maximum?

b. For what number the mean squared error of the Bayes estimate will be a minimum?

Suppose that \({X_1},...,{X_n}\) form a random sample from a beta distribution for which both parameters α and β are unknown. Show that the M.L.E.’s of α and β satisfy the following equation:

\(\frac{{\Gamma '\left( \alpha \right)}}{{\Gamma \left( \alpha \right)}} - \frac{{\Gamma '\left( \beta \right)}}{{\Gamma \left( \beta \right)}} = \frac{1}{n}\sum\limits_{i = 1}^n {\log \frac{{{X_i}}}{{1 - {X_i}}}} \)

The uniform distribution on the interval [a, b], where the value of a is known and the value of b is unknown\(\left( {b > a} \right)\):\(T = \min \left\{ {{X_1},...{X_n}} \right\}\)

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