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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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