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Chapter 7: Q16E (page 442)

Question: Suppose that each of two statisticians A and B mustestimate a certain parameter \({\bf{\theta }}\) whose value is unknown(\({\bf{\theta }}\)> 0). Statistician A can observe the value of a randomvariable X, which has the gamma distribution with parameters\({\bf{\alpha }}\,\,{\bf{and}}\,\,{\bf{\beta }}\), where\({\bf{\alpha = 3}}\,\,{\bf{and}}\,\,{\bf{\beta = \theta }}\); statistician Bcan observe the value of a random variable Y, which hasthe Poisson distribution with mean \({\bf{\theta }}\). Suppose that thevalue observed by statistician A is X = 2 and the value observedby statistician B is Y = 3. Show that the likelihoodfunctions determined by these observed values are proportional,and find the common value of the M.L.E. of \({\bf{\theta }}\)obtained by each statistician.

Short Answer

Expert verified

\(\widehat \theta = \frac{3}{2}\)

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Step by step solution

01

Given information

Here, statistician A observes a random variable X which has a gamma distribution with parameters that is,\(X \sim {\rm{Gamma}}\left( {\alpha = 3\,,\,\,\beta = \theta } \right)\).

Statistician B observes a random variable Y which has a Poisson distribution with parameters that is,\(Y \sim P\left( \theta \right)\).

02

Calculate the likelihood of variable X

\(\begin{array}{c}X \sim Gamma\left( {\alpha = 3\,,\,\,\beta = \theta } \right)\\f\left( x \right) = \frac{{{\beta ^\alpha }}}{{\Gamma \alpha }}{x^{\alpha - 1}}{e^{ - \beta x}},x > 0\\ = \frac{{{\theta ^3}}}{{\Gamma 3}}{x^2}{e^{ - \theta x}}\end{array}\)

Applying log to both sides,

\(\ln f\left( x \right) = 3\ln \theta + 2\ln x - x\theta \ln e - \ln \Gamma 3\)

Differentiating both sides and equating it to 0, to obtain value of parameter such that it gives M.L.E.

\(\begin{array}{c}\frac{\partial }{{\partial \theta }}\ln f\left( x \right) = \frac{\partial }{{\partial \theta }}\left( {3\ln \theta + 2\ln x - x\theta \ln e - \ln \Gamma 3} \right)\\ = \frac{3}{\theta } - x\end{array}\)

Equating with 0,

\(\frac{3}{\theta } - x = 0\)

As for value of x=2 is given,

\(\begin{array}{l}\frac{3}{\theta } - 2 = 0\\\frac{3}{\theta } = 2\\\theta = \frac{3}{2}\end{array}\)

Therefore, \(\theta = \frac{3}{2}\)

03

Calculate the likelihood of variable Y

\(\begin{array}{l}Y \sim P\left( \theta \right)\\p\left( y \right) = \frac{{{e^{ - \theta }} \times {\theta ^y}}}{{y!}}\end{array}\)

Applying log to both sides,

\(\ln p\left( y \right) = - \theta \ln e + y\ln \theta - \ln y!\)

Differentiating both sides and equating it to 0, to obtain value of parameter such that it gives M.L.E.

\(\begin{array}{c}\frac{\partial }{{\partial \theta }}\ln p\left( y \right) = \frac{\partial }{{\partial \theta }}\left( { - \theta \ln e + y\ln \theta - \ln y!} \right)\\ = - 1 + \frac{y}{\theta }\end{array}\)

Equating with 0,

\(\begin{array}{c} - 1 + \frac{y}{\theta } = 0\\\frac{y}{\theta } = 1\end{array}\)

As for value of y=3 is given,

\(\begin{array}{l} \Rightarrow \frac{3}{\theta } = 1\\ \Rightarrow \theta = 3\end{array}\)

Therefore, \(\theta = 3\)

04

Find common value

Since the two values of \(\theta \) obtained are,

\(\theta = \frac{3}{2}\)and\(\theta = 3\)

The common value is

\(\begin{array}{c}\widehat \theta = \min \left( {\frac{3}{2},3} \right)\\ = \frac{3}{2}\end{array}\)

Therefore, the answer is \(\widehat \theta = \frac{3}{2}\)

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Most popular questions from this chapter

Suppose that the number of defects in a 1200-foot roll of magnetic recording tape has a Poisson distribution for which the value of the mean 胃 is unknown and that the prior distribution of 胃 is the gamma distribution with parameters 伪 = 3 and 尾 = 1. When five rolls of this tape are selected at random and inspected, the numbers of defects found on the rolls are 2, 2, 6, 0, and 3. Determine the posterior distribution of 胃.

Question: Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\) form a random sample from the uniform distribution on the interval [0, 胃], where the value of the parameter 胃 is unknown. Suppose also that the prior distribution of 胃 is the Pareto distribution with parameters \({{\bf{x}}_{\bf{0}}}\) and 伪 (\({{\bf{x}}_{\bf{0}}}\)> 0 and 伪 > 0), as defined in Exercise 16 of Sec. 5.7. If the value of 胃 is to be estimated by using the squared error loss function, what is the Bayes estimator of 胃? (See Exercise 18 of Sec. 7.3.)

The Pareto distribution with parameters\({{\bf{x}}_{\bf{0}}}\)and伪\(\left( {{{\bf{x}}_{\bf{0}}}{\bf{ > 0}}\;{\bf{and}}\;{\bf{\alpha > 0}}} \right)\)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, 胃), where the value of the endpoint胃is unknown.

Question: Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\) form a random sample from the Bernoulli distribution with parameter 胃, which is unknown, but it is known that 胃 lies in the open interval 0 <胃< 1. Show that the M.L.E. of 胃 does not exist if every observed value is 0 or if every observed value is 1.

Suppose that the number of defects on a roll of magnetic recording tape has a Poisson distribution for which the mean \(\lambda \)is either 1.0 or 1.5, and the prior p.f. of \(\lambda \) is as follows:

\(\xi \left( {1.0} \right) = 0.4\)and\(\xi \left( {1.5} \right) = 0.6\)

If a roll of tape selected at random is found to have three defects, what is the posterior p.f. of \(\lambda \) ?
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