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Suppose that\({X_1}...{X_n}\)form a random sample from the normal distribution with mean 0 and unknown standard deviation\(\sigma > 0\). Find the lower bound specified by the information inequality for the variance of any unbiased estimator of\(\log \sigma \).

Short Answer

Expert verified

Lower bound specified by the information inequality for the variance of any biased estimator of \(\log \sigma \) is \(\frac{1}{{2n}}\)

Step by step solution

01

Given information

\({X_1}...{X_n}\) form a random sample form a normal distribution with unknown mean and known variance \(\sigma > 0\)

02

Fisher information 

Fisher information \(I\left( \theta \right)\) in the random variable X is defined as

\(I\left( \theta \right) = {E_\theta }\left\{ {{{\left( {{\lambda ^{'}}\left( {x|\theta } \right)} \right)}^{2}}} \right\}\)

Or

\(I\left( \theta \right) = - {E_\theta }\left\{ {\left( {{\lambda ^{''}}\left( {x|\theta } \right)} \right)} \right\}\)

03

calculating the lower bound specified by the information inequality for the variance of any unbiased estimator of \(\log \sigma \)

Let \(m\left( \sigma \right) = \log \sigma \)

Then

\({m^{'}}\left( \sigma \right) = \frac{1}{\sigma }\)and\({\left( {{m^{'}}\left( \sigma \right)} \right)^{2}} = \frac{1}{{{\sigma ^2}}}\)

First calculate the fisher information\(I\left( \sigma \right)\)in X as follows:

Normal distribution with parameter\(\mu = 0\)and\({\sigma ^{2}}\)is

\(\begin{align}f\left( {x|\mu ,{\sigma ^{2}}} \right) &= \frac{1}{{\sqrt {2\pi \sigma } }}\exp \left( { - \frac{{{x^2}}}{{2{\sigma ^2}}}} \right)\\\lambda \left( {x|\sigma } \right) = \log f\left( {x|\sigma } \right)\\ &= \log \left( {\frac{1}{{\sqrt {2\pi \sigma } }}\exp \left( { - \frac{{{x^2}}}{{2{\sigma ^2}}}} \right)} \right)\end{align}\)

\(\lambda \left( {x|\sigma } \right) = - \log \sigma - \frac{{{x^2}}}{{2{\sigma ^2}}} + const\)

Differentiate \(\lambda \left( {x|\sigma } \right)\) with respect to \(\sigma \)

\(\begin{align}{\lambda ^{'}}\left( {x|\sigma } \right) &= \frac{\partial }{{\partial \sigma }}\left( { - \log \sigma - \frac{{{x^2}}}{{2{\sigma ^2}}} + const} \right)\\ &= - \frac{1}{\sigma } + \frac{{{x^2}}}{{{\sigma ^3}}}\end{align}\)

Again differentiate \({\lambda ^{'}}\left( {x|\sigma } \right)\) with respect to \(\sigma \)

\(\begin{align}{\lambda ^{''}}\left( {x|\sigma } \right) &= \frac{\partial }{{\partial \sigma }}\left( { - \frac{1}{\sigma } + \frac{{{x^2}}}{{{\sigma ^3}}}} \right)\\ &= \frac{1}{{{\sigma ^2}}} + \frac{{3{x^2}}}{{{\sigma ^4}}}\end{align}\)

Hence from the equation of fisher information \(I\left( \sigma \right)\) in X is

\(\begin{align}I\left( \sigma \right) &= - {E_\sigma }\left( {{\lambda ^{''}}\left( {x|\sigma } \right)} \right)\\ &= - {E_\sigma }\left( {\frac{1}{{{\sigma ^2}}} + \frac{{3{x^2}}}{{{\sigma ^4}}}} \right)\\ &= \frac{1}{{{\sigma ^2}}} - \frac{{3E\left( {{x^2}} \right)}}{{{\sigma ^4}}}\end{align}\)

\(\begin{align}I\left( \sigma \right) &= \frac{1}{{{\sigma ^2}}} - \frac{{3\left( {{\sigma ^2}} - 0 \right)}}{{{\sigma ^4}}}\\ &= \frac{2}{{{\sigma ^2}}}\end{align}\)

Hence if T is an unbiased estimator of \(\log \sigma \), it follows from equation of cramer rao inequality that is

\(\begin{align}Var\left( T \right) \ge \frac{{{{\left( {{m^{'}}\left( \theta \right)} \right)}^{2}}}}{{nI\left( \theta \right)}}\\ &= \frac{1}{{{\sigma ^{2}}}} \times \frac{{{\sigma ^{2}}}}{{2n}}\\ &= \frac{1}{{2n}}\end{align}\)

Therefore, lower bound specified by the information inequality for the variance of any biased estimator of \(\log \sigma \) is \(\frac{1}{{2n}}\)

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Most popular questions from this chapter

For the conditions of Exercise 5, use the central limit theorem in Sec. 6.3 to find approximately the size of a random sample that must be taken so that \(P\left( {{\bf{|}}{{{\bf{\bar X}}}_{\bf{n}}} - {\bf{p|}}} \right) \ge 0.95\) whenp=0.2.

Prove that the distribution of\({\hat \sigma _0}^2\)in Examples 8.2.1and 8.2.2 is the gamma distribution with parameters\(\frac{n}{2}\)and\(\frac{n}{{2{\sigma ^2}}}\).

Suppose that \({X_1},...,{X_n}\) form a random sample from the exponential distribution with unknown mean μ. Describe a method for constructing a confidence interval for μ with a specified confidence coefficient \(\gamma \left( {0 < \gamma < 1} \right)\)

Question: Suppose a random variable X has the Poisson distribution with an unknown mean \({\bf{\lambda }}\) (\({\bf{\lambda }}\)>0). Find a statistic \({\bf{\delta }}\left( {\bf{X}} \right)\) that will be an unbiased estimator of \({{\bf{e}}^{\bf{\lambda }}}\).Hint: If \({\bf{E}}\left( {{\bf{\delta }}\left( {\bf{X}} \right)} \right){\bf{ = }}{{\bf{e}}^{\bf{\lambda }}}\) , then \(\sum\limits_{{\bf{x = 0}}}^\infty {\frac{{{\bf{\delta }}\left( {\bf{x}} \right){{\bf{e}}^{{\bf{ - \lambda }}}}{{\bf{\lambda }}^{\bf{x}}}}}{{{\bf{x!}}}}} = {{\bf{e}}^{\bf{\lambda }}}\)

Multiply both sides of this equation by \({{\bf{e}}^{\bf{\lambda }}}\)expanding the right side in a power series in \({\bf{\lambda }}\), and then equate the coefficients of \({{\bf{\lambda }}^{\bf{x}}}\) on both sides of the equation for x = 0, 1, 2, . . ..

In the situation of Exercise 9, suppose that a prior distribution is used forθwith p.d.f.ξ(θ)=0.1 exp(−0.1θ)forθ >0. (This is the exponential distribution with parameter 0.1.)

  1. Prove that the posterior p.d.f. ofθgiven the data observed in Exercise 9 is

\({\bf{\xi }}\left( {{\bf{\theta }}\left| {\bf{x}} \right.} \right){\bf{ = }}\left\{ \begin{align}{l}{\bf{4}}{\bf{.122exp}}\left( {{\bf{ - 0}}{\bf{.1\theta }}} \right)\;{\bf{if}}\;{\bf{4}}{\bf{.8 < \theta < 5}}{\bf{.2}}\\{\bf{0}}\;{\bf{otherwise}}\end{align} \right.\)

  1. Calculate the posterior probability \(\left| {{\bf{\theta - }}{{{\bf{\bar X}}}_{\bf{2}}}} \right|{\bf{ < 0}}{\bf{.1}}\), which \({{\bf{\bar X}}_{\bf{2}}}\)is the observed average of the datavalues.
  2. Calculate the posterior probability thatθis in the confidence interval found in part (a) of Exercise 9.
  3. Can you explain why the answer to part (b) is so close to the answer to part (e) of Exercise 9? Hint:Compare the posterior p.d.f. in part (a) to the function in Eq. (8.5.15).
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