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Suppose that \({x_1},...,{x_n}\) from a random sample from an exponential distribution with parameter\(\theta \).Explain how to use the parametric bootstrap to estimate the variance of the sample average\(\overline X \).(No simulation is required.)

Short Answer

Expert verified

The first step would be estimating parameters\(\theta \) using a maximum likelihood estimator.

M.L.E:

Let random variables\({x_1},{x_2},...,{x_n}\) have joint pdf or pmb

\(f\left( {{x_1},{x_2},...,{x_n};{\theta _1},{\theta _2},...,{\theta _m}} \right)\)

\(\frac{{{{\overline X }^2}}}{n}\)

Step by step solution

01

definition of the random variable is a variable  

A random variableis a factor with an undetermined number or a program that gives numbers to each study's results.

The first step of estimating parameters\(\theta \) using the maximum is the estimator.

M.L.E:

Let random variables\({x_1},{x_2},...,{x_n}\) have joint pdf or pmb

\(f\left( {{x_1},{x_2},...,{x_n};{\theta _1},{\theta _2},...,{\theta _m}} \right)\)

Let random variables have joint pdf or pmb

where the parameters \({\theta _{i,}}i = 1,2,...,m\) are unknown. When function f is a function of parameters\({\theta _{i,}}i = 1,2,...,m\) , it is called the likelihood function. Values\(\widehat {{\theta _i}}\) that maximize the likelihood function are the maximum likelihood estimates or equally values \(\widehat {{\theta _i}}\)for

\(f\left( {{x_1},{x_2},...,{x_n};\widehat {{\theta _1}},{{\widehat \theta }_2},...,{\theta _m}} \right) \ge f\left( {{x_1},{x_2},...,{x_n};{\theta _1},{\theta _2},...,{\theta _m}} \right)\)

For every\({\theta _{i,}}i = 1,2,...,m\) . By substituting\({X_i}\,\,\) with\({x_i}\) , the maximum likelihood estimators are obtained.

The sampling is from the Exponential Distribution function is

\(f\left( {{x_1},{x_2},...,{x_n};\beta } \right) = {\beta ^n}\,{e^{\beta y}},\)

\(y = \sum\limits_{i = 1}^n {\,{x_i}} \)

By maximizing the log of the likelihood function, the maximum likelihood estimates are more accessible to

\(\begin{aligned}{l}L\left( \beta \right) &= \log \,ff\left( {{x_1},{x_2},...,{x_n};\alpha ,\beta } \right)\\ &= n\,\log \beta - \beta y\end{aligned}\)

02

Partial derivative

From which, by finding the partial derivative and equating it with zero, the maximum is obtained.

\(\frac{\partial }{{\partial \beta }}L\left( \beta \right) = n\frac{1}{\beta } - y\)

which yields

\(\widehat \beta = \frac{n}{y} = \frac{1}{{{{\overline x }_n}}}\)

The M.L. estimator

\(\widehat \beta = \frac{1}{{{{\overline x }_n}}} = \widehat {\theta .}\)

The distribution \(\widehat F\)in the bootstrap would be the exponential distribution with parameter \(1/\overline X \). To use bootstrap, let the simulation size be from the distribution\(\widehat F\) .

The variance of the sample average V of simulations would be the bootstrap estimate of the variance of \(\overline X \) .

The variance of a single observation from the distribution\(\widehat F\) is the variance of\({\overline X ^2}\) the sample average of V simulations is \(1/n\) times larger, and thus the bootstrap estimate is

\(\frac{{{{\overline X }^2}}}{n}\)

Hence,

\(\frac{{{{\overline X }^2}}}{n}\)

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Most popular questions from this chapter

In this problem, we shall outline a Bayesian solution to the problem described in Example 7.5.10 on page 423. Let \(\tau \)= 1/ฯƒ2 and use a proper normal-gamma prior to the form described in Sec. 8.6. In addition to the two parameters, ฮผ and \(\tau \), introduce n additional parameters. For i = 1, n, let Yi = 1 if Xi came from the normal distribution with mean ฮผ and precision \(\tau \), and let Yi = 0 if Xi came from the standard normal distribution

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Let \({\bf{f}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{ = cg}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) be a joint p.d.f for \(\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right){\bf{,}}\)each \({x_{2\,}}\)let\({{\bf{h}}_{{\bf{2}}\,}}\left( {{{\bf{x}}_{{\bf{1}}\,}}} \right){\bf{ = g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) that is what we get by considering \({\bf{g}}\left( {{{\bf{x}}_{{\bf{1}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\) as a function of \({{\bf{x}}_{{\bf{1}}\,}}\)for fixed \({{\bf{x}}_{2\,}}\)show that there is a multiplicative factor \({{\bf{c}}_{{\bf{2}}\,}}\)that does not depend on such that is the conditional p.d.f of \({{\bf{x}}_{{\bf{1}}\,}}\) given \(\left( {{{\bf{x}}_{{\bf{2}}\,}}{\bf{,}}{{\bf{x}}_{{\bf{2}}\,}}} \right)\)

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