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Question:Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the uniform distribution on the interval (0, θ), where the value of the parameter θ is unknown; and let\({{\bf{Y}}_{\bf{n}}}{\bf{ = max}}\left( {{{\bf{X}}_{\bf{1}}}{\bf{,}}...{{\bf{X}}_{\bf{n}}}} \right)\). Show that\(\left( {\frac{{\left( {{\bf{n + 1}}} \right)}}{{\bf{n}}}} \right){{\bf{Y}}_{\bf{n}}}\) is an unbiased estimator of θ.

Short Answer

Expert verified

Proved. \(\left( {\frac{{\left( {n + 1} \right)}}{n}} \right){Y_n}\) is an unbiased estimator of \(\theta \)

Step by step solution

01

Given information

\({X_1},...,{X_n}\)are a random sample from a uniform distribution on the interval\(\left[ {0,\theta } \right]\)

\({Y_n} = \max \left( {{X_1},...{X_n}} \right)\)

02

Finding the unbiased estimator

The probability density function of X is,

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

For,\(0 < y < \theta \)

The c.d.f. of\({Y_n}\)is,

\(\begin{aligned}{}F\left( {y\left| \theta \right.} \right) &= P\left( {Y \le y\left| \theta \right.} \right)\\ &= P\left( {{X_1} \le y,...,{X_n} \le y\left| \theta \right.} \right)\\ &= {\left( {\frac{y}{\theta }} \right)^n}\end{aligned}\)

The p.d.f. of\({Y_n}\)is,

\(\begin{aligned}{}f\left( {y\left| \theta \right.} \right) &= \frac{d}{{dy}}F\left( {y\left| \theta \right.} \right)\\ &= \frac{d}{{dy}}{\left( {\frac{y}{\theta }} \right)^n}\\ &= \frac{{n{y^{n - 1}}}}{{{\theta ^n}}}\end{aligned}\)

Then,

\(\begin{aligned}{}{E_\theta }\left( {{Y_n}} \right) &= \int_0^\theta {yf\left( {y\left| \theta \right.} \right)} dy\\ &= \int_0^\theta {y\frac{{n{y^{n - 1}}}}{{{\theta ^n}}}dy} \\ &= \frac{n}{{{\theta ^n}}}\int_0^\theta {{y^n}dy} \end{aligned}\)

\(\begin{aligned}{} &= \frac{n}{{{\theta ^n}}}\left[ {\frac{{{y^{n + 1}}}}{{n + 1}}} \right]_0^\theta \\ &= \frac{n}{{{\theta ^n}}} \times \frac{{{\theta ^{n + 1}}}}{{\left( {n + 1} \right)}}\\ &= \frac{n}{{\left( {n + 1} \right)}}\theta \end{aligned}\)

Hence,

\({E_\theta }\left[ {\left( {\frac{{\left( {n + 1} \right)}}{n}} \right){Y_n}} \right] = \theta \)

This means that,

Hence, \(\left( {\frac{{\left( {n + 1} \right)}}{n}} \right){Y_n}\) is an unbiased estimator of \(\theta \)

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Most popular questions from this chapter

Suppose that a point(X, Y )is to be chosen at random in thexy-plane, whereXandYare independent random variables, and each has the standard normal distribution. If a circle is drawn in thexy-plane with its center at the origin, what is the radius of the smallest circle that can be chosen for there to be a probability of 0.99 that the point(X, Y )will lie inside the circle?

In the situation of Example 8.5.11, suppose that we observe\({{\bf{X}}_{\bf{1}}}{\bf{ = 4}}{\bf{.7}}\;{\bf{and}}\;{{\bf{X}}_{\bf{2}}}{\bf{ = 5}}{\bf{.3}}\).

  1. Find the 50% confidence interval described in Example 8.5.11.
  2. Find the interval of possibleθvalues consistent with the observed data.
  3. Is the 50% confidence interval larger or smaller than the set of possibleθvalues?
  4. Calculate the value of the random variable\({\bf{Z = }}{{\bf{Y}}_{\bf{2}}}{\bf{ - }}{{\bf{Y}}_{\bf{1}}}\)as described in Example 8.5.11.
  5. Use Eq. (8.5.15) to compute the conditional probability that\(\left| {{{{\bf{\bar X}}}_{\bf{2}}}{\bf{ - \theta }}} \right|{\bf{ < 0}}{\bf{.1}}\)givenZ isequal to the value calculated in part (d).

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\({\bf{y = }}\frac{{\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{m}}}}}{{{\bf{1 + }}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{m}}}}}\)

Compare the integral with the p.d.f. of a beta distribution. Alternatively, use Exercise 21 in Sec. 5.7.

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the exponential distribution with unknown parameter β. Show that if n is large, the distribution of the M.L.E. of β will be approximately a normal distribution with mean β and variance\(\frac{{{{\bf{\beta }}^{\bf{2}}}}}{{\bf{n}}}\).

Assume thatX1, . . . , Xnfrom a random sample from the normal distribution with meanμand variance \({\sigma ^2}\). Show that \({\hat \sigma ^2}\)has the gamma distribution with parameters \(\frac{{\left( {n - 1} \right)}}{2}\)and\(\frac{n}{{\left( {2{\sigma ^2}} \right)}}\).

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