Question

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

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

Step-by-step solution

Given information

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

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

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 \(\theta \)