Question

Let X1. . . . Xn be i.i.d Uniform over the interval (θ, θ + 1].Show that X(1)+X(n) )/2- 1/2 is also an unbiased estimator of θ, whereX(1) is the minimum order statistic and X(n) is the maximum order statistic. If X - 1/2 is also an unbiased estimator of θ which of the two estimators would you prefer to use.

Answer 1

Answer:-

Given That:-

Let X1. . . . Xn be i.i.d Uniform over the interval (θ, θ + 1].Show that X(1)+X(n) )/2- 1/2 is also an unbiased estimator of θ, whereX(1) is the minimum order statistic and X(n) is the maximum order statistic. If X - 1/2 is also an unbiased estimator of θ which of the two estimators would you prefer to use.

$X_{1}, X_{2}, ...X_{n}\sim U(\theta, \theta+1)$

Let

,   

X;~ U(0,6 +1)

$\theta < X_{i} < \theta + 1$

$\Rightarrow \theta < X_{i} - \theta < 1$

X; -O-U(0,0+1)=0 < ; -A<1

Put

$Y_{i} = X_{i} - \theta$

then

$Y_{i} \sim U(0, 1),$

Result:-

If   are order statics of  , $y_{i} \sim U(0, 1)$

then r th order statistic  

Y(r) beta (r, n-r+1)

Therefore  

Y(r) beta (r, n-r+1)

Therefore $y_{(1)}\sim \beta (1, n)$

$y_{(n)}\sim \beta (n, 1)$

If $X \sim beta\, \, \, (\alpha, \beta)$

$E(X) = \frac{\alpha}{\alpha + \beta}$

$Var(X) = \frac{\alpha \beta}{(\alpha + \beta)^{2}(\alpha + \beta + 1)}$

$\therefore E(y_{(1)}) = \frac{1}{n + 1}$

Put

$y_{(1)} = X_{(1)} - \theta$

$\Rightarrow E(X_{(1)} - \theta) = \frac{1}{n + 1}$

= E(X(1) = n +1

$E(y_{(n)}) = \frac{n}{n+1} \Rightarrow E(X_{(n)} - \theta) = \frac{n}{n+1}$

$\Rightarrow E(X_{(n)}) = \frac{n}{n+1} + \theta$

$\Rightarrow \frac{X_{(1)} + X_{(n)}}{2} - \frac{1}{2}$

$\Rightarrow E\left ( \frac{X_{(1)} + X_{(n)}}{2} - \frac{1}{2} \right ) = \frac{1}{2} \left [ E(X_{(1)} + E(X_{(n)})) \right ] - \frac{1}{2}$

$= \frac{1}{2}\left [ \frac{1}{n+1} + \theta + \frac{n}{n+1}+\theta \right ] - \left [ \frac{1}{2} \right ]$

$\Rightarrow \frac{1}{2}\left [ \frac{1}{n+1} [n+1] + 2\theta \right ] - \left [ \frac{1}{2} \right ]$

$\Rightarrow \frac{1}{2} + \theta - \frac{1}{2} = \theta$

$(1) + X(n) 1
is an unbaised estimation for $\theta$

$X\sim U(\theta, \theta + 1)$

$E(X) = \frac{\theta + \theta + 1}{2}$

$= \frac{2\theta + 1}{2}$

$E(X - \frac{1}{2})= \frac{2\theta + 1}{2} = \theta$

From both of these Statistic will choose the statistic which have minimum variance.

$\therefore Var\left ( \frac{X_{(1)} + X_{(n)}}{2} - \frac{1}{2} \right ) = \frac{1}{4}\left [ Var(X_{(1)}) + Var(X_{(n)}) \right ]$

  $= \frac{1}{4}\left [ \frac{n}{(n + 1)^{2}(n + 2)} + \frac{n}{(n + 1)^{2}(n + 2)} \right ]$

$= \frac{1}{2}\left ( \frac{n}{(n+1)^{2}(n + 2)} \right )$

$\therefore Var(X - \frac{1}{2}) = \frac{1}{12}$

$\therefore Var\left ( \frac{X_{(1)} + X_{(n)}}{2} - \frac{1}{2} \right )$ have minimum variance will prefer these estimator.

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