Question

Let \(X_1,X_2\) be a random sample from a distribution having the population density function
\[ f(x)=\begin{cases}\dfrac{1}{\theta}&\text{if }0<x<\theta\\0&\text{otherwise,}\end{cases} \]
where \(\theta\in(0,\infty)\). Let \(X_{(2)}=\max\{X_1,X_2\}\) and
\[ \psi(\theta)=P_\theta(X_1+X_2<1),\quad\theta>0. \]
Let \(\delta\big(X_{(2)}\big)\) be an unbiased estimator of \(\psi(\theta)\) that depends on observations \(X_1\) and \(X_2\) only through \(X_{(2)}\). If \(\delta(t)\) is a continuous function on \((0,\infty)\), then the value of \(18\,\delta\!\left(\dfrac{3}{4}\right)\) equals ______ (answer in integer).

Hint:

Differentiate \(\theta^2\psi(\theta)=\int_0^\theta 2t\,\delta(t)\,dt\) with respect to \(\theta\) to recover \(\delta(\theta)\) piece by piece.

Answer 1

Correct Answer: 6

Step 1: Setup.
\(X_1,X_2\) iid Uniform(0,\(\theta\)). Need \(\delta(t)\) with \(E[\delta(X_{(2)})]=\psi(\theta)\).

Step 2: Compute \(\psi(\theta)\) piecewise.
\(\theta\leq1/2\): \(\psi=1\). \(1/2\leq\theta\leq1\): \(\psi(\theta)=1-\frac{(2\theta-1)^2}{2\theta^2}\). \(\theta\geq1\): \(\psi(\theta)=\frac{1}{2\theta^2}\).

Step 3: Density of X(2).
\[ f_{X_{(2)}}(t)=\frac{2t}{\theta^2},\quad 0<t<\theta \]

Step 4: Unbiasedness integral equation.
\[ \int_0^\theta 2t\,\delta(t)\,dt=\theta^2\psi(\theta)=h(\theta) \]
Differentiate: \(\delta(\theta)=h'(\theta)/(2\theta)\).

Step 5: Piecewise delta.
\(\theta\leq1/2\): \(\delta=1\). \(1/2\leq\theta\leq1\): \(\delta(\theta)=\frac{1-\theta}{\theta}\). \(\theta\geq1\): \(\delta=0\). Continuous at joins.

Step 6: Evaluate at 3/4.
\[ \delta(3/4)=\frac{1/4}{3/4}=\frac13 \]

Step 7: Final.
\[ 18\times\frac13=6 \]

Final Answer: \[ \boxed{6} \]