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: Rao-Blackwell approach.
$\delta(X_{(2)})=E[I(X_1+X_2<1)\mid X_{(2)}]$ stays unbiased and depends only on $X_{(2)}$.

Step 2: Conditional law of min given max.
Given $X_{(2)}=v$, $X_{(1)}\sim$Uniform$(0,v)$.

Step 3: Write delta(t).
\[ \delta(t)=P(X_{(1)}<1-t\mid X_{(2)}=t) \]

Step 4: Cases.
$t\geq1$: $\delta=0$. $t\leq1/2$: $\delta=1$. $1/2<t<1$: $\delta(t)=\frac{1-t}{t}$.

Step 5: Continuity check.
Matches at boundaries $t=1/2,1$.

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

Final Answer: \[ 18\times\frac13=6 \]
\[ \boxed{6} \]