Question:
Let \(X\) be a single observation from a distribution having probability density function
\[
f_\theta(x)=\begin{cases}\dfrac{2x}{\theta^2}&\text{if }0<x<\theta\\0&\text{otherwise,}\end{cases}
\]
where \(\theta\in(0,\infty)\). For testing \(H_0:\theta\leq1\) against \(H_1:\theta>1\), at level of significance \(0.05\), let \(\beta_1\) be the size of the uniformly most powerful test and \(\beta_2\) be the power of the uniformly most powerful test at \(\theta=2\). Then \(10\beta_1+40\beta_2\) equals ______ (answer in integer).
Let \(X\) be a single observation from a distribution having probability density function
\[ f_\theta(x)=\begin{cases}\dfrac{2x}{\theta^2}&\text{if }0<x<\theta\\0&\text{otherwise,}\end{cases} \]
where \(\theta\in(0,\infty)\). For testing \(H_0:\theta\leq1\) against \(H_1:\theta>1\), at level of significance \(0.05\), let \(\beta_1\) be the size of the uniformly most powerful test and \(\beta_2\) be the power of the uniformly most powerful test at \(\theta=2\). Then \(10\beta_1+40\beta_2\) equals ______ (answer in integer).
\[ f_\theta(x)=\begin{cases}\dfrac{2x}{\theta^2}&\text{if }0<x<\theta\\0&\text{otherwise,}\end{cases} \]
where \(\theta\in(0,\infty)\). For testing \(H_0:\theta\leq1\) against \(H_1:\theta>1\), at level of significance \(0.05\), let \(\beta_1\) be the size of the uniformly most powerful test and \(\beta_2\) be the power of the uniformly most powerful test at \(\theta=2\). Then \(10\beta_1+40\beta_2\) equals ______ (answer in integer).
Show Hint
The UMP test for this monotone likelihood ratio family rejects for large \(X\); fix the cutoff using the boundary \(\theta=1\), then use it to find the power at \(\theta=2\).
Updated On: Aug 3, 2026
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Verified By Collegedunia
Correct Answer: 31
Solution and Explanation
Step 1: MLR family.
\(f_\theta(x)=2x/\theta^2\) has MLR in \(x\), so UMP test rejects for \(X>c\).
Step 2: Power function increasing in theta.
Worst case at \(\theta=1\) boundary.
Step 3: Find c.
At \(\theta=1\): \(f_1(x)=2x\), \(P_1(X>c)=1-c^2=0.05\Rightarrow c^2=0.95\).
Step 4: Size beta1.
\[ \beta_1=0.05 \]
Step 5: General power.
\[ P_\theta(X>c)=1-\frac{c^2}{\theta^2} \]
Step 6: Power at theta=2.
\[ \beta_2=1-\frac{0.95}{4}=0.7625 \]
Step 7: Combine.
\[ 10(0.05)+40(0.7625)=0.5+30.5=31 \]
Final Answer: \[ \boxed{31} \]
\(f_\theta(x)=2x/\theta^2\) has MLR in \(x\), so UMP test rejects for \(X>c\).
Step 2: Power function increasing in theta.
Worst case at \(\theta=1\) boundary.
Step 3: Find c.
At \(\theta=1\): \(f_1(x)=2x\), \(P_1(X>c)=1-c^2=0.05\Rightarrow c^2=0.95\).
Step 4: Size beta1.
\[ \beta_1=0.05 \]
Step 5: General power.
\[ P_\theta(X>c)=1-\frac{c^2}{\theta^2} \]
Step 6: Power at theta=2.
\[ \beta_2=1-\frac{0.95}{4}=0.7625 \]
Step 7: Combine.
\[ 10(0.05)+40(0.7625)=0.5+30.5=31 \]
Final Answer: \[ \boxed{31} \]
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