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Part BCSIR NET June 2025integrating-the-squared-density-raises-the-exponent-so-the-tails-are-cubes

Integrating the squared density raises the exponent so the tails are cubes

Let X be a random sample of size 1 from the probability density function f(x|θ)=(3/θ3)(θ−x)2\theta) = (3/\theta^{3})(\theta - x)^{2} if 0<x<θ0 < x < \theta, and 0 otherwise. If (X/(1−λ1),X/(1−λ2))(X/(1 - \lambda_{1}), X/(1 - \lambda_{2})) is a confidence interval for θ\theta with confidence coefficient 1−α1 - \alpha, where λi∈(0,1),i=1,2,λ1<λ2\lambda_{i} \in (0, 1), i = 1, 2, \lambda_{1} < \lambda_{2}, and α∈(0,1)\alpha \in (0, 1), then which of the following statements is true?

  1. A.λ22−λ12=1−α\lambda_{2}^{2} - \lambda_{1}^{2} = 1 - \alpha
  2. B.λ23−λ13=1−α\lambda_{2}^{3} - \lambda_{1}^{3} = 1 - \alpha✓ Correct
  3. C.λ22−λ12=4(1−α)\lambda_{2}^{2} - \lambda_{1}^{2} = 4(1 - \alpha)
  4. D.λ23−λ13=9(1−α)\lambda_{2}^{3} - \lambda_{1}^{3} = 9(1 - \alpha)

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: If X and Y are each normal and uncorrelated then they are independent

More on this topic

The chapter behind this: Joint distributions, transformations and order statistics — free to read

From Probability › Joint distributions, transformations, order statistics

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