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Let $X_1 \dots X_n$ denote a random sample from a uniform $(0, \theta$) distribution.

PROBLEM: Compute the coverage probability for the CI: $$\left(\frac{X_{(n)}}{0.95}, \frac{X_{(n)}}{0.25}\right)$$ using $W = X_{(n)}/\theta$ which has density $F_w = w^n$

My Work $$ \begin{split} CI &= \left(\frac{X_{(n)}}{0.95} \leq \theta \leq \frac{X_{(n)}}{0.25}\right)\\ &\iff \left((\frac{X_{(n)}}{0.95} \leq X_{(n)}/w \leq \frac{X_{(n)}}{0.25}\right)\\ &\iff \left(\frac{1}{0.95} \leq 1/w \leq \frac{1}{0.25}\right) \end{split} $$ And this is just $P\big(w \in [0.25, 0.95]\big) = F_w(0.95) - F_w(0.25) = 0.95^n - 0.25^n$.

Am I going about this the right way?

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    $\begingroup$ Please do not delete questions after having received an answer. $\endgroup$ – quid May 13 at 11:14
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The coverage probability is the probability that $\theta\in \left(\frac{X_{(n)}}{0.95},\frac{X_{(n)}}{0.25}\right)$.

That is, because $\theta>0$,

\begin{align} P\left[\frac{X_{(n)}}{0.95}<\theta<\frac{X_{(n)}}{0.25}\right]&=P\left[0.25<\frac{X_{(n)}}{\theta}<0.95\right] \\&=P\left[\frac{X_{(n)}}{\theta}<0.95\right]-P\left[\frac{X_{(n)}}{\theta}<0.25\right] \\&=0.95^n-0.25^n \end{align}

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