Solved – Confidence Interval, uniform distribution

confidence intervalinferenceself-study

Let $X_1,…X_n$ random sample from
$f(x;\theta)=I_{[\theta-\frac{1}{2};\theta+\frac{1}{2}]}(x)$.
i) Show that
$(X_{(1)},X_{(n)})$ is a confidence interval for $\theta$.ii) find
your confidence level.

The first item I do not know how to show that a range is confidence interval.

In the second item I tried $$P(X_{(1)}\leq \theta \leq X_{(n)})=P(X_{(n)}\geq\theta)-P(X_{(1)}\leq \theta)$$.

But the solutions say that $P(X_{(1)}\leq \theta \leq X_{(n)})=P(X_{(n)}\geq\theta)-P(X_{(1)}\geq \theta)$

But I don't understand this inequality,where $X_{(n)}$ is the max and $X_{(1)}$ is the minimum

Best Answer

  1. Have you been provided with a definition of "confidence interval"? You generally answer such questions by reference to a definition (does the thing you're given satisfy the definition?).

  2. The second can be easily seen by drawing a picture. Loosely speaking, the region between (green) is the region to the left of $X_{(n)}$ (blue) minus the region to the left of $X_{(1)}$ (red). So the set of cases where the region between the two encompasses the parameter is the set of cases where the parameter is left of $X_{(n)}$ minus those cases where it's also left of $X_{(1)}$.

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