Confidence interval and symmetric distribution

Question

Let $X_1,...,X_n$ random sample of $X$~$U[\theta-\frac{1}{2};\theta+\frac{1}{2}]$.Consider $[X_{(1)};X_{(n)}]$ a confidence interval for $\theta$. Find their confidence level and show that result is valid for any distribution symmetric around $\theta$

I know that $f(x)=1I_{[\theta-\frac{1}{2};\theta+\frac{1}{2}]}$ and $F_X(x)=\int_{\theta-\frac{1}{2}}^x 1 dx=x-\theta+\frac{1}{2}$

$X_{(n)}$ is the maximum then $F_{X_{(n)}}(x)=(x-\theta+\frac{1}{2})^n$

$X_{(1)}$ is the minimum then $F_{X_{(1)}}(x)=1-(\frac{1}{2}-x+\theta)^n$

then $$P(X_{(1)}\leq\theta\leq X_{(1)})=\gamma\Rightarrow P(X_{(n)}\geq\theta)-P(X_{(1)}\geq\theta)=\gamma$$ $$1-P(X_{(n)}\leq\theta)-1+P(X_{(1)}\leq\theta)=\gamma$$ $$P(X_{(1)}\leq\theta)-PX_{(n)}\leq\theta)=\gamma$$ Making $F_{X_{(n)}}(\theta)$ and $F_{X_{(1)}}(\theta)$ I get that $\gamma=1-(\frac{1}{2})^{n-1}$

I know that if $X$ is a random variable of any symmetric distribution, then $P(X\leq\theta)=P(X\geq\theta)=\frac{1}{2}$ but I do not know how to prove it in fact, perhaps using characteristic function.

Can anyone help me?

Answer 1

If $X$'s are i.i.d. and their density is symmetric around $\theta$, then $$ \mathsf{P}(X_{(1)}\ge \theta)=\mathsf{P}(X_{(n)}\le \theta)=\left(\mathsf{P}(X_1\le \theta)\right)^n=\left(\frac{1}{2}\right)^n. $$ Thus, $$ 1-\gamma=\mathsf{P}(X_{(1)}\ge \theta)+\mathsf{P}(X_{(n)}\le \theta)=2\times\left(\frac{1}{2}\right)^n. $$