Question

I) Lt X be a random sample from the tUimifom (o, ) distribution Consider max( Xz. X) a and b are the random interval (ay,' b%) around 0, where },-max(Xi, X. . X.), a and b are constants such that 1 sasb Find the confidence level of this interval

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Answer 1

$The~cdf~of~Y_n~is\\ F_n(y)=P(Y_n\leq y)=P(X_1\leq y,X_2\leq y,...,X_n\leq y)\\ =\prod_{i=1}^nP(X_i\leq y),~~since~X_i's~are~independent\\ =\prod_{i=1}^n\int_0^y\frac{1}{\theta}dx=\frac{y^n}{\theta^n},~0<y<\theta\\ Now,~The~pdf~of~Y_n~is\\ f_n(y)=\frac{\mathrm{d} F_n(y)}{\mathrm{d} y}=n\frac{y^{n-1}}{\theta^n},~0<y<\theta\\=0~otherwise.$

$Transform~Y_n\rightarrow Z~where\\ z=-2n\log(y/\theta)~then~0<z<\infty\\\\ y=\theta\exp\left(-\frac{z}{2n} \right )\\ \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{\theta}{2n}\exp\left(-\frac{z}{2n} \right ),~|Jacobian|=\frac{\theta}{2n}exp\left(-\frac{z}{2n} \right )\\ The~pdf~of~Z~is\\ f_Z(z)=\frac{1}{2}\exp(-z/2),~z>0\\ =0~otherwise\\ Hence~Z\sim~\chi^2_2.\\ Hence~we~consider~-2n\log(Y_n/\theta)~as~pivot.$

$Now,~P\left(\chi^2_{1-\alpha/2,2}<-2n\log(Y_n/\theta)<\chi^2_{\alpha/2,2}\right)=1-\alpha\\ or,~P\left(\exp\left(\frac{\chi^2_{1-\alpha/2,2}}{2n} \right )<\frac{\theta}{Y_n}<\exp\left(\frac{\chi^2_{\alpha/2,2}}{2n} \right ) \right )=1-\alpha\\ Hence~100(1-\alpha)\%~Confidence~interval~of~\theta~is\\ \left(Y_n \exp\left(\frac{\chi^2_{1-\alpha/2,2}}{2n} \right ),~Y_n \exp\left(\frac{\chi^2_{\alpha/2,2}}{2n} \right )\right )$