Question

Please answer all the parts neatly with all details.

8. Let X1, X2, ... be an i.i.d. with Xn Let Y min(X,... , Xn) + 1 and Zn = max(X1,... , Xn) - 1. (a) Show that Y, - 0 and 0 - Z, have the same distribution uniform(0 1,0+ 1) (b) Show that Y, -P>0. (c) Show that n(Yn - Zn) converges in distribution and specify the limit distribution

Answer 1

$(a)~CDF~of~U_n=Y_n-\theta~is\\ F_{U_n}(u)=P(U_n\leq u)=P(Y_n-\theta\leq u)=P(\min\{X_1,X_2,...,X_n\}\leq u&plus;\theta-1)\\ =1-P(\min\{X_1,X_2,...,X_n\}>u&plus;\theta-1)\\ =1-P(X_1>u&plus;\theta-1,~X_2>u&plus;\theta-1,...,X_n>u&plus;\theta-1)\\ =1-\prod_{i=1}^nP(X_i>u&plus;\theta-1)=1-\left(\int_{u&plus;\theta-1}^{\theta&plus;1} \frac{1}{2}dx\right )^n=1-\frac{(2-u)^n}{2^n};~0<u<2\\ =1~if~u\geq 2\\ Hence~CDF~of~U_n~is\\ F_{U_n}(u)=\left\{\begin{matrix} 0~if~u\leq 0~~~~~~~~~~~~~~~~\\ 1-\frac{(2-u)^n}{2^n};~0<u<2\\ 1~if~u\geq 2~~~~~~~~~~~~~~~~~~ \end{matrix}\right.\\$

$CDF~of~V_n=\theta-Z_n~is\\ F_{V_n}(v)=P(V_n\leq v)=P(\theta-Z_n\leq v)=P(\max\{X_1,X_2,...,X_n\}\geq \theta&plus;1-v)\\ =1-P(\max\{X_1,X_2,...,X_n\}\leq \theta&plus;1-v)\\ =1-P(X_1\leq \theta&plus;1-v,~X_2\leq \theta&plus;1-v,...,X_n\leq \theta&plus;1-v)\\ =1-\prod_{i=1}^nP(X_i\leq \theta&plus;1-v)=1-\left(\int_{\theta-1}^{ \theta&plus;1-v} \frac{1}{2}dx\right )^n=1-\frac{(2-v)^n}{2^n};~0<v<2\\ =1~if~v\geq 2\\ Hence~CDF~of~V_n~is\\ F_{V_n}(v)=\left\{\begin{matrix} 0~if~v\leq 0~~~~~~~~~~~~~~~~\\ 1-\frac{(2-v)^n}{2^n};~0<v<2\\ 1~if~v\geq 2~~~~~~~~~~~~~~~~~~ \end{matrix}\right.\\ Hence~Y_n-\theta~and~\theta-Z_n~have~the~same~distribution.\\ (b)~P(|Y_n-\theta|\leq \epsilon )=P(-\epsilon \leq Y_n-\theta\leq \epsilon )=F_{U_n}(\epsilon )-F_{U_n}(-\epsilon )\\ =1-\frac{(2-\epsilon)^n }{2^n}~[Since~F_{U_n}(-\epsilon )=0~for~\epsilon >0]\\ =1-\left(1-\frac{\epsilon }{2} \right )^n\rightarrow 1~as~n\rightarrow \infty(since~0<1-\frac{\epsilon }{2}<1~for~\epsilon >0)\\ Hence~Y_n\overset{p}{\rightarrow}\theta.\\$

$(c)~CDF~of~n(Y_n-Z_n)~is\\ F_{n(Y_n-Z_n)}(x)=P(n(Y_n-Z_n)\leq x)=P\left(\min(X_1,...,X_n)&plus;1-\max(X_1,X_2,...,X_n)&plus;1\leq \frac{x}{n} \right )\\ =P(X_{(n)}-X_{(1)}\geq 2-x/n)~where,~R=\max(X_1,X_2,...,X_n)-\min(X_1,X_2,...,X_n).....(1)\\ The~joint~pdf~of~X_{(1)}=\min(X_1,X_2,...,X_n)~and~X_{(n)}=\max(X_1,X_2,...,X_n)~is\\ f_{1,n}(x_1,x_2)=n(n-1)\left[F(x_2)-F(x_1) \right ]^{n-2}\\ =n(n-1)\left[ \frac{x_2-x_1}{2}\right ]^{n-2}\frac{1}{2^2};~\theta-1<x_1<x_2<\theta&plus;1\\ where,~F(x)=\int_{\theta-1}^x\frac{1}{2}dx=\frac{x&plus;1-\theta}{2}~if~\theta-1<x<\theta&plus;1\\ Transform~(X_{(1)},X_{(n)})\rightarrow (R,X_{(1)})~where\\ r=x_2-x_1\Rightarrow |Jacobian|=1,~x_2=r&plus;x_1\Rightarrow \theta-1<r&plus;x_1<\theta&plus;1\\ ~or,~\theta-1-r<x_1<\theta&plus;1-r;~since~r>0~and~x_1>\theta-1~so~\theta-1<x_1<\theta&plus;1-r\\ The~pdf~of~R~is\\ f_R(r)=\frac{n(n-1)r^{n-2}}{2^n}\int_{\theta-1}^{\theta&plus;1-r}dx_1=\frac{n(n-1)r^{n-2}(2-r)}{2^n}~if~0<r<2\\ =0~otherwise\\$

$From~(1),\\ F_{Y_n-Z_n}(x)=P(R\geq 2-x/n)=\frac{n(n-1)}{2^n}\int_{2-x/n}^2(2r^{n-2}-r^{n-1})dr \\=\frac{n(n-1)}{2^n}\left[2\left[\frac{1}{n-1}r^{n-1} \right ]^2_{2-x/n}-\left[\frac{1}{n}r^{n} \right ]^2_{2-x/n} \right ]\\ =1-\left(1-\frac{x}{2n} \right )^{n-1}&plus;\frac{n-1}{2n}\left(1-\frac{x}{2n} \right )^{n-1}\\ \rightarrow 1-e^{-\frac{x}{2}}&plus;\frac{1}{2}e^{-\frac{x}{2}}=1-\frac{1}{2}e^{-\frac{x}{2}}~as~n\rightarrow \infty.\\ Hence~n(Y_n-Z_n)\overset{d}{\rightarrow}X~where~X\sim Exponential~distribution~with\\~mean=2.$