Let $\xi$ - random variable with probability density function is $$p(x;\theta):=\left\{\begin{array}{ll}e^{\theta-x},&x\geq\theta \\ 0,&x<\theta\end{array}\right.,$$ $\tilde\theta=\tilde\theta(x_1,\ldots,x_n)=\min(x_1,\ldots,x_n)$ - estimator for $\theta$, $\tilde\theta_1:=\tilde\theta-\frac1{n}$
I need proof $D\tilde\theta_1=\frac1{nI(\theta)}$
I try this. Note $\varphi(x)$ - density function for $\tilde\theta$ $$F_{\xi}(x):=1-e^{\theta-x}$$ $$ 1-F_{\tilde\theta}(x)=P\{\min(x_1,\ldots,x_n)>x\}=P\{x_1>x,\ldots,x_n>x\}=\prod_{i=1}^n(1-F_{\xi}(x))=e^{n(\theta-x)}\Rightarrow{F}_{\tilde\theta}(x)=1-e^{n(\theta-x)}\Rightarrow\varphi(x)=F'_{\tilde\theta}(x)=ne^{n(\theta-x)} $$ $$ E\tilde\theta=\int_{\theta}^{\infty}x\varphi\left(x\right)dx=\int\limits_{\theta}^{\infty}nxe^{n(\theta-x)}dx=e^{n\theta}\int\limits_{\theta}^{\infty}nxe^{-nx}dx=e^{n\theta}\frac1{n}\int\limits_{n\theta}^{\infty}te^{-t}dt= -\frac{e^{n\theta}}{n}\int\limits_{n\theta}^{\infty}tde^{-t}=-\frac{e^{n\theta}}{n}\left(\left.te^{-t}\right|_{n\theta}^{\infty}-\int\limits_{n\theta}^{\infty}e^{-t}dt\right)= -\frac{e^{n\theta}}{n}\left(-n\theta{e}^{-n\theta}-e^{-n\theta}\right)=\theta+\frac1{n} $$ $$ E\tilde\theta^2=\int\limits_{\theta}^{\infty}x^2\varphi(x)dx=\int\limits_{\theta}^{\infty}nx^2e^{n(\theta-x)}dx=\frac{e^{n\theta}}{n}\int\limits_{\theta}^{\infty}(nx)^2e^{-nx}dx= -\frac{e^{-n\theta}}{n^2}\int\limits_{n\theta}^{\infty}t^2de^{-t}=\frac{e^{n\theta}}{n^2}\left(\left.t^2e^{-t}\right|_{n\theta}^{\infty}-2\int\limits_{n\theta}^{\infty}te^{-t}dt\right)= -\frac{e^{n\theta}}{n^2}\left(-(n\theta)^2e^{-n\theta}-2e^{-n\theta}(n\theta+1)\right)=\theta^2+\frac{2n}{\theta}+\frac2{n^2}= \left(\theta+\frac1{n}\right)^2+\frac1{n^2}=(E\tilde\theta)^2+\frac1{n^2} $$ So $$D\tilde\theta_1=D\tilde\theta=E\tilde\theta^2-(E\tilde\theta)^2=\frac1{n^2}$$ $$I\left(\theta\right)=\int_{\theta}^{\infty}\left(\frac{\partial\ln{p}(x;\theta)}{\partial\theta}\right)^2p\left(x;\theta\right)dx=\int_{\theta}^{\infty}p\left(x;\theta\right)dx=1$$ Thus $$\frac1{n^2}=D\tilde\theta_1\ne\frac1{nI(\theta)}=\frac1{n}.$$
Is there mistake?