Show $\min(X_1,...,X_n)-1/n$ is an unbiased estimator of $\theta$.

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Let $X_1,...,X_n$ be independent copies of a real-valued random variable $X$ where $X$ has Lebesgue density

\begin{align*} p_\theta(x) = \begin{cases} \exp(\theta-x),\quad x>\theta \\ 0, \quad\quad\quad\quad\;\ x\leq \theta, \end{cases} \end{align*} where $\theta\in \mathbb{R}$ is an unknown parameter. Let $S:=\min(X_1,...,X_n)$.

claim: $S-1/n$ is an unbiased estimator of $\theta$.

I tried the following: to show that $\text{bias}_\theta(S-1/n)=0$, by definition, we need to show that $\mathbb{E}_\theta[S-1/n]=\theta$, so

$\mathbb{E}_\theta[S-1/n]=\mathbb{E}_\theta[X_{(1)}]-1/n=\int_\theta^{\infty}x\exp{(\theta-x)}dx-1/n=e^\theta(\theta+1)e^{-\theta}-1/n=\theta+1-1/n$,

can someone help me and tell me what I did wrong?

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You considered the pdf of $\min\{X_1,\cdots,X_n\}$ as the same of $X_i$. In fact:$$F_{\min\{X_1,\cdots,X_n\}}(x)=1-(1-P_\theta(x))^n\\ f_{\min\{X_1,\cdots,X_n\}}(x)=n\cdot p_\theta(x)(1-P_\theta(x))^{n-1} $$