$X_1,X_2, .. ,X_n$ is a random sample of an exponential distribution with mean $\theta$. Show $nX_{(1)}$ is an unbiased estimator of $\theta$
When I calculated $X_{(1)}$, there's a $n$ in the numerator. There's no way $nX_{(1)}$ is unbiased since there's a $n^2$ in the numerator..
Can someone check if the problem is correct and give me links for the relevant formulas (if the problem is correct)?
The distribution of the first (minimum) order statistic $X_{(1)} = \min(X_1, \ldots, X_n)$ of an iid sample $(X_1, \ldots X_n)$ drawn from an exponential distribution with mean $\theta$ may be computed as follows. Consider the survival function of $X_{(1)}$, namely $$\begin{align*} S_{X_{(1)}}(x) &= \Pr[X_{(1)} > x] \\ &= \Pr\left[ \bigcap_{i=1}^n (X_i > x) \right] \\ &\overset{\text{ind}}{=} \prod_{i=1}^n \Pr[X_i > x] \\ &\overset{\text{id}}{=} \Pr[X > x]^n \\ &= (e^{-x/\theta})^n \\ &= e^{-nx/\theta}. \end{align*}$$ The equality with 'ind' on top holds because the $X_i$ are independent; the equality with 'id' on top holds because the $X_i$ are identically distributed. Consequently $X_{(1)}$ has probability density $$f_{X_{(1)}}(x) = -S'_{X_{(1)}}(x) = \frac{n}{\theta} e^{-nx/\theta}$$ over the same support as $X$. It is immediately obvious that $X_{(1)}$ is itself exponentially distributed with mean $\theta/n$; therefore, the expectation $$\operatorname{E}[nX_{(1)}] = \theta$$ as claimed.