Let $X_1 \cdots X_n$ be a random sample from a distribution $f(x) = e^{-(x-\theta)}, x>\theta, -\infty<\theta<\infty$. Let $Y_1 = \rm{min}\{X_1 \cdots X_n\}$.
Is $Y_1$ unbiased for $\theta$?
I'm stuck on finding the expected value of $Y_1$. I have the pdf as $f_{Y_1}(y) = n e^{-n(y-\theta)}$, but cannot find the integral $\int_\theta^\infty y n e^{-n(y-\theta)}$. I believe that integral is correct, but it appears to be undefined. Any idea where I may have gone wrong?
Posting an answer as it looks like-based on comments-the OP solved the problem. Let $y$ denote the minimum order statistic.
Take out the constant term to get:
$ne^{n\theta} \int_\theta^\infty ye^{-ny}dy$.
Then, integration by parts once gives you:
$ne^{n\theta}\left[\frac{\theta e^{-n\theta} }{n} + \frac{1}{n}\int_\theta^\infty e^{-ny}dy\right]$.
Note: the first term will give you a $ye^{-ny}$ term evaluated from $\theta$ to $\infty$, the limit as $y \to \infty$ is 0, you can check this by applying l'hospital's rule. Then integrate this second integral to get:
$ne^{n\theta}\left[\frac{\theta e^{-n\theta} }{n} + \frac{e^{-ny}}{n^2} \right]$.
Simplifying this expression gives:
$\theta + \frac{1}{n}$.
This indicates the estimator of $\theta$ is biased but is asymptotically unbiased. Alternately, you would know the value of $n$ (your sample size), so you could make the estimator unbiased by subtracting $\frac{1}{n}$.
