Given $Y_1,\ldots, Y_n$ are i.i.d random variables with pdf $\ f(y|\lambda) = e^{\lambda - y}$ for $y\geq \lambda$, and $\ f(y|\lambda) = 0$ otherwis, and $\lambda\in (-\infty, \infty)$. Find the Pitman estimator for a location-parameter $\lambda$ and compute the $MSE$ of that estimator.
My attempt: First, we see that the likelihood function $L(y|\lambda) = e^{-n(\overline{y} - \lambda)}$ for $y_{(1)}\geq \lambda$ ($y_{(1)} = min_{i=1,\ldots,n} y_i)$, and $L(y|\lambda) = 0$ otherwise. Now, using the formula to compute Pitman estimator of location, we have:
${\hat{\lambda}}(y) = \frac{\int_{-\infty}^{y_{(1)}} \lambda e^{-n(\overline{y} - \lambda)}d\lambda}{\int_{-\infty}^{y_{(1)}} e^{-n(\overline{y} - \lambda)}d\lambda} = \frac{\frac{y_{(1)}}{n}e^{-n(\overline{y} - y_{(1)})} - \frac{e^{-n(\overline{y} - y_{(1)})}}{n^2}}{\frac{1}{n}e^{-n(\overline{y} - y_{(1)})}} = y_{(1)} - \frac{e^{-n}}{n}$
Now, we have: $$MSE({\hat{\lambda}}(y)) = E(\hat{\lambda}(y)-\lambda)^2 = E(y_{(1)}^2) - 2(\lambda + \frac{e^{-n}}{n})E(y_{(1)}) + (\lambda + \frac{e^{-n}}{n})^2. $$
Now, since $E(y_{(1)}^2) = \frac{2}{n^2}$ and $E(y_{(1)}) = \lambda + \frac{1}{n}$ (are these results correct?), we plug them into the formula above to obtain
$$MSE({\hat{\lambda}}(y)) = \frac{1}{n^2} (2-2e^{-n} + e^{-2n}) -\frac{1}{n}(\lambda^2+2\lambda)$$ (I am skeptical about this result, because it's too ugly;p)
My question: Could someone kindly confirm if my Pitman estimator and particularly the MSE above are correct? If it is not, could you help point out the mistake?
You messed up in calculating the Pitman Estimator. Since every term has exp(-n(y-bar-y(1)) you can just factor it out. Which leaves the Pitman Estimator as y(1) - 1/n. This will make the MSE much nicer looking.