# Expected value of minimum of independent identically distributed random variables

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?

• You could try integration by parts. – Jarle Tufto Feb 15 '17 at 20:32
• $f(x)$ is not a valid density except for $\theta =0$. – Dilip Sarwate Feb 15 '17 at 20:36
• Sorry, I meant $x>\theta$. Fixed it above. – honeyBunchesOfFloats Feb 15 '17 at 20:50
• Integration by parts does not yield a defined answer. – honeyBunchesOfFloats Feb 15 '17 at 20:50
• @honeyBunchesOfFloats What if you have sample of size 1? Try to calculate expected value in this simple case. Your integral can be calculated by parts – Łukasz Grad Feb 15 '17 at 22:26

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 $\frac{1}{n}$.