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?

  • 1
    $\begingroup$ You could try integration by parts. $\endgroup$ – Jarle Tufto Feb 15 '17 at 20:32
  • 1
    $\begingroup$ $f(x)$ is not a valid density except for $\theta =0$. $\endgroup$ – Dilip Sarwate Feb 15 '17 at 20:36
  • 1
    $\begingroup$ Sorry, I meant $x>\theta $. Fixed it above. $\endgroup$ – honeyBunchesOfFloats Feb 15 '17 at 20:50
  • $\begingroup$ Integration by parts does not yield a defined answer. $\endgroup$ – honeyBunchesOfFloats Feb 15 '17 at 20:50
  • 2
    $\begingroup$ @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 $\endgroup$ – 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 $\frac{1}{n}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.