Unbiased estimators in an exponential distribution

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$E(Y_1) = \theta$, so unbiased; - $Y_1\sim \text{Expo}(\lambda)$ and $\text{mean}=\frac{1}{\lambda}$

$E(\overline Y)=E\left(\frac{Y_1 + Y_2 + Y_3}{3}\right)= \frac{E(Y_1) + E(Y_2) + E(Y_3)}{3}=\frac{\theta + \theta + \theta}{3}= \theta$, so unbiased

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segfault.py

Updated on July 21, 2022

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  • segfault.py
    segfault.py less than a minute

    We have $Y_{1}, Y_{2}, Y_{3}$ a random sample from an exponential distribution with the density function $ f(y) = \left\{ \begin{array}{ll} (1/\theta)\mathrm{e}^{-y/\theta} & y \gt 0 \\ 0 & elsewhere. \end{array} \right.$

    I'm suppose to find which of the following estimators are unbiased: $\hat{\theta_{1}} = Y_{1}, \hat{\theta_{2}} = (Y_{1} + Y_{2}) / 2, \hat{\theta_{3}} = (Y_{1} + 2Y_{2})/3, \hat{\theta_{4}} = \bar{Y}$.

    As far as I can tell none of these estimators are unbiased. For example

    $ E(\hat{\theta_{1}}) \\ = E(Y_{1}) \\ = Y_{1}\int_0^\infty (1/\theta)\mathrm{e}^{-y/\theta}\,\mathrm{d}y \\ = \left.Y_{1}(-\mathrm{e}^{y/\theta}) \right|_0^\infty \\ = Y_1(0 + 1) = Y_1 $

    and

    $E(\hat{\theta_{4}}) \\ = E(\bar{Y}) \\ = \int_0^\infty (1/\theta^2)\mathrm{e}^{-2y/\theta}\,\mathrm{d}y \\ = \left.(1/2\theta)(-\mathrm{e}^{-2y/\theta}) \right|_0^\infty \\ = (1/2\theta)(0 + 1) = 1/2\theta$

    So it looks like none of these are unbiased. I imagine the problem exists because one of $\hat{\theta_{1}}, \hat{\theta_{2}}, \hat{\theta_{3}}, \hat{\theta_{4}}$ is unbiased.

    What am I doing wrong?

    • fgp
      fgp over 8 years
      Your first derivation can't be right - $Y_1$ is a random variable, not a real number, and thus saying $E(\hat{\theta}_1)$ makes no sense. I think you meant $\int y (1/\theta) \ldots$ where you wrote $Y_1\int (1/\theta) \ldots$
    • André Nicolas
      André Nicolas over 8 years
      Calculate $\int_0^\infty \frac{y}{\theta}e^{-y/\theta}\,dy$. (Use integration by parts.) That is the only integral calculation that you will need to do for the entire problem.
    • Aaron
      Aaron over 8 years
      Using linearity of expectation, all of these estimators will have the same expected value.
    • fgp
      fgp over 8 years
      @AndréNicolas Or do as I did, recognize this as an exponential distribution, and after spending a half a minute or so trying to remember whether the expectation of $\lambda e^{-\lambda x}$ is $\lambda$ or $\lambda^{-1}$ go look it up on wikipedia ;-)
    • André Nicolas
      André Nicolas over 8 years
      In almost all situations you will be right. The way most courses are organized, the exponential distribution would have been discussed before one talks about estimators.
    • segfault.py
      segfault.py over 8 years
      Agh I think I got it. I was mistakenly using $Y_{i}$'s as constants. On $\theta_{4}$ we're using the fact that $E(\bar{Y}) = E(Y)$, right? I can't recall why this is true. Y'all have a link or explanation?
  • user88595
    user88595 about 8 years