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I got given a pdf:

$$f(x)=\tau x \exp\left(\frac{-\tau x^2}{2}\right)$$ $x,\tau >0$ I found $$E(X)=\sqrt{\frac{\pi}{2\tau}}$$ And I used the method of moments method to find: $$\hat{\tau}=\frac{\pi}{2\bar{x}^2}$$ Now, from what I found out is that an estimator is unbiased if $E(\hat{\tau})=\tau$, but I have: $$E(\hat{\tau})=E\left(\frac{\pi}{2\bar{x}^2}\right)=\frac{\pi}{2}E\left(\frac{1}{\bar{x}^2}\right)$$ and I have no idea how to calculate $E\left(\frac{1}{\bar{x}^2}\right)$

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  • $\begingroup$ sorry of course its pdf!! $\endgroup$ – Scavenger23 Jan 27 '18 at 21:22
  • $\begingroup$ I mean think of the special case where there is only one sample, in which case $E\left(\frac{1}{\bar x^2}\right)$ becomes $E\left(\frac{1}{x^2}\right)$, which is not convergent. So this might not be a very good estimator. $\endgroup$ – Mathematical Jan 27 '18 at 21:42
  • $\begingroup$ in terms of quantitative value it worked perfect. For the $\tau=0.2$ that I have been given to test $\hat{\tau}=0.208$. $\endgroup$ – Scavenger23 Jan 27 '18 at 21:45
  • $\begingroup$ Cool, I think maybe there is some asymptotic behavior but I'm really guessing. $\endgroup$ – Mathematical Jan 27 '18 at 21:48
  • $\begingroup$ I can include my r code if that helps my cause. Maybe I am trying too hard, but I have been given a generic task of "commenting on the estimated parameter compared with a real parameter" $\endgroup$ – Scavenger23 Jan 27 '18 at 22:09
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By Jensen's inequality with convex function $\varphi(x)=\dfrac{1}{x^2}$ for $x>0$, $E\left[\varphi\left(\bar{x}\right)\right]\geq \varphi\left(E\left[\bar{x}\right]\right)$. Therefore $$ E(\hat{\tau})=\frac{\pi}{2}E\left(\frac{1}{\bar{x}^2}\right) \geq \frac{\pi}{2}\frac{1}{\left(E\left[\bar{x}\right]\right)^2}=\frac{\pi}{2}\frac{1}{\left(E\left[X\right]\right)^2}=\tau$$

Note that the inequality is strict since the function $\varphi$ is not linear and the distributon of $\bar{x}$ is not degenerate. So, the estimator is not unbiased: $E[\hat\tau]>\tau$.

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