Timeline for Unbiased estimator for $\mu_1/\mu_2$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2022 at 23:01 | vote | accept | RiXaTorAgu | ||
| Aug 15, 2022 at 1:17 | comment | added | Sextus Empiricus | I think I made a mistake in the comment and should use a minus $$ \int_\mathbb{R} t(z) e^{-(z-\mu)^2/2}\,\mathrm{d}z = \int_0^\infty \left(t(z) e^{-(z-\mu)^2/2}-t(-z) e^{-(-z-\mu)^2/2}\right)\,\mathrm{d}z$$ and end up with $$\int_0^\infty \left( g(z) e^{\mu z}-g(-z) e^{-\mu z} \right) \,\mathrm{d}z + \int_0^\infty \left( g(z) - g(-z) \right) e^{-\mu^2/2} \,\mathrm{d}z$$ but that is still not the same. | |
| Aug 15, 2022 at 1:11 | comment | added | whuber♦ | @Sextus I don't obtain that additional term. Please note that $e^{-\mu^2/2}$ is a factor of the original integral and therefore must be a factor of whatever you come up with. | |
| Aug 15, 2022 at 1:09 | comment | added | Sextus Empiricus | I have this additional term $\int_0^\infty \left( g(z) + g(-z) \right) e^{-\mu^2/2} \,\mathrm{d}z $ is this integral $\int_\mathbb{R} g(z) \,\mathrm{d}z$ equal to zero? | |
| Aug 15, 2022 at 1:08 | comment | added | Sextus Empiricus | $$ \int_\mathbb{R} t(z) e^{-(z-\mu)^2/2}\,\mathrm{d}z = \int_0^\infty \left(t(z) e^{-(z-\mu)^2/2}+t(-z) e^{-(-z-\mu)^2/2}\right)\,\mathrm{d}z = \int_0^\infty \left( t(z) e^{-(z^2+\mu^2-2\mu z)/2}+t(-z) e^{-(z^2+\mu^2+2\mu z)^2/2}\right)\,\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) \,\mathrm{d}z = \int_0^\infty \left( g(z) e^{-\mu^2/2+\mu z}+g(-z) e^{-\mu^2/2-\mu z} \right) \,\mathrm{d}z = \int_0^\infty \left( g(z) e^{\mu z}+g(-z) e^{-\mu z} \right) \,\mathrm{d}z + \int_0^\infty \left( g(z) + g(-z) \right) e^{-\mu^2/2} \,\mathrm{d}z $$ | |
| Aug 15, 2022 at 0:50 | comment | added | whuber♦ | @Sextus (1) Rao-Blackwellization, (2) Split into positive and negative parts and change the variable in the negative part. | |
| Aug 14, 2022 at 18:52 | comment | added | Sextus Empiricus | The change with the integral domain from $\int_\mathbb{R}$ to $\int_0^\infty$ is not so clear. | |
| Aug 14, 2022 at 18:42 | comment | added | Sextus Empiricus | This is a nice answer, but at the points where it states that an unbiased estimator must be a function of the sufficient statistic it goes too fast for me. It seems intuitive, but is there also some theorem or question/proof that states that an unbiased estimator does not exist if we can not construct an unbiased estimator based on the sufficient statistic? | |
| Aug 14, 2022 at 17:38 | comment | added | mlofton | wow. amazing and thanks because I was wondering how one could obtain the correct estimator but had no clue. | |
| Aug 14, 2022 at 16:09 | history | answered | whuber♦ | CC BY-SA 4.0 |