2
$\begingroup$

Let $X,Y,Z$ be three unfair coins. We consider the coin $\Gamma=XI(Z=1)+YI(Z=0).$ We are given the sample $S=\{(\Gamma_i,Z_i)| i \leq n \}$ Let $k=\sum_{i \leq n}I(Z_i=1)$. For every pair of $(\Gamma_i,Z_i)$ we know either the value $X_i$ or $Y_i$.
The means over these respective values shall be denoted by $\overline{X_{n-k}}$ and $\overline{Y_{k}}.$ The standard estimator for the coin $\Gamma$ would be $\overline{\Gamma_n}$. Is there a general decision rule (dependent on $k$ and $n$) when $E[|\overline{\Gamma_n}-\Gamma|]$ will be smaller than $E[|\overline{X_{n-k}}I(Z=1)+\overline{Y_{k}}I(Z=0)-\Gamma|]$?
I am also happy with assuming that $Z$ is a fair coin.

$\endgroup$
7
  • $\begingroup$ I don't understand what the Z with no subscript is in the last equation $\endgroup$
    – user83457
    Commented Jan 3, 2017 at 12:58
  • $\begingroup$ It's the same Z as in the first and second line. $\endgroup$ Commented Jan 3, 2017 at 16:22
  • $\begingroup$ You flip a Z type coin n times to determine whether you sample from X or Y, and you know the results of those tosses, otherwise you couldn't decide if you are getting an X or Y. Then you flip the Z coin one more time and use I(Z = 1) as, sort of, an estimate of the P(Z=1) ? $\endgroup$
    – user83457
    Commented Jan 3, 2017 at 20:15
  • $\begingroup$ Yes the idea is that i have the information of n Gamma and Z tosses and now I am left with only a Z toss and try to find the better estimator for the Gamma coin which I know is dependent on the Z coin. The idea is that if I have a lot of samples with the same value for Z I probably should take the respective X/Y mean but when I have little experience for the result of Z I might be better of with the mean of Gamma. $\endgroup$ Commented Jan 3, 2017 at 20:21
  • $\begingroup$ This is very very important to me, if you have a hint for me please help me, it really is highly appreciated. $\endgroup$ Commented Jan 3, 2017 at 20:23

0

You must log in to answer this question.

Browse other questions tagged .