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Let $X_1$, $X_2$,..., $X_n$, $n>2$, be a random sample from the binomial distribution $b(1,\theta)$ and let $Y_2=(X_1+X_2)/2$.

I know how to show that $Y_1$=$X_1+X_2$ +$\dots$+ $X_n$ is a complete sufficient statistic for $\theta$ and I can find out $Y_1/n$ is the MVUE of $\theta$.

My question is how can I determine $\mathbb{E} \{ Y_2 |Y_1=y_1 \}$. I have trouble dealing this type of conditional probability problem.

Thanks.

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  • $\begingroup$ Is $Y_1=X_1+\dots +X_n$? $\endgroup$ – d.k.o. Jun 21 '15 at 6:41
  • $\begingroup$ Yes, I have corrected it, thank you! $\endgroup$ – wanchihsin Jun 21 '15 at 6:45
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Let $S_n=X_1+\dots +X_n$. Then

$$\mathbb{E}[X_1|S_n]=\mathbb{E}[X_2|S_n]=...=\mathbb{E}[X_n|S_n]$$

Then $\mathbb{E}[X_1|S_n]+\cdots+\mathbb{E}[X_n|S_n]=\mathbb{E}[S_n|S_n]=S_n$ so that $$\mathbb{E}[X_i|S_n]=\frac{S_n}{n} \text{, }i=1,...,n$$

and $$\mathbb{E}[Y_2|S_n]=\frac{1}{2}\{\mathbb{E}[X_1|S_n]+\mathbb{E}[X_2|S_n]\}=\frac{S_n}{n}$$

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  • $\begingroup$ Thank you for your help. I'm wondering another question. Given $X_1,X_2$ be iid $N(\theta,1)$ and let $Y=(X_1+X_2)/2$. I want to find $\mathbb{E} [ Y |X_1 ]$ and $\mathbb{E} [X_1|Y]$. Is the solution similar to this question? $\endgroup$ – wanchihsin Jun 21 '15 at 7:19
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    $\begingroup$ $\mathbb{E}[Y|X_1]=\frac{1}{2}[X_1+\mathbb{E}[X_2]]$ and the second is similar. $\endgroup$ – d.k.o. Jun 21 '15 at 7:21
  • $\begingroup$ $\mathbb{E} [ Y |X_1 ]$ = $1/2\mathbb{E}[X_1+X_2|X_1]$=$1/2\{\mathbb{E}[X_1|X_1]+\mathbb{E}[X_2|X_1]\}$=$1/2\{X_1+\mathbb{E}[X_2|X_1]\}$ I can't find out which step is wrong. $\endgroup$ – wanchihsin Jun 21 '15 at 7:31
  • $\begingroup$ $X_1$ and $X_2$ are independent... $\endgroup$ – d.k.o. Jun 22 '15 at 2:38

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