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An enterprise has workers divided in 3 groups. $p_i$ is a proportion of the $i$-th group to all, $p_1+p_2+p_3=1$.

Select independently and equiprobably with replacement $n$ workers out of all, $X_i$ workers belong to $i$-th group. $n=X_1+X_2+X_3$. Use $Y_i=X_i/n$ as unknown proportion $p_i$. After find the joint probability function $p(y_1,y_2,y_3)$. If I have large size of sample, does the random variable $Y_i$ converge to the true value $p_i$? If $y_1>p_1$, which is more probable, $Y_2>p_2$ or $p_2>Y_2$?and what is a conditional mean E($Y_i/y_1$),(i=2,3)?

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    sorry, I will lear$n$ how to type correctly.a$n$d,thank enzotib!2012-07-30

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The first part can be settled by just thinking about the situation. If $y_1>p_1$, that means that our sample has a larger than "expected" number of workers from group $1$. That means that the total number of workers from groups $2$ and $3$ is less than expected, so with probability $\gt 1/2$ we will have $Y_2\lt p_2$.

For the conditional expectation, please see one of the answers to this previous question of yours.

There the conditional distribution of $X_2$ given that $X_1=x_1$ is shown to be binomial, number of trials $n-x_1$, probability of "success" $p_2'=\frac{p_2}{p_2+p_3}=\frac{p_2}{1-p_1}$.

So you can find the conditional expectation of $X_2$ given $X_1=x_1=ny_1$ by using the ordinary formula for the expectation of a binomial. Then multiply by $1/n$ to find the conditional expectation of $Y_2$.

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    Too little space! For first (and second) can do binomil coefficient manipulation. But much easier to use following. Let $W_k=1$ if (say) picked from third group on $k$-th pick, $W_k=0$ otherwise. Then $X_3=W_1+W_2+\cdots +W_n$. So $E(X_3)=E(\sum W_k)=\sum E(W_k)=np_3$ (Method of Indicator functions).2012-07-31