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Suppose $Y \sim \mathrm{Bin}(n,p)$ given $\Theta=(p,q)$ and $Z \sim \mathrm{Bin}(y,q)$ given $Y=y$ and $\Theta=(p,q)$. Now I want to determine the variance of $Z-Y$, but I don't know how. I know $\operatorname{Cov}(Z,Y)=E(ZY)-E(Z)E(Y)$, but I'm stuck on the first term.

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We'll use the result $\mathbb{E}[X] = \mathbb{E}[ \mathbb{E}[X\mid Y] ]$

It's easy to see that

$$\mathbb{E}[Z\mid Y] = Y q $$

Now,

\begin{align} \mathbb{E}[ZY ] &= \mathbb{E} [ \mathbb{E}[ZY \mid Y] ] \\ &= \mathbb{E}[ Y \mathbb{E}[Z\mid Y] ] \\ &= \mathbb{E}[ q Y^2 ]\\ &= npq ( np + 1 - p) \end{align}

You'll also need the second moment of $Z$, can you figure out how to calculate it now?

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  • $\begingroup$ Thanks! Only have one question, why it's E[Z|Y]=Yq? $\endgroup$ – Roos Jansen Oct 20 '15 at 19:49
  • $\begingroup$ Just the mean of a $(Y,q)$ binomial random variable. Perhaps it's clearer if I write $\mathbb{E}[ Z ~| Y=y] = yq$, and so $\mathbb{E}[Z ~| Y]$ is a well defined random variable, the value of which is $qY$. It's generally important to remember that conditional expectations with respect to a $\sigma$-field are themselves random variables in that $\sigma$ field. $\endgroup$ – stochasticboy321 Oct 21 '15 at 0:45

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