Let $X\sim\textrm{Bin}_{n_x,p_x}$ and $Y\sim\textrm{Bin}_{n_y,p_y}$ be two independent random variables. Now Define $Z|_{X=x, Y=y}\sim\textrm{Bin}_{x\cdot y,p_z}\enspace$.

Is there a simple argument that \begin{align}\mathbb{E}[Z]=\mathbb{E}[X]\cdot\mathbb{E}[Y]\cdot p_z\enspace.\enspace\enspace\enspace(1)\end{align} I mean of course, it holds that $\mathbb{E}[X\cdot Y]=\mathbb{E}[X]\cdot\mathbb{E}[Y]\enspace$, cause $X$ and $Y$ are independent, but does this directly imply $(1)$?

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos May 17 '18 at 13:49

General rule:$$\mathbb EU=\mathbb E[\mathbb E[U\mid V]]\tag1$$

Further we have $\mathbb E[Z\mid X=x,Y=y]=xyp_z$ so that $\mathbb E[Z\mid (X,Y)]=XYp_z$.

So we can apply the rule in order to find that $$\mathbb EZ=\mathbb E[XYp_z]=p_z\mathbb E[XY]=p_z\mathbb EX\mathbb EY=p_zn_xp_xn_yp_y$$

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