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If we know $X|Y$ is a normal random variable with mean $Y$ and variance $2$, and $Y$ is a binomial distribution with success probability $0.3$ and the number of trials $5$, why is $\mathbb E(Y) = 0.3\times5 =\mathbb E(X)?$

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    $\begingroup$ since $E(X)=EE(X|Y)$ so $E(X)=EE(X|Y)=E(Y)$ $\endgroup$ – masoud 2 days ago
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$E(X|Y)=Y$ so (by taking expectation on both sides) we get $EX=EY$. Since $Y\sim B(0,3.5)$ we have $EY=(0.3)(5)$.

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    $\begingroup$ if these information, can we find out what is var(x)? $\endgroup$ – 2028istheway 2 days ago
  • $\begingroup$ Yes, Use the following: $E(X^{2}|Y)=2+Y^{2}$ and $EX^{2}=2+EY^{2}$. $\endgroup$ – Kavi Rama Murthy 2 days ago
  • $\begingroup$ $\newcommand{\V}{\operatorname{Var}}$To find $\V(X)$, you can try using the fact that $\V(X) = \Bbb{E}[\V(X\mid Y)]+ \V(\Bbb{E}[X\mid Y])$. $\endgroup$ – Minus One-Twelfth 2 days ago

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