# Why is $\mathbb E(X) =\mathbb E(Y)$ in this joint probability distribution?

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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• since $E(X)=EE(X|Y)$ so $E(X)=EE(X|Y)=E(Y)$ – masoud 2 days ago

## 1 Answer

$$E(X|Y)=Y$$ so (by taking expectation on both sides) we get $$EX=EY$$. Since $$Y\sim B(0,3.5)$$ we have $$EY=(0.3)(5)$$.

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