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Hi Michelle!
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{\displaystyle \mathrm {Var} (X)=\mathrm {E} [X^{2}]-(\mathrm {E} [X])^{2}\,}
See if you can prove it!
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{\displaystyle \mathrm {E} [\mathrm {E} [X|Y]]=\mathrm {E} [X]\,}
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{\displaystyle X\,}
Let
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{\displaystyle \mathbb {X} }
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{\displaystyle X\,}
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{\displaystyle \mathbb {Y} }
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{\displaystyle Y\,}
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{\displaystyle \mathrm {E} [\mathrm {E} [X|Y]]=\sum _{y\in \mathbb {Y} }p_{Y}(y)\mathrm {E} [X|Y=y]}
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{\displaystyle =\sum _{y\in \mathbb {Y} }p_{Y}(y)\sum _{x\in \mathbb {X} }xp_{X|Y}(x|y)}
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{\displaystyle =\sum _{y\in \mathbb {Y} }p_{Y}(y)\sum _{x\in \mathbb {X} }x{\frac {p_{X,Y}(x,y)}{p_{Y}(y)}}}
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{\displaystyle =\sum _{y\in \mathbb {Y} }\sum _{x\in \mathbb {X} }xp_{Y}(y){\frac {p_{X,Y}(x,y)}{p_{Y}(y)}}}
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{\displaystyle =\sum _{y\in \mathbb {Y} }\sum _{x\in \mathbb {X} }xp_{X,Y}(x,y)}
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{\displaystyle =\sum _{x\in \mathbb {X} }\sum _{y\in \mathbb {Y} }xp_{X,Y}(x,y)}
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{\displaystyle =\sum _{x\in \mathbb {X} }x\sum _{y\in \mathbb {Y} }p_{X,Y}(x,y)}
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{\displaystyle =\sum _{x\in \mathbb {X} }xp_{X}(x)}
=
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{\displaystyle =\mathrm {E} [X]\,}
![{\displaystyle \mathrm {Var} (X)=\mathrm {E} [X^{2}]-(\mathrm {E} [X])^{2}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2c9fea098e64547f51394ba25fda19c62c9e0db)
![{\displaystyle \mathrm {E} [\mathrm {E} [X|Y]]=\mathrm {E} [X]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/552c817f1597d279d2aab84bbadbfcb4e25b44c5)
![{\displaystyle \mathrm {E} [\mathrm {E} [X|Y]]=\sum _{y\in \mathbb {Y} }p_{Y}(y)\mathrm {E} [X|Y=y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d760e7fb18d09cc41e5ed8fee9ca60abb2ac4715)
![{\displaystyle =\mathrm {E} [X]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f62aef99098080ba54c218613682cd11827c19d5)
