User talk:Chundt

Hi Michelle!

Here is a helpful formula:

$\mathrm {Var} (X)=\mathrm {E} [X^{2}]-(\mathrm {E} [X])^{2}\,$

See if you can prove it!

Here is the proof that $\mathrm {E} [\mathrm {E} [X|Y]]=\mathrm {E} [X]\,$ for any $X\,$:

Let $\mathbb {X}$ be the range of $X\,$ and $\mathbb {Y}$ the range of $Y\,$. Then

$\mathrm {E} [\mathrm {E} [X|Y]]=\sum _{y\in \mathbb {Y} }p_{Y}(y)\mathrm {E} [X|Y=y]$

$=\sum _{y\in \mathbb {Y} }p_{Y}(y)\sum _{x\in \mathbb {X} }xp_{X|Y}(x|y)$

$=\sum _{y\in \mathbb {Y} }p_{Y}(y)\sum _{x\in \mathbb {X} }x{\frac {p_{X,Y}(x,y)}{p_{Y}(y)}}$

$=\sum _{y\in \mathbb {Y} }\sum _{x\in \mathbb {X} }xp_{Y}(y){\frac {p_{X,Y}(x,y)}{p_{Y}(y)}}$

$=\sum _{y\in \mathbb {Y} }\sum _{x\in \mathbb {X} }xp_{X,Y}(x,y)$

$=\sum _{x\in \mathbb {X} }\sum _{y\in \mathbb {Y} }xp_{X,Y}(x,y)$

$=\sum _{x\in \mathbb {X} }x\sum _{y\in \mathbb {Y} }p_{X,Y}(x,y)$

$=\sum _{x\in \mathbb {X} }xp_{X}(x)$

$=\mathrm {E} [X]\,$