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Given

$E[|X|]<\infty$, $E[|Y|]<\infty$, $E[X|\sigma(Y)]=Y$, $E[Y|\sigma(X)]=X$,

I need to show that out of $E[X^2]<\infty$ and $E[Y^2]<\infty$ follows also $E[(X-Y)^2]<\infty$.

I started with using the linearity of the expected value to my advantage like this:

$E[(X-Y)^2]=E[X^2-2XY+Y^2]=E[X^2]+E[Y^2]-2E[XY]$

Here is where I'm getting stuck. I know the first two terms are smaller than infinity, but how do I show that $2E[XY]<\infty$?

I also thought of using the formula for conditional expected value and replacing the $Y$ in $2E[XY]$ with $E[X|\sigma(Y)]$ but I'm struggling with continuing.

Can anyone help me with the proof or at least give some hint?

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    Why you don't use: $2XY \leqslant X^2 + Y^2$ and taking expected values both sides? – 2017-02-27

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By tower property, $E[XY]=E[E[XY|Y]]=E[YE[X|Y]]=E[Y^2]$

You can also use Cauchy-Schwarz inequality.

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    Thanks! Ended up using the Cauchy-Schwarz inequality, because "proofing" the tower property was kind of the next step in the exercise. – 2017-02-28