Conditional Expected Value $E[(X-Y)^2] < \infty $ Given 

$E[|X|]&lt;\infty$, $E[|Y|]&lt;\infty$, $E[X|\sigma(Y)]=Y$, $E[Y|\sigma(X)]=X$,

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

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 $2E[XY]&lt;\infty$?

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

You can also use Cauchy-Schwarz inequality.
By tower property, 
$E[XY]=E[E[XY|Y]]=E[YE[X|Y]]=E[Y^2]$

You can also use Cauchy-Schwarz inequality.