How would I do this question by hand? I know I integrate from $-\infty$ to $\infty$ for $f_{x,y}$, but I have no idea how to do it by hand! My algebra soup is bad, can someone please help me?
P.S I heard some of my friends talking about some 'trick' you can do with the exponential part of the equation to solve it quicker.. but I don't know what they were talking about.
Thank you
"Complete the square" to get the equivalent expression: $$f_{X,Y}(x,y)=\frac{1}{\pi \sqrt{2}}exp[-(x+y\sqrt{2}/2)^2]exp[-{y^2}/2]$$ Now for a fixed value of $y$ the first exp term shows the conditional distribution of $X$ is $N(-y\sqrt{2}/2,1/2).$ If we integrate out $x$ the first exp term disappears leaving only a constant and no $y$ term. So the second exp term shows $Y$ is $N(0,1).$
Computing $E(XY)=E[E(XY|Y)]=E(YE(X|Y))=-E(Y^2)\sqrt{2}/2=-\sqrt{2}/2$
The previous line is equivalent to performing the iterated integral to get $E(XY)$: First we integrate $xf(x,y)$ over $x$ to get $$-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} y^2\sqrt{2}/2exp[-y^2/2]dy$$
which then gives $-\sqrt{2}/2.$