How to calculate with events defined in terms of products of random variables? $X,Y  \overset{i.i.d.}{\sim} N(0,1)$. Define $Z=X+Y$. So that $X$ and $Z$ are correlated. I want to calculate:
$$P(XZ>0)$$
The obvious approach is
$$P(XZ>0)=P(X>0)=1/2$$
But this is obviously wrong as the right side equals $1/2$, but the left side has to be larger. Why can't I divide by Z but I can always subtract?
 A: Numerically, the problem is straighforward: 0.75. See the following R code.
X = rnorm(10000000)
Y = rnorm(10000000)
Z = X + Y
mean((X*Z>0))

If you want to get some expressions, then you need to use the total probability law:
$$P(XZ>0) =  P(X^2+XY>0) = P(X^2+XY>0\vert X>0)P(X>0) + P(X^2+XY>0\vert X< 0)P(X<0),$$
You know that $P(X>0)=P(X<0)=0.5$ and $P(X=0)=0$. Now, you need to calculate the conditional probabilities by integrating:
$$P(X^2+XY>0\vert X>0) = P(Y>-X\vert X>0) = \int_0^{\infty}\int_{-x}^{\infty} \phi(y)\phi(x)dx$$
where $\phi$ is the standard normal density. Similarly for the other conditional probability term. There is probably a closed form expression for this, which is left to the reader :).
A: There is no rule of algebra that says (for real numbers $x$ and $z$) that $xz \gt 0$ implies $x \gt 0$, because when $z$ is negative, this would be exactly the wrong conclusion to draw.  Instead, let's draw a picture.
Contours of constant $Z=X+Y$ form lines slanting diagonally down (with slopes of $-1$, or--equivalently--an angle of 45 degrees).  This makes it easy to figure out where $XZ$ is positive, shown in these darker shaded regions:
 
This is the union of two semi-infinite wedges, evidently comprising three-fourths of a circle.  Because the distribution of $(X,Y)$ is symmetric under rotations about the origin, the desired chance must be three-fourths of the total (of 100%), or $3/4$ exactly.
