The probability of y < x * z, if x,y, z independent and normally distributed I'm trying to calculate the probability of $Y < XZ$ if $X \sim N(0, 1), Y \sim N(0, 1), Z\sim N(0, 1)$ are independent.
If I understand correctly the distribution function of $xz$ is calculated in this way:
$$
f(u) = \int^{\infty}_{-\infty}p_x(x)p_y(\frac{u}{x})\frac{1}{|x|}dx = \frac{K_0(|u|)}{\pi},
$$
where $K_0$ is a modified Bessel function of the second kind.
Then we say that for every $y$ we want to find $P(u < y)$:
$$
P(U < Y) = \int_{-\infty}^{y}f(u)du = \int_{-\infty}^{y}\frac{K_0(|u|)}{\pi}du = \text{It does not converge :(}
$$
And then (if the previous step is right) I will just integrate over $y$:
$$
P(XZ < Y) = P(U < Y) = \int^{\infty}_{-\infty}p_y(y)dy\int_{-\infty}^{y}f(u)du
$$
I don't really understand what's wrong here :-(

Conditional on $Z=z$,
$$
P(X<YZ|Z=z) =P(X-Yz<0|Z=z) =1/2
$$
since $X-Yz$ is normally distributed with mean zero.
Hence, using the law of total probability,
$$
P(X<YZ)=E(P(X<YZ|Z))=E(1/2) = 1/2.
$$
