Integrating a function of density function and distribution function

Question

I was trying to solve the problem of finding $P(X> 4Y)$ when $X$ and $Y$ are standard normal IID RVs. I know the answer is $1/2$ due to the symmetry of the distribution around $0$.

I wanted to solve it in another way as follows:

$$P(X> 4Y) = \int_{y =-\infty}^{\infty}\int_{x=4y}^{\infty}f_{XY}(x,y)dxdy\\ = \int_{y =-\infty}^{\infty}\int_{x=4y}^{\infty}f_X(x)f_Y(y)dxdy\\=\int_{y =-\infty}^{\infty}f_Y(y)[1- F_X(4y)]dy\\=1-\int_{y =-\infty}^{\infty}f_Y(y)F_Y(4y)dy $$

I am unable to write the next step from here. I know if I take $u =F_Y(4y)$, then $du/dy =4f_y(4y)$, $but this doesn't seem to simplify much.

Can someone help me proceed with this? How can I reach to the answer $1/2$ from here.

Answer 1

Okay! Finally, I managed to solve it.

$$P{(X>4Y)}=\int_{y =-\infty}^{\infty}f_Y(y)[1- F_X(4y)]dy\\=1-\int_{y =-\infty}^{\infty}f_Y(y)F_Y(4y)dy $$

Also, $$P{(X>4Y)}=\int_{y =-\infty}^{\infty}f_Y(y)[1- F_X(4y)]dy\\=\int_{y =-\infty}^{\infty}f_Y(-y)F_Y(-4y)dy \\= \int_{y =-\infty}^{\infty}f_Y(y)F_Y(4y)dy$$ (Due to symmetry around $y=0$ being zero mean RVs)

Thus, $$P{(X>4Y)}= 1-P{(X>4Y)} \implies P{(X>4Y)}=1/2.$$

Answer 2

An easier "calculation" starts with the observation that $X-4Y$ is also a zero-mean normal random variable (with variance $17$). Hence $$P(X>4Y) = P(X-4Y > 0) = \Phi(0) = \frac 12.$$