I'm stuck on this problem, and I dont know how to continue from here. It would be great if someone could just give me a hint of how to proceed from here.
The problem is,
"Let $X,Y,Z$ be independent and $N(1,1)$. Find $P(X+Y \geqq 3Z)$."
Now, the way I see it there are two obvious ways to proceed. First off, one can simply condition on the different variables and because of their independence obtain
$P(X+Y \geqq 3Z) = \displaystyle \int_{-\infty}^{\infty}P(X+Y \geqq 3Z | Z=z)f_Z(z)dz = \int_{-\infty}^{\infty}P(X+Y \geqq 3z)f_Z(z)dz $ $ (1)$
and by doing the exact same reasoning we get $P(X+Y \geqq 3z) = \int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dy$
and thus (with (1))
$P(X+Y \geqq 3Z) = \int_{-\infty}^{\infty}f_Z(z)\int_{-\infty}^{\infty}F_X(3z-y)f_Y(y)dydz$.
This is problematic however, since $X\sim N(1,1)$ means that we cannot express $F_X(3z-y)$ explicitly. If the variables would have had some other distribution with known cdf, the calculation could continue in a straight forward way from here. This is not the case right now, however. I did, however, try to use the fact that $F_X(3z-y) = \int_{0}^{3z-y}f_X(t)dt$. But then I couldnt make sense of the integration limits. So I doubt that this is the right way to go in solving this particular problem.
I was also thinking about using the fact that $X+Y =: W$ then $W\sim N(1+1,1+1)$ and would simplify the calculation above but yield the same problem. Maybe there is a nice trick one can use in calculating the integral above. Either way, I would really appreciate if someone could give me some hints as to how to continue from here.
Thanks!
HINT 1: Recall (or prove) that if $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$ are independent, then $X + Y \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$.
HINT 2: Recall (or prove) that if $X \sim N(\mu, \sigma^2)$ and $a$ is constant, then $aX\sim N(a\mu, a^2\sigma^2)$.
HINT 3: Rearrange the equation $X + Y \geq 3Z$ so there's just one random variable on one side of the equation and $0$ on the other.