$\mathbb{P}(X\le2Y)$ for independent normal distribution Let $ X,Y\sim\mathcal{N}\left(8,10\right) $
$X,Y$ are independent.
Find $\mathbb{P}(X\le 2Y)$.
If these were discrete variables I would do :$$ \mathbb{P}\left(X\le2Y\right)=\mathbb{P}\left(X\le t\cap2Y=t\right)=\mathbb{P}\left(X\le t\right)\cdot\mathbb{P}\left(Y=\frac{t}{2}\right) $$
In this case I cannot do it as the probability of a continuous random variable at any point is $0 $ .
I thought about subtracting them and letting $ Z=2Y-X $ and calculating $$ \mathbb{P}(0\le Z) = 1-\mathbb{P}(Z<0)=1-\Phi(0)$$
But even if I did it , I am unsure how to proceed,especially considering I don't know $Z's$ distribution.
 A: Follow the hint provided by @Henry. Let $X,Y\sim \mathcal{N}(\mu,\sigma^2)$. By using characteristic functions and independence (recall if $X\perp Y$ then $E[f(X)g(Y)]=E[f(X)]E[g(Y)]$ for Borel $f,g$), we obtain
$$E[e^{ib(X-2Y)}]=E[e^{ibX}]E[e^{(-2b)iY}]=e^{ib\mu-\frac{b^2\sigma^2}{2}}e^{-2ib\mu-\frac{4b^2\sigma^2}{2}}=e^{ib(\mu-2\mu)-\frac{b^2}{2}(\sigma^2+4\sigma^2)}$$
Therefore $X-2Y\sim \mathcal{N}(-\mu,5\sigma^2)$. So
$$P(X-2Y\leq 0)=P\bigg(\underbrace{\frac{(X-2Y)+\mu}{\sqrt{5}\sigma}}_{\sim\mathcal{N}(0,1)}\leq \frac{\mu}{\sqrt{5}\sigma}\bigg)=\Phi\bigg(\frac{\mu}{\sqrt{5}\sigma}\bigg)$$
Equivalently, $2Y-X=-(X-2Y)\sim \mathcal{N}(\mu,5\sigma^2)$ thus
$$P(2Y-X\geq 0)=P\bigg(\underbrace{\frac{(2Y-X)-\mu}{\sqrt{5}\sigma}}_{\sim \mathcal{N}(0,1)}\geq -\frac{\mu}{\sqrt{5}\sigma}\bigg)=1-\Phi\bigg(-\frac{\mu}{\sqrt{5}\sigma}\bigg)=\Phi\bigg(\frac{\mu}{\sqrt{5}\sigma}\bigg)$$
A: A sum $Z = X + (-2Y)$ of two independent random variables has itself a normal distributions. Your only mistake is operation with variances. Namely,
$$
\Bbb EZ = \Bbb E X - 2\Bbb EY = -8
$$
you have computed correctly. Yet,
$$
\Bbb VZ = \mathrm{Cov}(X - 2Y, X - 2Y) = \mathrm{Cov}(X,X) - 4\cdot\mathrm{Cov}(X, Y) + 4\cdot \mathrm{Cov(Y,Y)}  =\Bbb V X + 4\Bbb V Y
$$
since the term $\mathrm{Cov}(X, Y)$ vanishes due to independence. That is, in general
$$
\sigma^2_{aX + bY} = a^2\sigma^2_X + b^2\sigma^2_Y
$$
unlike the formula for variances you've written in this comment of yours.
This way you should get $Z \sim \mathcal N(-8, 10\sqrt 5)$ and you are asked to compute $\Bbb P(Z \leq 0)$. I assume that in the OP $\sigma_X = \sigma_Y = 10$, though I'm more familiar with the notation $\mathcal N(\mu, \sigma^2)$.
