One random variable bigger/smaller than another independent Suppose $X$ and $Y$ are independent. To be concrete, suppose $X$ is $N(0,a)$ and $Y$ is $N(0,b)$. For any scalar c,d with c < d, is there a way to bound the following probability from above in terms of a,b,c,d?
$P(cX<Y<dX)$
 A: Here are some ideas for an explicit computation of the integral. Define $\Delta:=\{(x,y)\in\mathbb R^2,cx\lt y\lt dx\}$. Then 
$$P(cX\lt Y\lt dX)=(2\pi ab)^{-1}\iint_\Delta\exp\left(-\frac{y^2}{2b^2}\right)\exp\left(-\frac{x^2}{2a^2}\right)\mathrm dy\mathrm dx.$$
The inner integral is $\int_{cx}^{dx}\exp\left(-\frac{y^2}{2b^2}\right)\mathrm dy$, which can be rewritten as $x\int_c^d\exp\left(-\frac{x^2t^2}{2b^2}\right)\mathrm dt$ after a substitution. 
Then switch the integrals.
A: Now that the homework is past due, here is the solution suggested in @whuber's
comments.
Assuming that $a$ and $b$ are the standard deviations of the independent
zero-mean normal random variables, that $c < d$ as the OP says,
$$\begin{align}
P\{cX < Y < dX\} &= P\left\{\frac cb X < \frac Yb < \frac db X\right\}\\
&= P\left\{\left(\frac {ac}{b}\right) \frac Xa < \frac Yb < \left(\frac {ad}{b}\right) \frac Xa\right\}\\
&= P\left\{\alpha \hat{X} < \hat{Y} < \beta \hat{X}\right\}
\end{align}$$
where $\alpha = ac/b < \beta = ad/b$ and $\hat{X}$ and $\hat{Y}$ are
independent standard normal random variables. This is the probability that
the random point $\left(\hat{X}, \hat{Y}\right)$ lies above the line
$ y = \alpha x$ and below the line $y = \beta x$ in the $x$-$y$ plane,
which is a wedge-shaped region between the two lines of slopes $\alpha$
and $\beta$. But, since
$\left(\hat{X}, \hat{Y}\right)$ have a circularly symmetric joint
distribution, this probability is just
$$P\{cX < Y < dX\} = P\left\{\alpha \hat{X} < \hat{Y} < \beta \hat{X}\right\}
 = \frac{\arctan(\beta) - \arctan(\alpha)}{2\pi}
= \frac{\arctan(ad/b) - \arctan(ac/b)}{2\pi}.$$
Note that the probability cannot exceed $\frac 12$.
