Find the probability that the following equation has real roots Let $(X_1,X_2$) have bivariate normal distribution with means 0, variance $\sigma_1^2, \sigma_2^2$ respectively, and with correlation coefficient $-1<\rho <1$. 
(i) Determine the distribution of $aX_1 + bX_2$, where $a,b \in \mathbb R $, such that $a^2 + b^2> 0$.
(ii) Find constants $b$ such that $X_1 + bX_2$ is independent of $X_1$.
(iii) Find the probability that the following equation has real roots:
$$X_1 x^2 -2X_1x - bX_2=0 $$ where b is the constant found in part (ii)
Attempt: 
(i) Using the transformation $U=aX_1, V = bX_2$, we have the joint distribution $$f_{U,V}= \frac{1}{2\pi \sigma_1 \sigma_2 a b \sqrt{1-\rho^2}}exp(\frac{-1}{2(1-\rho^2)}[\frac{u^2}{\sigma_1^2}+\frac{v^2}{\sigma_2^2}-\frac{2\rho uv}{ab\sigma_1 \sigma_2}]) $$
(ii) Not sure how to approach this, $b=0$ is a trivial solution, but are there any other solutions?
edit: $b=0$ is not allowed since $a^2 + b^2> 0$
(iii) Equation has real roots if $4X_1^2 + 4bX_1X_2\ge 0 $, which simplifies to $X_1(X_1+bX_2) \ge 0$,so we are interested in $P(X_1(X_1+bX_2) \ge 0 )$ which can be evaluated with a double integral. 
 A: Part (i).
You can avoid working with the density by looking at properties of the Gaussian distribution.


*

*Hint: $aX_1 + bX_2$ is (univariate) Gaussian. (Why?)


 This is an important property of the bivariate Gaussian distribution, and more generally of the multivariate Gaussian distribution.


*Given the previous hint, it suffices to find the mean and variance of $aX_1 + bX_2$, call them $\tilde{\mu}$ and $\tilde{\sigma}^2$. Then $aX_1 + bX_2 \sim N(\tilde{\mu}, \tilde{\sigma}^2)$.


Part (ii).


*

*$(X_1+b X_2, X_1)$ is also bivariate Gaussian (why?),  so it suffices to compute $\text{Cov}(X_1 + bX_2, X_1)$ and check which values of $b$ make it zero.


 $\text{Cov}(X_1 + bX_2, X_1) = \sigma_1^2 + b \rho \sigma_1 \sigma_2$

Part (iii).


*

*If you want to show $X_1(X_1 + bX_2) \ge 0$, then note that by part (b), $X_1$ and $X_1 + bX_2$ are independent.
\begin{align}
P(X_1(X_1 + bX_2) \ge 0) &= P(X_1 \ge 0) P(X_1+bX_2 \ge 0) + P(X_1 \le 0)P(X_1+bX_2 \le 0)\\
&= \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}.
\end{align}

