# 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.

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}
• Im not aware of the properties of univariate gaussian, will look into them. There was a typo in my deductions, apologies for that! Thanks for the hints, really appreciate them :) – Timothy May 2 '17 at 3:53
• In order to leverage part (ii) when solving part (iii), the discriminant should look like the product of the random variables in part (ii). I suspect there's a typo somewhere, or maybe I have made some mistake... – angryavian May 2 '17 at 4:15
• You are right, there was indeed a typo in the question – Timothy May 2 '17 at 10:57