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Let $X_1$ and $X_2$ have a bivariate normal distribution with parameters $\mu_1 = \mu_2 = 0, \sigma_1 = \sigma_2 = 1, \rho = 0.5$.
Find the probability that all of the roots of the following equation are real:
$X_1x^2 + 2X_2x + X_1 = 0$

Hint: Answer the following two questions first: Are $X_1 + X_2$ and $X_1 - X_2$ bivariate normally distributed? Are they mutually independent?

My answer to the hint is yes they are bivariate normally distribution (closure property of normal distributions).
I mean I know that $\rho = 0.5$ means that the 2 variables within $X_1$ and $X_2$ themselves are not independent so that means that $X_1$ and $X_2$ are not mutually independent right?

For the roots to be real means, I need to evaluate $P[(2X_2)^2-4X_1^2 \ge 0] = P[4X_2^2-4X_1^2 \ge 0] = P[X_2^2-X_1^2 \ge 0] = P[X_2^2 \ge X_1^2]$
Now since $X_1$ and $X_2$ are the same, therefore the probability evaluates to 0.5? Also I don't get where the hint is suppose to lead me?

Please help and thanks!

Edit:
Change of Variable
Let transformation be $U = X_1 + X_2$ and $V = X_1 - X_2$
Inverse Transformation: $X_1 = \frac{U+V}{2}$ and $X_2 = \frac{U-V}{2}$
Jacobian det = -2
$f_{U,V} = \frac{1}{\left|-2\right|}f_{X1,X2}(\frac{u+v}{2},\frac{u-v}{2})$

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  • $\begingroup$ $X_1$ and $X_2$ are not same They are dependent random variables $\endgroup$
    – Qwerty
    Nov 26, 2016 at 12:33
  • $\begingroup$ but I thought they share the exact same parameters? $\endgroup$
    – matthew.j
    Nov 26, 2016 at 12:34
  • $\begingroup$ Yes, but(by not same , I meant) you cannot say $P[X_1^2>X_2^2]=1/2$ because they are not independent, $\endgroup$
    – Qwerty
    Nov 26, 2016 at 12:37
  • $\begingroup$ oh ok.. erm then how do I evaluate that probability? $\endgroup$
    – matthew.j
    Nov 26, 2016 at 12:39
  • $\begingroup$ Try to understand the hint $\endgroup$
    – Qwerty
    Nov 26, 2016 at 12:51

1 Answer 1

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Without the hint you wont be able to solve the problem.

Check that $U(=X_1+X_2)$ and $V=(X_1-X_2)$ follow bivariate normal with parameters $(0,0,2\times1.5=3,2\times0.5=1,0 )$

So $$P[X_1^2-X_2^2>0]=P[UV>0]=P[U>0,V>0]+P[U<0,V<0]$$

Now you need to find the joint distribution of $U$ and $V$ .Use the bivariate normal distribution.$$f_{U,V}(u,v)={1\over 2\pi\sqrt{|\Sigma|}}e^{1\over 2}\{(u \ v)\Sigma^{-1}(u\ v)^T\}$$

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  • $\begingroup$ how did you get those parameters? $\endgroup$
    – matthew.j
    Nov 26, 2016 at 13:14
  • $\begingroup$ Var$(U)=2\sigma^2(1+\rho)$ and Var$(V)=2\sigma^2(1-\rho)$ $\endgroup$
    – Qwerty
    Nov 26, 2016 at 13:22
  • $\begingroup$ I add some more workings. Is it something like this? $\endgroup$
    – matthew.j
    Nov 26, 2016 at 14:12
  • $\begingroup$ @matthew.j Absolutely. Only $\rho=0$ will make calculations easier! $\endgroup$
    – Qwerty
    Nov 26, 2016 at 14:26
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    $\begingroup$ Var$(X_1+X_2)=$ Var$(X_1)$+Var$(X_2)$+$2$Covar$(X_1,X_2)=\sigma_1^2+\sigma_2^2+2\rho \sigma_1\sigma_2\\$. There you go! $\endgroup$
    – Qwerty
    Nov 26, 2016 at 14:35

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