0
$\begingroup$

I have the following normal distribution that all of the parameter are known $$\begin{pmatrix} X_1\\ X_2\\ \end{pmatrix} \sim N\left[\begin{pmatrix} \mu_1 \\ \mu_2 \\ \end{pmatrix},\begin{pmatrix}\sigma^2 & \rho \\ \rho & \sigma^2\\\end{pmatrix} \right]$$

$1.$ $\Bbb P(X_1\le\mu_2) = \phi(\frac{\mu_2 - \mu_1}{\sigma})$ is that correct?

$2.$ How do I calculate $\Bbb P(X_1< X_2)$

$\endgroup$

1 Answer 1

1
$\begingroup$
  1. Correct!

  2. It is easy to verify that the marginal distributions are

$X_1\sim N(\mu_1;\sigma^2)$

$X_2\sim N(\mu_2;\sigma^2)$

with $Cov(X_1;X_2)=\rho$

Then it is immediate to calculate

$\mathbb{P}[X_1<X_2]=\mathbb{P}[X_1-X_2<0]$

via distribution of $Z=X_1-X_2$ that is known

$\endgroup$
5
  • $\begingroup$ so $X_1 - X_2 \sim N(\mu_1 - \mu_2 ;\rho \sigma^2)$ ? $\endgroup$
    – MC1325
    Jun 26, 2020 at 12:00
  • $\begingroup$ Why is the variance $\rho \sigma^2$ and not $\rho $? $\endgroup$
    – MC1325
    Jun 26, 2020 at 12:02
  • 1
    $\begingroup$ @MC 1325 : sorry, it was a typo...I amend my answer but the procedure is the same $\endgroup$
    – tommik
    Jun 26, 2020 at 12:16
  • $\begingroup$ thank you!. is the variance of $X_1 + X_2$ is also $\rho$? $\endgroup$
    – MC1325
    Jun 26, 2020 at 12:21
  • 1
    $\begingroup$ @MC1325 : the variance of $X_1\pm X_2=2(\sigma^2\pm\rho)$ $\endgroup$
    – tommik
    Jun 26, 2020 at 12:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .