# Probabilities of Bivariate Normal Distribution

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 $$\Bbb P(X_1< X_2)$$

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

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

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