# How to calculate the expected value of bivariate normal distribution?

Let $$X= (X_1,X_2)$$ be a random vector with bivariate normal distribution $$X\sim N(\mu,\Sigma)$$ such that $$X_1$$ and $$X_2$$ are positively correlated and we also have to: $$P(X_1<1) = 0,84134$$, $$P(X_2>6) = 0,02275$$, $$Var[X_1] = 1$$ and $$Var[X_2] = 2$$. How can I calculate the expected value $$E[X_1\mid X_2= 6]$$ and the covariance $$Cov(X_1,X_2)$$?.

I have tried to do the integral according to the expected value and Conditional Variance, but I have not been able to arrive at something concrete. Is there a miraculous property that I don't know about and that works for me?

We know $$X_1$$ and $$X_2$$ have variances $$\sigma_1^2 = 1$$ and $$\sigma_2^2 = 2$$ respectively. $$0.84134 = P(Z < 1)$$ and $$0.02275 = P(Z > 2)$$ for a standard normal random variable $$Z$$. That lets you figure out $$\mu_1 = E[X_1]$$ and $$\mu_2 = E[X_2]$$. But you have given us no information about the covariance $$\Sigma_{12}$$ except that it is positive.
• Thanks for the answer! I forgot to put this: $Var[X1|X2=x2] = 0,75$. I know that $μ1=E[X1]$, but how can it help me to find the value of $E[X1|X2=6]$?. Oct 28, 2020 at 21:56
• The conditional distribution of $X_1$ given $X_2 = x_2$ is normal with mean $\mu_1 + \frac{\sigma_1}{\sigma_2} \rho (x_2 - \mu_2)$ and variance $(1 - \rho^2) \sigma_1^2$, where $\rho = \Sigma_{12}/(\sigma_1 \sigma_2)$ is the correlation coefficient of $X_1$ and $X_2$. See e.g. Wikipedia. Oct 28, 2020 at 23:35