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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?

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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.

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  • $\begingroup$ 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 $E[X1|X2=6]$?. $\endgroup$
    – Johan C
    Oct 28, 2020 at 21:56
  • $\begingroup$ 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. $\endgroup$
    – Robert Israel
    Oct 28, 2020 at 23:35

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