# Multivariate normal distribution conditional probability question.

$\newcommand{\Cov}{\operatorname{Cov}}$$\newcommand{\Var}{\operatorname{Var}}$$\newcommand{\E}{\mathbb{E}}$$\newcommand{\P}{\mathbb{P}}We have that X and Y are random variables with a multivariate normal density with \E[X]=2, \E[Y]=-3, \Var[X]=4, \Var[Y]=25 and \Cov[X,Y]=-3. And they ask me for \P[X≤3|Y=1]. So what I did first was to get the conditional expectation value for X and the conditional variance. This is what I got:$$\E[X|Y=1]=2+(-3/10)(2/5)(1-(-3))=1.52\Var[X|Y=1]=4(1-(-3/10))=3.64$$Then I got that$Z=(3-1.52)/3.64=0.40659$, and that$\P[X≤3|Y=1]=0.6591$. But the correct answer is$.76. What was my mistake? Here are the formulas that I used to get the variance and the expectation value. \begin{align} \E[X_2\mid X_1=x_1] &= \mu_2 + \rho\frac{\sigma_2}{\sigma_1}(x_1-\mu_1)\\ \Var[X_2\mid X_1=x_1] &= \sigma_2^2(1-\rho^2). \end{align} • You might want to explain how these formulas forE[X|Y=1]$and$Var[X|Y=1]\$ are computed -- then help should come. – Did Jul 23 '16 at 15:52
• post edited with the formula :D – neto333 Jul 23 '16 at 16:12
• Picture unreadable. In general, please avoid pictures. – Did Jul 23 '16 at 16:23
• @Did I am able to read it. Perhaps you need to have your vision checked ;) – Math1000 Jul 23 '16 at 17:42
• thanks @Math1000 for taking your time to edit my post. I was out so i wasn't able to read the comments. – neto333 Jul 23 '16 at 21:03