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$\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}

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    $\begingroup$ You might want to explain how these formulas for $E[X|Y=1]$ and $Var[X|Y=1]$ are computed -- then help should come. $\endgroup$ – Did Jul 23 '16 at 15:52
  • $\begingroup$ post edited with the formula :D $\endgroup$ – neto333 Jul 23 '16 at 16:12
  • $\begingroup$ Picture unreadable. In general, please avoid pictures. $\endgroup$ – Did Jul 23 '16 at 16:23
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    $\begingroup$ @Did I am able to read it. Perhaps you need to have your vision checked ;) $\endgroup$ – Math1000 Jul 23 '16 at 17:42
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    $\begingroup$ thanks @Math1000 for taking your time to edit my post. I was out so i wasn't able to read the comments. $\endgroup$ – neto333 Jul 23 '16 at 21:03

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