$\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: $$\E[X|Y=1]=2+(-3/10)(2/5)(1-(-3))=1.52$$


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}

  • 1
    $\begingroup$ Perhaps you divided by the variance instead of the standard deviation while computing $Z$? $\endgroup$ Commented Jul 24, 2016 at 1:54
  • $\begingroup$ As others have pointed out, you divided by variance rather than square root to get Z. I calculated the probability in question as 0.7810, not 0.76. Also note the typo in your formula for $Var[X|Y = 1]$ in which you forgot the square on $(-3/10)$,. even though the result is correct. $\endgroup$ Commented Jul 24, 2016 at 2:00

1 Answer 1


One mistake is, the $Z$-score should be

$$Z = \dfrac{X - \mu}{\sigma} = \dfrac{3 - 1.52}{\sqrt{3.64}} = .7757 $$


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