$(x_1, x_2, x_3)$~$N(0, \Sigma(\sigma_{ij}))$

then how to calculate $$E[x_2| x_1\leq a, x_3\leq b]$$

  • $\begingroup$ What happened to $x_3$? or is it a 2D problem? $\endgroup$ – Alex C Jan 17 '17 at 3:16
  • $\begingroup$ sorry, I made a mistake, pls see the updated version @Alex C $\endgroup$ – user6703592 Jan 17 '17 at 3:21
  • $\begingroup$ equivalently, I want to know how to solve $E[z_1 | z_1\leq a, z_2+z_1 \leq b]$ if $z_1, z_2$ are iid $N(0,1).$ $\endgroup$ – user6703592 Jan 17 '17 at 3:23

Your expectation is given by $$ \begin{align*} E[x_2 \:|\: x_1 \leq a, x_3 \leq b] &= \int_{-\infty}^\infty x_2 f(x_2 \:|\: x_1 \leq a, x_3 \leq b) \:dx_2 \end{align*} $$

To solve this problem you first need the pdf of $x_2 \:|\: x_1 \leq a, x_3 \leq b$. This is given by $$ f(x_2 \:|\: x_1 \leq a, x_3 \leq b) = \frac{\int_{-\infty}^a\int_{-\infty}^bf_x(x_1,x_2,x_3) \:dx_3dx_1}{P[x_1 \leq a, x_3 \leq b]} $$ where $$ P[x_1 \leq a, x_3 \leq b] = \int_{-\infty}^a\int_{-\infty}^\infty\int_{-\infty}^bf_x(x_1,x_2,x_3)\:dx_3dx_2dx_1 $$ and the joint probability distribution $f_x$ is given by $$ f_x(x_1,x_2,x_3) = \frac{\mathrm{exp}\left(-\frac{1}{2}x^T\Sigma^{-1}x\right)}{\sqrt{(2\pi)^3|\Sigma|}} $$

| improve this answer | |
  • $\begingroup$ so, it can not be represented as the explicit formula of (a,b)? @nonanull $\endgroup$ – user6703592 Jan 17 '17 at 6:19
  • $\begingroup$ There are indeed no analytical solution for this problem. $\endgroup$ – Matthieu Brucher Jan 17 '17 at 16:24
  • 1
    $\begingroup$ @user6703592 No, there's no closed form solution for the integral. However, there are implementations which implement the multivariate normal cdf, which you can use. For example, MATLAB's mvncdf. Then perform a numerical integration of the first integral over $x_2$. $\endgroup$ – msitt Jan 18 '17 at 0:37

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