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I am working on a paper in which I'd need to use the two following conditional expectations:

  1. $E(X_{1}|a \leq X_{2} \leq b)$

  2. $E(X_{1}|a \leq X_{2} \leq b, a \leq X_{3} \leq b)$

where $X_{1}, X_{2}, X_{3}$ come from a trivariate normal distribution and $a$ and $b$ are some real numbers ($X_{2}$ and $X_{3}$ have the same bounds).

I have checked several textbooks and references on the web, including those mentioned in previous related posts (e.g., Tallis 1961), but did not find anything relevant. Could anyone help me derive these conditional expectations, or else point me to any good reference(s) I may have missed?

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By definition if $P(A)>0$

$$E(X|A)=\frac{E(X1_A)}{p(A)}$$

so

$$E(X_1|a<X_2<b)=\frac{E(X_1 1_{a<X_2<b})}{p(a<X_2<b)} =\frac{\int_a^b \int x_1 f(x_1,x_2) dx_1 dx_2}{p(a<X_2<b)}$$

similarity define $A=\{X_2\in (a,b)\}$ and $B=\{X_3\in (a,b)\}$

$$E(X_1|a<X_2<b,a<X_2<b)=E(X_1|AB)= \frac{E(X_1 1_{AB})}{p(AB)}=\frac{E(X_1 1_{a<X_2<b} 1_{a<X_3<b})}{p(AB)} =\frac{\int_a^b \int_a^b \int x_1 f(x_1,x_2,x_3) dx_1 dx_2 dx_3}{\int_a^b \int_a^b f(x_2,x_3) dx_2 dx_3}$$

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Let's address the questions in turn.

The first one concerns the bivariate Normal distribution of $(X_1,X_2).$ This is determined by its mean $(\mu_1,\mu_2),$ the variances $\sigma_1\gt 0$ and $\sigma_2\gt 0,$ and the correlation coefficient $\rho = \rho_{12}.$ The theory of least-squares regression in the bivariate Normal setting teaches us that the distribution of $Z_1=(X_1-\mu_1)/\sigma_1$ conditional on $Z_2=(X_2-\mu_2)/\sigma_2$ is Normal with mean $\rho Z_2.$ Since $Z_2$ has a standard Normal distribution, we may directly compute

$$\begin{aligned} E[X_1\mid a \le X_2 \le b] &= E[\mu_1+\sigma_1 Z_1\mid a \le \mu_2 + \sigma_2 Z_2 \le b]\\ &= \mu_1 + \sigma_1\,E[Z_1 \mid (a-\mu_2)/\sigma_2 \le Z_2 \le (b-\mu_2)/\sigma_2]\\ &= \mu_1 + \sigma_1 \,E[\rho Z_2 \mid (a-\mu_2)/\sigma_2 \le Z_2 \le (b-\mu_2)/\sigma_2]\\ &= \mu_1 + \sigma_1\rho \,E[Z_2 \mid (a-\mu_2)/\sigma_2 \le Z_2 \le (b-\mu_2)/\sigma_2]. \end{aligned}$$

Abbreviate things by setting

$$\alpha=(a-\mu_2)/\sigma_2\text{ and }\beta=(b-\mu_2)/\sigma_2$$

for the standardized interval endpoints. Letting $\phi$ be the standard Normal PDF and recalling that $\mathrm{d}\phi(z) = -z\,\phi(z)\mathrm{d}z,$ it is easy to compute this conditional expectation as

$$\begin{aligned} E[Z_2 \mid \alpha\le Z_2 \le \beta] &= \frac{1}{\Phi(\beta)-\Phi(\alpha)}\int_\alpha^\beta z \phi(z)\,\mathrm{d}z \\ &= \frac{1}{\Phi(\beta)-\Phi(\alpha)}\int_\alpha^\beta -\mathrm{d}\phi(z) \\ &= \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}. \end{aligned}$$


This result suggests we should tackle the trivariate problem by first making similar simplifications: that is, standardize the variables and perform linear regression. Generalize the preceding notation to include a third subscript for the third variable. As before,

$$\begin{aligned} E[X_1\mid a_2 \le X_2 \le b_2,\, a_3 \le X_3 \le b_3] &= E[\mu_1+\sigma_1 Z_1\mid a_2 \le \mu_2 + \sigma_2 Z_2 \le b_2, \ldots]\\ &= \mu_1 + \sigma_1\,E[Z_1 \mid (a_2-\mu_2)/\sigma_2 \le Z_2 \le (b_2-\mu_2)/\sigma_2, \ldots]. \end{aligned}$$

(This handles the case where the value of $(X_2,X_3)$ is conditioned on a rectangle rather than a square; it's no more difficult to solve and the notation makes the pattern a little clearer.)

The regression of $Z_1$ on $(Z_2,Z_3)$ is found by finding coefficients $(\gamma_2,\gamma_3)$ that minimize $$E[(Z_1 - \gamma_2Z_2 - \gamma_3Z_3)^2] = 1 - 2\rho_{12}\gamma_2 - 2\rho_{13}\gamma_3 + 2\rho_{23}\gamma_2\gamma_3 + \gamma_2^2 + \gamma_3^2.$$ Assuming $(X_1,X_2,X_3)$ has a nondegenerate distribution, its unique critical point (where the gradient vanishes), which must be the global minimum, therefore occurs when

$$\pmatrix{\rho_{12}\\\rho_{13}} = \pmatrix{1 & \rho_{23} \\ \rho_{23} & 1}\pmatrix{\gamma_2\\\gamma_3}$$

with solution

$$\pmatrix{\gamma_2\\\gamma_3} = \frac{1}{1-\rho_{23}^2} \pmatrix{\rho_{12} - \rho_{23} \rho_{13} \\ \rho_{13} - \rho_{23}\rho_{12}}.$$

Thus $$E[Z_1 \mid (Z_2,Z_3)] = \gamma_2 E[Z_2 \mid (Z_2,Z_3)]+ \gamma_3 E[Z_3\mid (Z_2,Z_3)]$$ and we may proceed as before to find this via integration. Again, to simplify the notation, write

$$\alpha_i = (a_i - \mu_i)/\sigma_i,\ \beta_i = (b_i - \mu_i)/\sigma_i$$

for $i=2, 3,$ and define the standard bivariate Normal conditional expectation functions

$$\begin{aligned} \Psi_i([a,b],\,[c,d]\mid \rho) &= E[Z_i\mid Z_2\in[a,b],\ Z_3\in[c,d]] \\ &= \frac{\iint_{[a,b]\times[c,d]} z_i\,\phi(z_2,z_3\mid \rho)\,\mathrm{d}z_2\mathrm{d}z_3}{\iint_{[a,b]\times[c,d]} \phi(z_2,z_3\mid \rho)\,\mathrm{d}z_2\mathrm{d}z_3} \end{aligned}$$

where

$$\phi(z_2,z_3\mid \rho) = \frac{1}{2\pi(1-\rho^2)} \exp\left(-\frac{z_2^2-2\rho z_2z_3+z_3^2}{2(1-\rho^2)}\right)$$

is the standard bivariate Normal PDF (for variables with correlation $\rho$). Our result (at the top of this section) then is

$$\begin{aligned} &E[X_1\mid a_2 \le X_2 \le b_2,\, a_3 \le X_3 \le b_3] \\ &= \mu_1 + \sigma_1\,E[Z_1 \mid (a_2-\mu_2)/\sigma_2 \le Z_2 \le (b_2-\mu_2)/\sigma_2, \ldots] \\ &= \mu_1 + \sigma_1 \left(\gamma_2 \Psi_2([\alpha_2,\beta_2],[\alpha_3,\beta_3]\mid\rho_{23}) + \gamma_3 \Psi_3([\alpha_2,\beta_2],[\alpha_3,\beta_3]\mid\rho_{23})\right). \end{aligned}$$

The integrals in the definition of the $\Psi_i$ are over a rectangle within the domain of the bivariate Normal distribution of $(Z_2,Z_3).$ Unless $\rho_{23}=0$ (when $Z_2$ and $Z_3$ are independent) the formula for this is messy; unless you really need a formula for further analysis, numerical integration may be most appropriate.

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