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Unfortunately nobody seem to know the answer to my first question...does anyone know how to compute a conditional expectation on absolute value?

Let $$\boldsymbol y = \begin{bmatrix} \boldsymbol y_{a}^{\top} \\ \boldsymbol y_{b}^{\top} \end{bmatrix} $$

$$\boldsymbol y \sim \mathcal{N}({\boldsymbol 0},{\Sigma_{y}})$$

where $$ \Sigma_{y} = \begin{bmatrix} \Sigma_{aa} & \Sigma_{ab}\\ \Sigma_{ba} & \Sigma_{bb} \end{bmatrix}$$

we know that $E[y_a|y_b]=\Sigma_{ab}\Sigma_{bb}^{-1} \boldsymbol y_b $

does anyone know how to compute the following $E[y_a|abs(y_b)]$ ?

That would be very helpful for me.

Thanks a lot!

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The symmetry of the distribution around $0$ shows that you can reduce the calculation to expectations not involving absolute values: $y_b=c$ and $y_b=-c$ are equally likely when $|y_b|=c$ is given. If you intend $y_a$ and $y_b$ to be vector-valued, a similar approach ought to work (but will be more complicated to compute). – whuber♦ Oct 28 '12 at 20:03
Many thanks for this - but would you be able to elaborate as I’m still struggling with this. A simpler looking version of this formula is in web.mit.edu/wangj/www/pap/LlorenteMichaelySaarWang02.pdf .... where do the terms $$ f_+=exp( -(\Sigma^{-1}_{22})_{yz}y|z| )$$ and $$\frac{f_{+}-f_{-}}{f_{+}+f_{-}} $$ come from? – Matus Nov 25 '12 at 0:09
Look at the argument of $\exp$ in the pdf of the trivariate normal distribution: it has one term that is a multiple of $(x-\beta_{xy}y-\beta_{xz}z)^2$: this determines the (unconditional) expectation of $x$. Among the remaining terms, the only one that changes when $z$ is negated is the coefficient of $yz$. Because this is exponentiated, this change multiplies the PDF by twice the exponential of this coefficient: that's where $f_{+}$ and $f_{-}$ are coming from. The rest follows from the definition of conditional expectation. – whuber♦ Nov 25 '12 at 16:29

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