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Suppose I have random variable $Z$ which is normally distributed $N(\mu,\sigma^2)$. I also have $Z_E=Z+aE$ and $Z_p=Z+bH$ were $E$ and $H$ are $N(0,1)$.

How can I find the expectation of a function of $Z$ given that $Z_E$ and $Z_p$ are greater than a certain threshold i.e. $\mathbf{E}[f(Z)|Z_p>a_1, Z_E>a_2]$. Thank you.

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2 Answers 2

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(Hint) Write the joint distribution $f_{Z, Z_e, Z_p}$ and the marginal distribution $f_{Z_e, Z_p}$ (both are multivariate normal), then $$ E \left[ f \left( Z \right) |Z_p > a_1, Z_e > a_2 \right] = \frac{\int_{- \infty}^{+ \infty} \int_{a_1}^{\infty} \int_{a_2}^{\infty} f \left( z \right) f_{Z, Z_e, Z_p} \left( z, z_e, z_p \right) \mathrm{d} z_e \mathrm{d} z_p \mathrm{d} z}{\int_{a_1}^{\infty} \int_{a_2}^{\infty} f_{Z_e, Z_p} \left( z_e, z_p \right) \mathrm{d} z_e \mathrm{d} z_p} $$

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By definition, assuming without loss of generality that $a$ and $b$ are positive, $$\mathbb P(Z-bH\geqslant v,Z-aK\geqslant u\mid Z)=\mathbb P(H\leqslant (Z-v)/b,K\leqslant (Z-u)/a\mid Z), $$ hence $$\mathbb P(Z\pm bH\geqslant v,Z\pm aK\geqslant u)=\mathbb E(\Phi((Z-v)/b)\Phi((Z-u)/a)). $$ Likewise, assuming without loss of generality that $Z$ is standard normal, $$ \mathbb E(g(Z)\mid Z+bH\geqslant v,Z+aK\geqslant u)=\frac{\mathbb E(G(Z)\Phi((Z-v)/|b|)\Phi((Z-u)/|a|))}{\mathbb E(\Phi((Z-v)/|b|)\Phi((Z-u)/|a|))}. $$

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