Derivation Of Mean-Squared Prediction Error For Lognormal Kriging

Question

I wonder how that equation was derived from $$E(\varepsilon(s_0))^2=\mu_\varepsilon^2(exp(\sigma^2_Y) -1)$$ for $\varepsilon=Z(s_0)-\hat p_Z(Z;s_0)$ and $Y=log \varepsilon$.

Answer 1

I've tried to figure out my answer. Let's say $Y=\ln Z$ and $\hat p_Y=\ln \hat p_Z$.

$E(Z(s_0))=v_Z$ and $E(\hat p_Z)=v_(\hat p_Z)$

$$E(Z(s_0)-\hat p_Z(Z;s_0))^2=\sigma_Z^2-2cov(Z,\hat p_Z)+var(\hat p_Z) $$ $$=v_Z^2(exp(\sigma_Y^2)-1)-2v_Zv_(\hat p_Z)(exp(\sigma_(Y\hat p_Y)-1)+v^2_(\hat p_Z)(exp(var(\hat p_Y))-1)$$

To satisfy unbiasedness assumption, so $E(Z(s_0)=E(\hat p_Z(Z;s_0))=v$. $$E(Z(s_0)-\hat p_Z(Z;s_0))^2=v^2(exp(\sigma_Y^2)-1)-2v^2(exp(\sigma_(Y\hat p_Y)-1)+v^2(exp(var(\hat p_Y))-1)$$ $$=v^2(exp(\sigma_Y^2)-1-2exp(\sigma_(Y\hat p_Y)+2+exp(var(\hat p_Y))-1)$$ $$=v^2(exp(\sigma_Y^2)-2exp(\sigma_(Y\hat p_Y)+exp(var(\hat p_Y)))$$ $$=(exp(2\mu_Y+\sigma^2_Y(s_0))^2(exp(\sigma_Y^2)-2exp(\sigma_(Y\hat p_Y)+exp(var(\hat p_Y)))$$