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Now $p_{\theta ,\eta }(x,y)=v(x)f(y-g_{\theta }(x))$. With $h_{\theta }={\partial }_{\theta }{g}_{\theta }$ and the location score $\rho (e)=-f^{\prime }(e)/f(e)$, differentiation of the log-likelihood givesTake the usual regularity conditions $h_{\theta }\in L^2(v)$ and $\rho \in L^2(f)$ so that this score function belongs to L2 space.

For the specified nuisance statistical path, differentiating $\log (1+t\gamma (e))$ at zero gives the score function $\gamma (e)$. Normalization requires $E_f\gamma =0$. There is a second constraint: the statistical path must preserve the model's zero error expected value. ThusThis constraint follows from being a path through the model, rather than from normalization alone. For a centered symmetric logistic distribution, for example, $\gamma (e)=\tanh (e)$ is bounded and has zero expected value, but $E[\epsilon \tanh (\epsilon )]>0$; it would not preserve the mean.

For any $\psi \in L^2(v)$, independence impliesAll these products are integrable by the Cauchy-Schwarz inequality, since $\epsilon \psi (X)$ is square-integrable. Thus every such error-density score function is orthogonal to every $e\psi (x)$. Covariate-density nuisance score functions $b(X)$ are also orthogonal to these functions, since $E\epsilon =0$. This is the mean-preserving error tangent space orthogonality used in the final calculation.