Denote the observable E ( f$_{z}$x$_{k}$ | z ) by w$_{2}$$_{k}$, k = 1 , ...d. For example 2 Denote here E ( x$_{k}$y | z ) by w$_{2}$$_{k}$, k = 1 , ...d. Thus for all the examples we need to solve the system of convolution equations It is advantageous to consider the functions as generalized functions. The interest is often in distributions of the unobservables and there is no reason to restrict those to be absolutely continuous; density may not necessarily exist but can be represented as a generalized derivative of the distribution function rather than an ordinary function. Since solving the convolution equations is done via Fourier transforms restricting regression functions in Example 2 to have ordinary Fourier transforms excludes binary choice or polynomial regression and can be overcome by using generalized functions. Also, if some variables have singular distributions, or if only some variables are subject to measurement error, or there is a mass point in the error distribution (e.g. in measurement error from surveys with some portion of true responses) 5 g ∗ f = w$_{1}$ ; (6) x$_{k}$g ∗ f = w$_{2}$$_{k}$, k = 1 , ...d. E ( x$_{k}$y | z ) = E ( x ∗ $_{k}$g ( x $^{∗}$) | z ) = ∫ ( z$_{k}$ − u$_{k}$ ) g ( z − u ) f ( u ) du.