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Thread: Smearing Transformation

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    Smearing Transformation




    I've found several papers on the smearing transformation for ln(x). However, I"m curious if there is one for ln\frac{x}{1-x}. I can't find anything, but thought I would ask here before giving up.

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    Re: Smearing Transformation

    I can't answer your question, but I can give some insights into smearing estimators and why they are desirable, for those who aren't familiar with it. Consider we have a model of the following form:

    ln(y_i)=\alpha+\beta x_i+\varepsilon_i.

    If this is the true model, we can attain unbiased estimates of \alpha and \beta, a=\hat{\alpha} and b=\hat{\beta}. Then we can get unbiased predictions of ln(y_j) as follows:

    \hat{ln(y_j)}=a+bx_j. It is easy to show that \hat{ln(y_j)} is unbiased,

    E[\hat{ln(y_j)}]=E[a+bx_j+e_i]=E[a]+E[b]x_j+E[e_i]=\alpha+\beta{x_j}

    Since we assume that E[\varepsilon_j]=0 \forall j this term vanishes. But what happens if we would want an unbiased prediction of y_j instead of its logarithm? Due to Jensen's inequality, we have that g(E[X]) \leq E[g(X)]. So we do not get an unbiased estimate of y_j by taking e^{ln(y_j)}=e^{a+bx_j}. Now, due to Jensen's inequality,

    E[e^{a+bx_j+\varepsilon_j}]=E[e^{a+bx_j}]E[e^{\varepsilon_j}] \neq E[e^{a+bx_j}]

    since E[e^{\varepsilon_j}] \geq e^{E[\varepsilon_j]}=1. So this is the reason to use smearing estimators, such as Duan's smearing estimator and other similar techniques.

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    Re: Smearing Transformation

    Thanks - I was debating whether or not to explain it - I got lazy.

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    Re: Smearing Transformation


    You could run a simple simulation to get a better grip of the problem. Estimate a model, solve for x and repeat simulation M times.

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