# Smearing Transformation

#### Jrb599

##### New Member
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.

#### Englund

##### TS Contributor
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,

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

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.

#### Jrb599

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

#### Englund

##### TS Contributor
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.