Propensity Score can be used as a covariate in regression? I have treated and control groups with a problem of selection in the treatment group. 
I am interested in the identification of the following model: $y= exp(X^\prime\beta + \alpha\cdot T)$ where $T$ denotes the assignment to the treatment. 
To address the problem of selection, I was planning to estimate the propensity score, i.e., $P(T=1|X)$, in a separate regression and use the predicted values as the only control in the regression (eventually as a set of dummies for its quantiles). 
Would this strategy work? Or should I rather rely on more complicated matching estimators?
 A: This would be the standard propensity score estimator. For a binary treatment the conditional independence assumption (CIA) states that
$$
\newcommand\independent{\protect\mathpalette{\protect\independenT}{\perp}}
\def\independent#1#2{\mathrel{\rlap{$#1#2$}\mkern2mu{#1#2}}}
T_i\perp\hspace{-0.28cm}\perp (Y_{i0}, Y_{i1})|X_i
$$
i.e. the treatment is independent of the outcomes conditional on the observed covariates. If you are looking for the average treatment effect (ATE), the estimator would be
$$\widehat{ATE} = \frac{1}{n}\sum^n_{i=1}\frac{Y_i(T_i-\widehat{P}(X_i))}{\widehat{P}(X_i)(1-\widehat{P}(X_i))}$$
because you need to do some weighting of treated and non-treated observations. For instance, under the CIA $E(Y_i|T_i=1, X_i = x) = E(Y_1|X_i = x)$, so observations with $T_i = 1, X_i = x$ are representative for all observations with $X_i = x$. However, for recovering $E(Y_1)$ from $E(Y_i|T_i=1,X_i=x)$ you need to weight the observations in the cell $X_i=x$ by $P(X_i=x)$ which is their weight in the total population. In that sense, the above propensity score estimator will give you the ATE by weighting the mean outcome for the treated and non-treated in order to take their difference like
$$\begin{align}
\newcommand\given[1][]{\:#1\vert\:}
E(Y_{i1}-Y_{i0}) &= E\left(Y_i\frac{P(T_i=1)}{P(X_i)}\given[\huge]\normalsize T_i=1\right) - E\left(Y_i\frac{1-P(T_i=1)}{1-P(X_i)}\given[\huge]\normalsize T_i=0\right) \newline
&= E\left( \frac{Y_iT_i}{P(X_i)} - \frac{Y_i(1-T_i)}{1-P(X_i)} \right) \newline
&= E\left( \frac{Y_i(T_i-P(X_i))}{P(X_i)(1-P(X_i)}\right)
\end{align}$$
which is the population equivalent to the estimator given above. So simply including the propensity score in a regression will not do the appropriate weighting. It will also be easier when you consider a log transform of your model. When you do the probit model to get $\widehat{P}(X_i)$ you should also include some polynomials of $X_i$ in that regression in order to capture also non-linear effects of the covariates on treatment choice.
