Estimating logistic regression coefficients in a case-control design when the outcome variable is not case/control status Consider sampling data from a population of size $N$ in the following way: For $k=1, ..., N$ 


*

*Observe individual $k$'s "disease" status

*If they have the disease, include them in the sample with probability $p_{k1}$ 

*If they do not have the disease, include them with probability $p_{k0}$. 
Suppose you observed a binary outcome variable $Y_i$ and predictor vector ${\bf X}_i$, for $i=1, ..., n$ subjects sampled this way. The outcome variable is not the "disease" status. I want to estimate the parameters of the logistic regression model: 
$$ \log \left( \frac{ P(Y_i = 1 | {\bf X}_i) }{ P(Y_i = 0 | {\bf X}_i) } \right)  = \alpha + {\bf X}_i {\boldsymbol \beta} $$
All I care about are the (log) odds ratios, ${\boldsymbol \beta}$. The intercept is irrelevant to me. 
My question is: Can I get sensible estimates of ${\boldsymbol \beta}$ by ignoring the sampling probabilities $\{ p_{i1}, p_{i0} \}$, $i=1, ..., n$ and fitting the model as if it were an ordinary random sample?

I am pretty much certain the answer to this question is "yes". What I'm looking for is a reference that validates this. 
There are two main reasons I'm confident about the answer:  


*

*I've done many simulation studies and none of them contradict this, and 

*It is straightforward to show that, if the population is governed by the model above, then the model governing the sampled data is 
$$ \log \left( \frac{ P(Y_i = 1 | {\bf X}_i) }{ P(Y_i = 0 | {\bf X}_i) } \right)  = \log(p_{i1}) - \log(p_{i0}) + \alpha + {\bf X}_i {\boldsymbol \beta} $$
If the sampling probabilities did not depend on $i$, then this would represent a simple shift to the intercept and the point estimate of ${\boldsymbol \beta}$ would clearly be unaffected. But, if the offsets are different for each person this logic does not quite apply since you will certainly get a different point estimate, although I suspect something similar does.   
Related: The classic paper by Prentice and Pyke (1979) says that logistic regression coefficients from a case-control (with disease status as the outcome) have the same distribution as those collected from a prospective study. I suspect this same result would apply here but I must confess I don't fully understand every bit of the paper. 
Thanks in advance for any comments/references. 
 A: This is a variation of the selection model in econometrics. The validity of the estimates
using only the selected sample here depends on the condition that
$\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)=\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right)$. Here $D_i$ is $i$'s disease status.
To give more details, define the following notations: $\pi_{1}=\Pr\left(D_{i}=1\right)$
and $\pi_{0}=\Pr\left(D_{i}=0\right)$; $S_{i}=1$ refers to the event
that $i$ is in the sample. Moreover, assume $D_{i}$ is independent
of $X_{i}$ for simplicity.
The probability of $Y_{i}=1$ for a unit $i$ in the sample is
\begin{eqnarray*}
\Pr\left(Y_{i}=1\mid X_{i},S_{i}=1\right) & = & \mathrm{{E}}\left(Y_{i}\mid X_{i},S_{i}=1\right)\\
 & = & \mathrm{{E}}\left\{ \mathrm{{E}}\left(Y_{i}\mid X_{i},D_{i},S_{i}=1\right)\mid X_{i},S_{i}=1\right\} \\
 & = & \Pr\left(D_{i}=1\mid S_{i}=1\right)\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1,S_{i}=1\right)+\\
 &  & \Pr\left(D_{i}=0\mid S_{i}=1\right)\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0,S_{i}=1\right),
\end{eqnarray*}
by the law of iterated expecation. Suppose conditional on the disease
status $D_{i}$ and other covariates $X_{i}$, the outcome $Y_{i}$
is independent of $S_{i}$. As a result, we have
\begin{eqnarray*}
\Pr\left(Y_{i}=1\mid X_{i},S_{i}=1\right) & = & \Pr\left(D_{i}=1\mid S_{i}=1\right)\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)+\\
 &  & \Pr\left(D_{i}=0\mid S_{i}=1\right)\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right).
\end{eqnarray*}
It is easy to see that
$$
\Pr\left(D_{i}=1\mid S_{i}=1\right)=\frac{\pi_{1}p_{i1}}{\pi_{1}p_{i1}+\pi_{0}p_{i0}}\mbox{ and }\Pr\left(D_{i}=0\mid S_{i}=1\right)=\frac{\pi_{0}p_{i0}}{\pi_{1}p_{i1}+\pi_{0}p_{i0}}.
$$
Here $p_{i1}$ and $p_{i0}$ are as defined your sampling scheme.
Thus,
$$
\Pr\left(Y_{i}=1\mid X_{i},S_{i}=1\right)=\frac{\pi_{1}p_{i1}}{\pi_{1}p_{i1}+\pi_{0}p_{i0}}\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)+\frac{\pi_{0}p_{i0}}{\pi_{1}p_{i1}+\pi_{0}p_{i0}}\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right).
$$
If $ $$\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)=\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right)$,
we have
$$
\Pr\left(Y_{i}=1\mid X_{i},S_{i}=1\right)=\Pr\left(Y_{i}=1\mid X_{i}\right),
$$
and you can omit the sample selection problem. On the other hand,
if $\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)\neq\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right)$,
$$
\Pr\left(Y_{i}=1\mid X_{i},S_{i}=1\right)\neq\Pr\left(Y_{i}=1\mid X_{i}\right)
$$
in general. As a particular case, consider the logit model,
$$
\Pr\left(Y_{i}=1\mid X_{i},D_{i}=1\right)=\frac{e^{X_{i}'\alpha}}{1+e^{X_{i}'\alpha}}\mbox{ and }\Pr\left(Y_{i}=1\mid X_{i},D_{i}=0\right)=\frac{e^{X_{i}'\beta}}{1+e^{X_{i}'\beta}}.
$$
Even when $p_{i1}$ and $p_{i0}$ are constant across $i$, the resulted
distribution will not keep the logit formation. More importantly,
the intepretations of the parameters would be totally different. Hopefully,
the above arguments help to clarify your problem a little bit.
It is tempted to include $D_{i}$ as an additional explanatory variable,
and estimate the model based on $\Pr\left(Y_{i}\mid X_{i},D_{i}\right)$.
To justify the validity of using $\Pr\left(Y_{i}\mid X_{i},D_{i}\right)$,
we need to prove that $\Pr\left(Y_{i}\mid X_{i},D_{i},S_{i}=1\right)=\Pr\left(Y_{i}\mid X_{i},D_{i}\right)$,
which is equivalent to the condition that $D_{i}$ is a sufficient
statistic of $S_{i}$. Without further information about your sampling
process, I am not sure if it is true. Let's use an abstract notation.
The observability variable $S_{i}$ can be viewed as random function
of $D_{i}$ and the other random variables, say $\mathbf{Z}_{i}$.
Denote $S_{i}=S\left(D_{i},\mathbf{Z}_{i}\right)$. If $\mathbf{Z}_{i}$
is independent of $Y_{i}$ conditional on $X_{i}$ and $D_{i}$, we
have $\Pr\left(Y_{i}\mid X_{i},D_{i},S\left(D_{i},\mathbf{Z}_{i}\right)\right)=\Pr\left(Y_{i}\mid X_{i},D_{i}\right)$
by the definition of independence. However, if $\mathbf{Z}_{i}$ is
not independent of $Y_{i}$ after conditioning on $X_{i}$ and $D_{i}$,
$\mathbf{Z}_{i}$ intuitively contains some relevant information about
$Y_{i}$, and in general it is not expected that $\Pr\left(Y_{i}\mid X_{i},D_{i},S\left(D_{i},\mathbf{Z}_{i}\right)\right)=\Pr\left(Y_{i}\mid X_{i},D_{i}\right)$.
Thus, in the 'however' case, the ignorance of sample selection
could be misleading for inference. I am not very familiar with the
sample selection literature in econometrics. I would recommend Chapter
16 of Microeconometrics: methods and applications' by Cameron
and Trivedi (especially the Roy model in that chapter). Also G. S.
Maddala's classic bookLimited-dependent and qualitative variables
in econometrics' is a systematic treatment of the issues about sample
selection and discrete outcomes.
