# Conditional Maximum Likelihood Estimation with Subsample

Suppose that we have an $$i.i.d$$ sample, $$\{Y_i,X_i\}_{i=1}^N$$, and a correctly specified conditional density of $$Y$$ given $$X$$, $$f(Y|X; \theta)$$, where $$\theta$$ is the parameters of the density.

Then, the MLE theory show that the true value of the parameters, $$\theta_0$$, maximize the expected value, $$\mathbb{E}\left[f(Y|X;\theta)\right]$$.

That is, $$\theta_0=\arg\max_\limits{\theta\in\Theta} \mathbb{E}\left[f(Y|X;\theta)\right]$$, where $$\Theta$$ is the compact parameter space.

Therefore, we use the maximum likelihood estimator, $$\widehat{\theta}_{MLE}=\arg\max_\limits{\theta\in\Theta}N^{-1}\sum_{i=1}^Nf(Y_i|X_i;\theta)$$.

Here, I am wondering what will happens if we use only a subsample that has a specific value of a random variable: $$\{Y_i,X_i:D_i=1\}$$, where $$D$$ is a random variable indicating whether $$i$$ is included in the subsample or not.

In particular, the resulting M-estimator will be $$\widehat{\theta}_{sub}=\arg\max_\limits{\theta\in\Theta}\sum_\limits{i:D_i=1}f(Y_i|X_i;\theta)$$.

In my opinion, $$\widehat{\theta}_{sub}$$ may converge to a parameter value $$\theta_*=\arg\max_\limits{\theta\in\Theta}\mathbb{E}\left[f(Y|X;\theta)|D=1\right]$$.

Since $$\mathbb{E}\left[f(Y|X;\theta)\right]=\Pr[D=1]\mathbb{E}[f(Y|X;\theta)| D=1]+\Pr[D=0]\mathbb{E}[f(Y|X;\theta)| D=0]$$, there is no reason $$\theta_*$$ is in fact the true value $$\theta_0$$.

Then, what is the condition that is required for $$\theta_0=\theta_*$$?

It is clearly sufficient that $$Y$$ is independent of $$D$$ given $$X$$, since then $$E[f(Y|X;\theta)|D]=E[f(Y|X;\theta)]$$ and $$\arg\max_\theta E[f(Y|X;\theta)|D]=\arg\max_\theta E[f(Y|X;\theta)]$$
A weaker condition will depend on the model. For example, if the model were a classical linear regression model with known variance, $$Y\sim N(X\beta,1)$$, it would work to choose $$D$$ so that the distributions $$Y|X=x$$ were all symmetrically trimmed of outliers. But while that would give $$\beta_0=\beta^*$$ in this model, it would no longer work after expanding the model to $$Y\sim N(X\beta,\sigma^2)$$ since $$\sigma^2$$ would be underestimated.
You can see from this example that preserving $$\beta^*$$ requires the distribution of $$D$$ to depend on $$\beta$$, not just on $$Y$$ and $$X$$ -- the observations you are removing are those with large values of $$|Y-X\beta|$$. If you assume that the relationship between $$(X,Y)$$ and $$D$$ does not depend on $$\theta$$ and you require $$\theta_0=\theta^*$$ for all values of $$\theta_0\in \Theta$$ then I think it's necessary to have $$D$$ (asymptotically) independent of $$Y$$ given $$X$$