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 $\theta_0=\theta_*$?


1 Answer 1


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 $X$

  • $\begingroup$ Thank you for your contribution. $\endgroup$ Commented Jun 27, 2023 at 10:37

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