# A random censored regression problem?

Assume I have random variables $X$ and $Y$. $X$ is observable; $Y$ is not. We also know that, if $X \leq Y$, $Z = f(X,Y)$ where $Z$ is a third random variable and $f(\cdot)$ is a known function; otherwise $Z$ is missing. $Z$ is also observable (missing or a real number).

Assume I know the distribution of $X$ and I can estimate the distribution of $Z$ from data. I wonder if I can estimate the distribution of $Y$.

My first question is: what is this problem? It is similar to but different from a textbook censored regression problem. Is it a random censored regression problem?

My next question is, is the problem identifiable? I realize that, for any given set of random samples $X_1, X_2, X_3, ...$, there may exist multiple sequences $Y_1, Y_2, Y_3, ...$ that yield the same output $Z_1, Z_2, Z_3,...$. So I guess it will be difficult to find the best solution unless some constraints are imposed. If my guess is right, what are typical constraints people apply?

If we can appropriate apply some constraints to the problem so that we can try to find optimal parameters that minimizes some pre-defined error metric. Hopefully we can "estimate" the distribution of $Y$.

Your suggestions, books, and papers are more than welcome.

Any idea?

Consider these two simple situations:

1. The joint distribution of $(X,Y)$ is uniform on $H^2 = [0,1] \times [0,1]$; $Z=X-Y$.

2. The joint distribution of $(X, \eta)$ is uniform on $H^2$. When $X \le \eta$, $Y=\eta$; otherwise, $Y=1$. Again, $Z=X-Y$.

What is common to both situations are (a) the distribution of $X$ (which is uniform) and (b) the distribution of $Z$ (given that $X \le Y\$). Yet--obviously--the distributions of $Y$ differ radically: in the first case $Y$ is uniform, implying $\Pr[Y=1]=0$, whereas in the second case $\Pr[Y=1]=1/2$.

The point is that we are free to modify $Y$ on the set $Y\gt X$ without changing any of the information in the problem. The information places a (crude) lower bound on the marginal CDF of $Y$, but that's all.

That answers all but the first question. But given that the solution is so indeterminate, I doubt that problems like this (with such limited information) have been investigated, which addresses the first question. To make progress, you have to make (parametric) assumptions about the distribution of $Y$ or the joint distribution of $(X,Y)$.

• (+1) A couple other simple examples to consider to see the wide disparity of information available: (1) $X \sim \mathcal U[0,1]$ and $Y \sim \mathcal U(1,2]$ with $f(x,y) = x$ and (2) $X$ and $Y$ with arbitrary distributions, but $f(x,y) = y$. – cardinal Oct 12 '11 at 14:24
• @cardinal Although it doesn't matter, I specified an $f$ that depends on both arguments rather than just one to make it clear that such degeneracies aren't the crux of the matter. Also, by applying the probability transform, we may as well assume $(X,Y)$ is supported on $H^2$. – whuber Oct 12 '11 at 15:04
• Since, as per the OP, $X$ is observable, then the choice of $Z$ in the answer is equivalent to having all the $(X_i, Y_i)$ pairs on $\{X_i \leq Y_i\}$. Regarding my examples, the first is to show an example in which even in the infinite-data setting, we know next to nothing about the distribution of $Y$. The purpose of the second is that it is essentially equivalent to a standard left-censored experiment if the $(X_i,Y_i)$ are iid and independent of one another, for which standard analysis can be brought to bear. (Actually by always knowing $X_i$, we're in slightly better shape than that.) – cardinal Oct 12 '11 at 16:38
• @whuber thank you for answering the question. I do have an additional question though: you mentioned that the information places a (crude) "lower bound" on the marginal CDF of $Y$". Are you using the world "lower bound" literally here? Or are you basically trying to say that the CDF is subject to constraint imposed by what happens when $X<Y$? – Haining Yu Oct 18 '11 at 15:09
• @Haining I mean it literally: for each value $y$ of $Y$, we are given a minimum value for the CDF $F_Y(y)$. – whuber Oct 18 '11 at 15:31