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Suppose $Y, X, Z$ are all discrete random variables. Then the conditional mean of $Y$ on levels of $X=x$ and $Z=Z$ is represented by

$E(Y\mid X=x,Z=z)$

now suppose that there are values of $x_1 \in X$ and $z_1 \in Z$ such that both cannot simultaneously exist. An example might be if $Y, X, Z$ were vectors of observations corresponding to people, and so such person had $x_1, z_1$ simultaneously. In such a case,

$E(Y\mid X=x_1,Z=z_1)$

is undefined. Is there a way define the conditional expectation in a notationally correct way? One thought is using the indicator function

$\mathbb{1}(X=x_1,Z=z_1)E(Y\mid X=x_1,Z=z_1)$

but $E(Y\mid X=x_1,Z=z_1)$ is still undefined and multiplying by zero seems to be a hacky fix.

Example:

For an example, consider $Y$ to be mortality, where $Y=1$ is death, $Y=0$ is living. Then, suppose $X$ is gender, such that $X=1$ is female, $X=0$ male and that $Z$ is whether one had ovarian cancer before ($Z=1$).

Then,

$$ E(Y\mid X=0, Z=1) $$

can't be computed because to have ovarian cancer implies one is female, therefore we cannot compute it since $X=0$ is male (making the assumption only females can get ovarian cancer).

Further Motivation:

I am also motivated by how the law of total probability might work under such cases. For example, suppose we wanted to find $E(Y\mid X) but only had information at the level of $X$ and $Z$. We can therefore use the law of total probability to have:

$$ E(Y\mid X) = \sum_z E(Y\mid X,Z=z)P(Z=z\mid X) $$

However, in such cases, sometimes, like the example above, $E(Y\mid X,Z=z)$ is undefined at certain values. How might the law of total probability work in this case?

I would appreciate any insights. Thanks!

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  • $\begingroup$ $\mathbb{E}(W|A)$ can be defined as long as $A$ is an event in the sample space. If $P(A) \neq 0$ then the usual way works, otherwise one has to resort to more sophisticated theory. If $A$ is an impossible event, meaning $A$ is not a subset of $\Omega$ then everything is meaningless. How does one condition on an event that isn’t in the experiment? Nevermind notations—*the event cannot happen*, so how can we condition on it? A less rhetorical question—what led you to wish for this? $\endgroup$ – Nap D. Lover May 10 at 1:47
  • $\begingroup$ I have updated my question with a concrete example. Thank you for response! $\endgroup$ – user321627 May 10 at 2:38
  • $\begingroup$ One way to resolve this is to take $Z$ to be binary on $\{0,1\}$ conditional on $X=1$ and $Z=0$ a.s. conditional on $X=0$ (i.e. deterministic!). Then $\{X=0, Z=1\}$ is in the sample space but has probability zero. In words, conditional on selecting a female from the study, there is a probability $p_Z=\mathbb{P}(Z=1|X=1)$ of them having ovarian cancer, and probability $q_Z=1-p_Z=\mathbb{P}(Z=0|X=1)$ of not having cancer, and conditional on selecting a male, almost surely the male does not have ovarian cancer: $P(Z=0| X=0)=1$. You should be able to see through applying LTE now. What do ya think? $\endgroup$ – Nap D. Lover May 10 at 23:18

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