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Say we want to fit some model to predict $\mathbf{E}(A | B)$, which is the expected value for some distribution (ex. Poisson).

What would be the benefit/loss of fitting this vs. computing $\mathbf{E}(A | C) * P(C | B)$ for some variable $C$, where we can fit both $\mathbf{E}(A | C)$ and $P(C | B)$ separately? Intuitively, I think if we get unbiased estimates for $\mathbf{E}(A | C)$ and $P(C | B)$, we should get an unbiased estimate for $\mathbf{E}(A | B)$. However, the variance in predictions may be significantly different, considering how well $\mathbf{E}(A | C)$ and $P(C | B)$ are fit and what each's variance and accuracy is.

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    $\begingroup$ I wonder what you mean, exactly, by "$P(C\mid B)$" and -- assuming it's what it looks like, which would be the full probability distribution of $C$ conditional on $B$ -- how you might possibly hope to obtain an unbiased or even reasonably precise estimate of it. Could you give an example? $\endgroup$
    – whuber ♦
    Feb 23 at 17:42
  • $\begingroup$ Let's say B represents if a person is a male, and C represents if a person drives a car, and A represents if a person bought a ticket. Then in our data table, P(C | B) would be the probability that a person drives a car given they are male. We can fit a logistic regression model for this value? $\endgroup$
    – Victor M
    Feb 23 at 19:04
  • $\begingroup$ You can, but the result is not something you can reasonably supply as an explanatory variable to a logistic regression, because (1) the result is a probability, not a category like drives/does not drive a car and (2) it likely has substantial random error (from the initial regression), whereas inputs to a logistic regression should be definite values or at least measured with inconsequential error. Why not directly model $A$ as a function of $B$ and $C$ together? $\endgroup$
    – whuber ♦
    Feb 23 at 22:17

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