# Fitting the conditional expectation?

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.

• 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?
– whuber
Feb 23 at 17:42
• 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? Feb 23 at 19:04
• 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?
– whuber
Feb 23 at 22:17