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I found the information about the likelihood of conditional logistic regression for paired data is few. The Wiki gives this answer, but I think it is wrong because event Yi1 and Yi2 are dependent, because Yi1 + Yi2 =1. (Yi is a binary random variable). Therefore, how do we solve the conditional likelihood based on the assumption of Yi1 + Yi2 = 1? I have not found any proof yet. Thanks.

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Updated on 10May2024

This is my calculation.

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No, Wikipedia is correct. $Y_1$ and $Y_2$ are dependent, but they are dependent only after conditioning on $Y_1+Y_2=1$. You start with $Y_{i1}$ and $Y_{i2}$ independent Bernoulli with $Y_{ij}$ having mean $\mathrm{expit}(\alpha_i+X_{ij}\beta)$ in the population.

At this point you either sample in such a way that $Y_{i1}+Y_{i2}=1$ (matched case-control sampling) or you sample pairs without regard to $Y$ and condition on $Y_{i1}+Y_{i2}=1$ in the analysis (because pairs where that isn't true aren't informative for $\beta$)

Either way, the conditional likelihood for pair $i$ is $P(Y_{i1}=1|Y_{i1}+Y_{i2}=1;\beta)$ or $P(Y_{i1}=0|Y_{i1}+Y_{i2}=1;\beta)$

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  • $\begingroup$ Hello, Thomas. Thank you for your reply. However, I do not quite understand the proof. I drew a flow of my understanding. $\endgroup$
    – Tom Hsiung
    Commented May 10 at 12:16
  • $\begingroup$ You're still treating $A$ and $B$ as dependent. They are independent Bernoulli prior to conditioning, so that eg $P(A=0|B=1)$ is just $P(A=0)$ $\endgroup$
    – Thomas Lumley
    Commented May 12 at 23:55
  • $\begingroup$ Hi, Thomas. Oh! I got your point. Hmmm, without the conditioning of A + B = 1, there should not have been the correlation between A and B. Thank you very much, pal. $\endgroup$
    – Tom Hsiung
    Commented May 17 at 12:19
  • $\begingroup$ Hi, Thomas. Here I have another question. We know that P(A and B) = P(A | B)*P(B) holds true regardless of the status of the correlation between A and B. When we compute P(A and B), how should we determine whether to treat it as independent between A and B, or to treat them as P(A and B) = P(A | B)*P(B). Thx $\endgroup$
    – Tom Hsiung
    Commented Jun 5 at 12:53

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