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I have a coxme object: (Age, status) ~ F + X1 + X2 + (1|R1) + (1|R2) + (1|R3) Where F is an individual's level of inbreeding, X1 & X2 are covariate factors, and R1, R2 & R3 are random factors denoting non-independence between individuals, and breeding facility.

I understand that the coxme output gives me the % change in the rate of the event occurring due to inbreeding increase, but I'd like to get point estimates of survival to the 5th year of age for each inbreeding level from this model. With this, I'd like to calculate the effect of increasing inbreeding on survival to a specific point in time (giving us a rate which can be fed into population models for this species).

Seeing as there is no simple method to extract this data from acoxme object (e.g., survfit), is there a process by which I can extract it manually for specific points in time from my coxme fit?

Alternatively, would it be more appropriate to extract this data from Kaplan Meier curves for each inbreeding strata of the model without the random factors (I don't feel like it would be accurate, but I'm not 100% sure)?

My cox models may not be appropriate to use in determining this rate, but it's something that I've worked with a lot so far and I'd like to know if it is at all possible.

Any discussion, comments, advice would be greatly appreciated, Thanks.

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  • $\begingroup$ A coxme object is a frailty model, a cox proportional hazards model with a random intercept. The subtleties around estimating survival curves from Cox models (prediction at the means) and frailty models have been discussed elsewhere on the site. $\endgroup$
    – AdamO
    Commented Jan 18, 2018 at 18:00

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In the documentation for the coxme package (http://cran.r-project.org/web/packages/coxme/coxme.pdf), there is a predict function, predict(object, newdata, type = c("lp", "risk")) that returns the linear prediction or risk at each time period in the model.

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    $\begingroup$ You can't estimate the survival function with the linear predictor. $\endgroup$
    – AdamO
    Commented Jan 18, 2018 at 18:02

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