hazard for weibullsurv

[Aria]

Hello,

I have a question regarding the hazard function using family= 'weibullsurv'. I derived a formula that differs from what I expected. I have attached my derivation in the following file. 

Thanks for any help!

[Helpdesk (Haavard Rue)]

the implementation follows the usual parameterisation, as discussed in

https://en.wikipedia.org/wiki/Weibull_distribution

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Håvard Rue
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[Aria]

Hi Havard,

Thank you for your reply. I'm wondering if my model is as I specified, i.e. exp(\beta*x) without the power \alpha. Do I need to multiply each coefficient estimate by \alpha?

[Helpdesk (Haavard Rue)]

the INLA implementation puts the covariates into the density, either using
variant=0 or 1. it seems like you want to have to covariates in the hazard
function ???

[Denis Rustand]

You can check the attached code for equivalence of INLA's Weibull variant 0 and 1 with Weibull proportional hazards and Weibull AFT.

[Janet Van Niekerk]

Yes for the hazard ratio's for example you need to multiply the coef with alpha. 

If you want the credible intervals for the hazard ratio's you need to sample from the posteriors, using inla.posterior.sample, and then calculate the product from where you get the posteriors of the hazard ratio's.

This is only for variant = 1. For variant = 0 it is the usual formulation - baseline*  exp(coef*x).

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