Expected value command R and JAGS

Question

Assuming this ís my Bayesian model, how can i calculate the expected value of my Weibull distribution? Is there a command for finding the expected value of the Weibull distribution in R and JAGS? Thanks

model{  
#likelihood function  
for (i in 1:n)   
    {  
        t[i] ~ dweib(v,lambda)#MTBF    

        }    

#Prior for MTBF  
v ~ dgamma(0.0001, 0.0001)   
lambda ~ dgamma(0.0001, 0.0001)     
  }  

  #inits
list(v=1, lambda=1,mu=0,tau=1)

#Data
list(n=10, t=c(5.23333333,8.95,8.6,230.983333,1.55,85.1,193.033333,322.966667,306.716667,1077.8)

Answer 1

The mean, or expected value, of the Weibull distribution using the moment of methods with parameters v and lambda, is:

lambda * Gamma(1 + 1/v)

JAGS does not have the Gamma function, but we can use a work around with a function that is does have: logfact. You can add this line to your code and track the derived parameter exp_weibull.

exp_weibull <- lambda * exp(logfact(1/v))

Gamma is just factorial(x - 1), so the mean simplifies a bit. I illustrate below with some R functions how this derivation is the same.

lambda <- 5
v <- 2

mu_traditional <- lambda * gamma(1 + 1/v)
mu_logged <- lambda * exp(lfactorial(1/v))
identical(mu_traditional, mu_logged)
[1] TRUE

EDIT: It seems like JAGS also has the log of the Gamma distribution as well: loggam. Thus, another solution would be

exp_weibull <- lambda * exp(loggam(1 + 1/v))

Answer 2

My understanding is that the parameterization of the Weibull distribution used by JAGS is different from that used by dweibull in R. I believe the JAGS version uses shape, v and rate lambda with an expected value of lambda^{-1/v}*gamma(1+1/v). Thus, I've implemented the expected value in JAGS as lambda^(-1/v)*exp(loggam(1+(1/v))). Interested if others disagree, admittedly I've had a tough time tracking which parameterization is used and how the expected value is formulated, especially give some of the interchangeability in symbols used for different parameters in different formulations!