Defining prior on variance and not precision

Question

I know that WinBugs uses precision as a parameter in dnorm instead of variance

 model {

    #Likelihood
        for(i in 1:N1){
            y1[i] ~dnorm(mu,tau)
        }

    sigma <- sqrt(1/tau)

    #Priors
    mu ~ dnorm(0,0.000001)
    tau~ dgamma(taumu, taus)
   }

My question is: if instead I want to specify the prior for sigma because I know its mean and variance would it be right to use the following model ?

  model {

    #Likelihood
        for(i in 1:N1){
            y1[i] ~dnorm(mu,tau)
        }

    tau <- sqrt(1/sigma)

    #Priors
    mu ~ dnorm(0,0.000001)
    sigma ~ dnorm(sigmamu, sigmas)
   }

Thanks in advance

Answer 1

If "sigma <- sqrt(1/tau)" as indicated in the model on the top, then how is "tau <- sqrt(1/sigma)"?

$\sigma = 1/\sqrt\tau \implies \sigma^2 = 1/\tau \implies \tau = 1/\sigma^2$.

Answer 2

In the second model, tau <- sqrt(1/sigma) is wrong; it should be tau <- 1/sigma^2. You also cannot give a normal distribution to a non-negative random variable, so you will need to choose a different distribution for $\sigma$. If you simply want to keep the same form as Model 1 but set your prior mean and variance for $\sigma$ to known values, this can be accomplished by checking the appropriate moments for the gamma prior and using some algebra to get the desired moments.