WinBUGS uses “precision” as a parameter in specifying a Normal distribution, instead of variance.

Precision = $\frac{1}{\text{Variance}}$

dnorm(0, 0.0001) is the same as a Normal distribution with mean 0 and variance $\frac{1}{0.0001} = 100^2$, or $\sim \mathcal{N}(0,100^2)$

Edit

In response to the question, I assume your model looks something like this...

model { 
   for (i in 1:n) { 
      y[i] ~ dnorm(mu[i], tau) 
      mu[i] <- a + b*x[i]
   }

   a ~ dnorm(0, 1.0E-6)
   b ~ dnorm(0, 1.0E-6)

   tau ~ dnorm(0,1.0E-6)

   sigma <- 1/sqrt(tau)
}

So, you simulate a $\tau$ value as $\sim \mathcal{N}(0, \frac{1}{1.0E-6})$, then use that $\tau$ value to calculate $y[i]$, such that $y[i] \sim \mathcal{N}(0, \frac{1}{\mathcal{N}(0, \frac{1}{1.0E-6})})$

Using a mean-zero Gaussian is standard practice for most people simulating random positive-or-negative errors. I do not know why you chose $1.0 \times 10^{-6}$ as your precision parameter. It is your model, not mine.

WinBUGS uses “precision” as a parameter in specifying a Normal distribution, instead of variance.

Precision = $\frac{1}{\text{Variance}}$

dnorm(0, 0.0001) is the same as a Normal distribution with mean 0 and variance $\frac{1}{0.0001} = 100^2$, or $\sim \mathcal{N}(0,100^2)$

Edit

In response to the question, I assume your model looks something like this...

model { 
   for (i in 1:n) { 
      y[i] ~ dnorm(mu[i], tau) 
      mu[i] <- a + b*x[i]
   }

   a ~ dnorm(0, 1.0E-6)
   b ~ dnorm(0, 1.0E-6)

   tau ~ dnorm(0,1.0E-6)

   sigma <- 1/sqrt(tau)
}

So, you simulate a $\tau$ value as $\sim \mathcal{N}(0, \frac{1}{1.0E-6})$, then use that $\tau$ value to calculate $y[i]$, such that $$y[i] \sim \mathcal{N}\left(0, \frac{1}{\mathcal{N}(0, \frac{1}{1.0E-6})}\right).$$

Using a mean-zero Gaussian is standard practice for most people simulating random positive-or-negative errors. I do not know why you chose $1.0 \times 10^{-6}$ as your precision parameter. It is your model, not mine.

WinBUGS uses “precision” as a parameter in specifying a Normal distribution, instead of variance.

Precision = $\frac{1}{\text{Variance}}$

dnorm(0, 0.0001) is the same as a Normal distribution with mean 0 and variance $\frac{1}{0.0001} = 100^2$, or $\sim \mathcal{N}(0,1)$

WinBUGS uses “precision” as a parameter in specifying a Normal distribution, instead of variance.

Precision = $\frac{1}{\text{Variance}}$

dnorm(0, 0.0001) is the same as a Normal distribution with mean 0 and variance $\frac{1}{0.0001} = 100^2$, or $\sim \mathcal{N}(0,100^2)$

Edit

In response to the question, I assume your model looks something like this...

model { 
   for (i in 1:n) { 
      y[i] ~ dnorm(mu[i], tau) 
      mu[i] <- a + b*x[i]
   }

   a ~ dnorm(0, 1.0E-6)
   b ~ dnorm(0, 1.0E-6)

   tau ~ dnorm(0,1.0E-6)

   sigma <- 1/sqrt(tau)
}

So, you simulate a $\tau$ value as $\sim \mathcal{N}(0, \frac{1}{1.0E-6})$, then use that $\tau$ value to calculate $y[i]$, such that $y[i] \sim \mathcal{N}(0, \frac{1}{\mathcal{N}(0, \frac{1}{1.0E-6})})$

Using a mean-zero Gaussian is standard practice for most people simulating random positive-or-negative errors. I do not know why you chose $1.0 \times 10^{-6}$ as your precision parameter. It is your model, not mine.