Hyperparameters for f(spde, control.group(rw1))

[Ruby Ji]

Dear INLA experts,

I recently noticed that there are only two hyperparamters (range and std) for this part:

f(s, model=spde, group=s.group, control.group=list(model="rw1", scale.model=TRUE))

Could you please explain why the precision parameter of the RW1 model is not treated as a separate hyperparameter?

Thanks,

Ruby

[Helpdesk (Haavard Rue)]

because its perfectly confounded with the one is the spde. its a product hence
we set the group one to 1 (or log.prec=0)

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[Ruby Ji]

Hi Haavard,

Thank you for your clarification! I have two additional questions regarding this topic:

First, regarding the scale.model = TRUE option in control.group: Does INLA first scale the precision matrix of the RW1 model (Q_rw1) and then compute the Kronecker product of the SPDE precision matrix (Q_spde) and the scaled Q_rw1?

Second, concerning the addition of a diagonal term: If I specify the model as follows:
f(s, model = spde, group = s.group, diagonal = 1e-06, control.group = list(model = "rw1", scale.model = TRUE))
Is the diagonal term added to the Kronecker product of Q_spde and Q_rw1_scaled? After adding the diagonal term, does INLA rescale the combined precision matrix to make the generalized variance = 1?

Thanks,
Ruby

[Helpdesk (Haavard Rue)]

On Sat, 2024-12-07 at 04:07 -0800, Ruby Ji wrote:
> First, regarding the scale.model = TRUE option in control.group: Does INLA
> first scale the precision matrix of the RW1 model (Q_rw1) and then compute the
> Kronecker product of the SPDE precision matrix (Q_spde) and the scaled Q_rw1?

the group models are always scaled no matter what scale.model= says

>
> Second, concerning the addition of a diagonal term: If I specify the model as
> follows:
> f(s, model = spde, group = s.group, diagonal = 1e-06, control.group =
> list(model = "rw1", scale.model = TRUE))
> Is the diagonal term added to the Kronecker product of Q_spde and
> Q_rw1_scaled? After adding the diagonal term, does INLA rescale the combined
> precision matrix to make the generalized variance = 1?

diagonal term is added after scaling and kronecker product

[Helpdesk (Haavard Rue)]

let me be more clear

Q_spde is already scaled, so not effect of 'scale.model'

Q_rw1 is scaled using and epsilon diagonal term. I *think* this not done using
the 'diagonal' argument, but then you need to check the code

then the kronecker product is formed and epsilon is added to its diagonal

On Sat, 2024-12-07 at 04:07 -0800, Ruby Ji wrote:

[Ruby Ji]

Thank you! Your clarification is really helpful!

[Ruby Ji]

Hi Haavard,

I came across a new question related to the RW1 model. The precision is a hyperparameter in the model: f(year, model="rw1", scale.model=TRUE).

The precision matrix, Q, is defined as precision * R, where R is the structure matrix reflecting the neighborhood structure of the model.

I scaled Q using inla.scale.model(Q, constr = list(A = matrix(1, 1, n), e = 0)) and noticed that, regardless of the precision value, the scaled Q (Q.scaled) remains the same. If this is the case, how is the precision estimated?

Thanks,

Ruby

[Ruby Ji]

I think I got the answer. Does the final Q equal prec*Q.scaled?

>library(INLA)
>df <- data.frame(x = 1, y = NA)

> Q <- 100 * (t(diff(diag(4))) %*% diff(diag(4)))
> (Q.scaled <- inla.scale.model(Q, constr = list(A = matrix(1, 1, n), e = 0)))
4 x 4 sparse Matrix of class "dgTMatrix"
                                           
[1,]  0.572822 -0.572822  .         .      
[2,] -0.572822  1.145644 -0.572822  .      
[3,]  .        -0.572822  1.145644 -0.572822
[4,]  .         .        -0.572822  0.572822
> fit_scaled <- inla(y ~ 0 + f(x, model = "rw1", scale.model= TRUE, values = 1:4, diagonal=0,
+                              hyper = list(prec = list(initial = log(100), fixed = TRUE))),
+                    data = df,
+                    control.inla = list(int.strategy = "eb"),
+                    control.compute = list(config = TRUE))
> (fit_scaled$misc$configs$config[[1]]$Q)
4 x 4 sparse Matrix of class "dgCMatrix"
                                           
[1,] 57.28219 -57.28219   .         .      
[2,]  .       114.56438 -57.28219   .      
[3,]  .         .       114.56438 -57.28219
[4,]  .         .         .        57.28219

[Helpdesk (Haavard Rue)]

yes

with

demo(Bym)

then

inla(Y ~ f(region.struct,model="besag",graph=g, scale.model=TRUE),
data=Germany,E=E,verbose=T,family="poisson")

you'll see in the output

inla_parse_ffield...
section=[region.struct]
dir=[random.effect0000000001]
scale.model[1]
connected component[0] size[544] scale[0.561134]
scale.model: prec_scale[0.561134]

so the constant which is used to scale is '0.56...'

[Ruby Ji]

This makes sense. Thank you for your clarification!