the meaning of theta1 and theta2 in model hyperparameters

[wanxinru]

Hi,

I'm using INLA and SPDE models, I wonder about the meaning of theta1 and theta2 in model hyperparameters ("w" is the name of SPDE here), I guess they represent the latitudinal and longitudinal effects here, is that a good guess?

> bri.hyperpar.summary(model)
                                                                  mean           sd        q0.025          q0.5        q0.975          mode
size for nbinomial zero-inflated observations             2.596977e+00 5.772638e-02  2.513831e+00  2.587014e+00  2.731514e+00  2.550184e+00
zero-probability parameter for zero-inflated nbinomial_1  7.286996e-05 1.284755e-06  7.058025e-05  7.277658e-05  7.559402e-05  7.251369e-05
Theta1 for w                                             -2.383328e+00 1.736765e-02 -2.420064e+00 -2.382164e+00 -2.352364e+00 -2.378464e+00
Theta2 for w                                             -5.064675e-02 1.884760e-02 -8.247626e-02 -5.260927e-02 -9.558912e-03 -5.875899e-02
SD for year                                               1.879538e-01 1.644374e-03  1.851764e-01  1.877814e-01  1.915551e-01  1.872396e-01
Rho for year                                              3.189336e-01 1.023554e-02  2.954582e-01  3.204829e-01  3.342899e-01  3.263286e-01
SD for inla.group(elev, n = 50)                           2.308091e-01 1.642305e-03  2.273502e-01  2.309126e-01  2.337639e-01  2.312336e-01
SD for inla.group(temp_u, n = 50)                         3.755271e-01 2.457009e-03  3.707036e-01  3.755226e-01  3.803532e-01  3.755080e-01

[finnlindgren]

It depends on precisely which SPDE model you have, but for inla.spde.2matern they are (by default; it's user-redefinable) $log(\tau)$ and $\log(\kappa)$. The function inla.spde.result() can be used to help convert them to range and variance values.
But I would recommend using inla.spde2.pcmatern() for the model specification instead, which automatically converts the internal parameters to range and standard deviation.

[wanxinru]

Thanks for your instant response, I notice the theta here has nothing to do with sd, I try using inla.spde2.pcmatern(), and the sd was given:

> bri.hyperpar.summary(modelx1.1)
                                                                 mean           sd       q0.025         q0.5       q0.975         mode
size for nbinomial zero-inflated observations            2.546205e+00 6.934500e-02 2.402976e+00 2.549560e+00 2.673498e+00 2.562767e+00
zero-probability parameter for zero-inflated nbinomial_1 2.632020e-05 1.418838e-05 9.696052e-06 2.273220e-05 6.305996e-05 1.755117e-05
Range for w                                              2.863417e-02 1.020038e-02 1.255022e-02 2.741963e-02 5.187632e-02 2.488063e-02
Stdev for w                                              6.333500e+01 2.463467e+01 3.066060e+01 5.804304e+01 1.250703e+02 4.915904e+01
SD for year                                              1.930504e-01 3.885804e-02 1.173057e-01 1.940607e-01 2.654096e-01 2.001488e-01
Rho for year                                             4.769190e-01 2.002061e-01 5.218779e-02 4.902290e-01 8.150698e-01 5.263982e-01
SD for inla.group(elev, n = 50)                          1.306181e+00 2.884957e-01 9.248413e-01 1.242303e+00 2.029116e+00 1.099775e+00
SD for inla.group(temp_u, n = 50)                        8.241273e-01 2.023346e-01 5.417480e-01 7.834320e-01 1.323970e+00 6.985269e-01

One last question, what's the meaning or full name for the Rho of Rho for year?

Best Regards

[finnlindgren]

I'm guessing the Rho is for some "ar1" component in your model? See INLA::inla.doc("ar1") for information.

[wanxinru]

yes, it's ar1, I see it is described in INLA::inla.doc("ar1")

theta2
hyperid 14102
name logit lag one correlation
short.name rho
prior pc.cor0
param 0.5 0.5
initial 2
xed FALSE
to.theta function(x) log((1 + x) / (1 - x))
from.theta function(x) 2 * exp(x) / (1 + exp(x)) - 1

[finnlindgren]

Yes. Note that the summary output for Rho is the actual rho, not the internal theta-representation of it.

[wanxinru]

Thanks for your kind reminder.

Best Regards,
Xinru