Dirichlet-Multinomial WinBUGS code - bayesian

I'm trying to code a dirichlet-multinomial model using BUGS.
Basically I have 18 regions and 3 categories per region. In example,
Region 1: 0.50 belongs to Low, 0.30 belongs to Middle, and 0.20 belongs to High. The list goes on to Region 18 of course with varying proportions.The only code I got is this
`model {
for (i in 1:N) {
x[1:3] ~ dmulti(p[],n[i])
p[1:3] ~ ddirch(alpha[])
}
for (k in 1:3) {
alpha[k] <- 1
}
}
DATA list(n=c(38483, 2259, 1900),x=c(29256.42719, 1857.431404, 1548.007808, 29256.42719, 1857.431404, 1548.007808, 29256.42719, 1857.431404, 1548.007808), N=3)`
I shortened it to 3 regions first just for example. It states 'Dirichlet36' after clicking 'gen inits'. Please help me to code this.


This may be helpful (source):
Learning about the parameters of a Dirichlet distribution
Suppose as part of a model there are J probability arrays p[j, 1:K], j = 1, ..., J, where K is the dimension of each array and sum(p[j, 1:K]) = 1 for all j. We give each of them a Dirichlet prior:
p[j, 1:K] ~ ddirch(alpha[])
and we would like to learn about alpha[]. However, the parameters alpha[] of a Dirichlet distribution cannot be stochastic nodes. The trick is to note that if delta[k] ~ dgamma(alpha[k], 1), then the vector with elements delta[k] / sum(delta[1:K]), k = 1, ..., K, is Dirichlet with parameters alpha[k], k = 1, ..., K. So the following construction should allow learning about the parameters alpha[]:
for (k in 1:K) {
p[j, k] <- delta[j, k] / sum(delta[j,])
delta[j, k] ~ dgamma(alpha[k], 1)
}
A prior can be put directly on the alpha[k]'s.

Related

How to sample from a sum of two distributions: binomial and poisson

Is there a way to predict a value from a sum of two distributions? I am getting a syntax error on rstan when I try to estimate y here: y ~ binomial(,) + poisson()
library(rstan)
BH_model_block <- "
data{
int y;
int a;
}
parameters{
real <lower = 0, upper = 1> c;
real <lower = 0, upper = 1> b;
}
model{
y ~ binomial(a,b)+ poisson(c);
}
"
BH_model <- stan_model(model_code = BH_model_block)
BH_fit <- sampling(BH_model,
data = list(y = 5,
a = 2),
iter= 1000)
Produces this error:
SYNTAX ERROR, MESSAGE(S) FROM PARSER:
error in 'model2c6022623d56_457bd7ab767c318c1db686d1edf0b8f6' at line 13, column 20
-------------------------------------------------
11:
12: model{
13: y ~ binomial(a,b)+ poisson(c);
^
14: }
-------------------------------------------------
PARSER EXPECTED: ";"
Error in stanc(file = file, model_code = model_code, model_name = model_name, :
failed to parse Stan model '457bd7ab767c318c1db686d1edf0b8f6' due to the above error.

Stan doesn't support integer parameters, so you can't technically do that. For two real variables, it'd look like this:
parameters {
real x;
real y;
}
transformed parameters {
real z = x + y;
}
model {
x ~ normal(0, 1);
y ~ gamma(0.1, 2);
}
Then you get the sum distribution for z. If the variables are discrete, it won't compile.
If you don't need z in the model, then you can do this in the generated quantities block,
generated quantities {
int x = binomial_rng(a, b);
int y = poisson_rng(c);
int z = x + y;
}
The drawback of doing this is that none of the variables are available in the model block. If you need discrete parameters, they need to be marginalized as described in the user's guide chapter on latent discrete parameters (also in the chapter on mixtures and HMMs). This is not so easy with a Poisson, because support isn't bounded. If the expectations of the two discrete distributions is small, then you can do it approximately with a loop over plausible values.
It looked from the example in the original post that z is data. That's a slightly different marginalization over x and y, but you only sum over the x and y such that x + y = z, so the combinatorics are greatly reduced.

An alternative is to substitute the Binomial with a Poisson, and use Poisson additivity:
BH_model_block <- "
data{
int y;
int a;
}
parameters{
real <lower = 0, upper = 1> c;
real <lower = 0, upper = 1> b;
}
model{
y ~ poisson(a * b + c);
}
"
This differs in that if b is not small, the Binomial has a lower variance than the Poisson, but maybe there is overdispersion anyhow?

rjags error Invalid vector argument to ilogit

I'd like to compare a betareg regression vs. the same regression using rjags
library(betareg)
d = data.frame(p= sample(c(.1,.2,.3,.4),100, replace= TRUE),
id = seq(1,100,1))
# I am looking to reproduce this regression with jags
b=betareg(p ~ id, data= d,
link = c("logit"), link.phi = NULL, type = c("ML"))
summary(b)
Below I am trying to do the same regression with rjags
#install.packages("rjags")
library(rjags)
jags_str = "
model {
#model
y ~ dbeta(alpha, beta)
alpha <- mu * phi
beta <- (1-mu) * phi
logit(mu) <- a + b*id
#priors
a ~ dnorm(0, .5)
b ~ dnorm(0, .5)
t0 ~ dnorm(0, .5)
phi <- exp(t0)
}"
id = d$id
y = d$p
model <- jags.model(textConnection(jags_str),
data = list(y=y,id=id)
)
update(model, 10000, progress.bar="none"); # Burnin for 10000 samples
samp <- coda.samples(model,
variable.names=c("mu"),
n.iter=20000, progress.bar="none")
summary(samp)
plot(samp)
I get an error on this line
model <- jags.model(textConnection(jags_str),
data = list(y=y,id=id)
)
Error in jags.model(textConnection(jags_str), data = list(y = y, id = id)) :
RUNTIME ERROR:
Invalid vector argument to ilogit
Can you advise
(1) how to fix the error
(2) how to set priors for the beta regression
Thank you.

This error occurs because you have supplied the id vector to the scalar function logit. In Jags inverse link functions cannot be vectorized. To address this, you need to use a for loop to go through each element of id. To do this I would probably add an additional element to your data list that denotes how long id is.
d = data.frame(p= sample(c(.1,.2,.3,.4),100, replace= TRUE),
id = seq(1,100,1), len_id = length(seq(1,100,1)))
From there you just need to make a small edit to your jags code.
for(i in 1:(len_id)){
y[i] ~ dbeta(alpha[i], beta[i])
alpha[i] <- mu[i] * phi
beta[i] <- (1-mu[i]) * phi
logit(mu[i]) <- a + b*id[i]
}
However, if you track mu it is going to be a matrix that is 20000 (# of iterations) by 100 (length of id). You are likely more interested in the actual parameters (a, b, and phi).

I have a code in OpenBUGS but the error is "variable CR is not defined"

model
{
for( i in 1 : N ) {
dgf[i] ~ dbin(p[i],n[i])
logit(p[i]) <- a[subject[i]] + beta[1] * CR[i]
}
for (j in 1:94)
{
a[j]~dnorm(beta0,prec.tau)
}
beta[1] ~ dnorm(0.0,.000001)
beta0 ~ dnorm(0.0,.000001)
prec.tau ~ dgamma(0.001,.001)
tau<-sqrt(1/prec.tau)
}
list(
n=c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1),
dgf=c(0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,1,1,1,0,0,1,1,0,0,0,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),
subject=c(1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,26,26,27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35,35,36,36,37,37,38,38,39,39,40,40,41,41,42,42,43,43,44,44,45,45,46,46,47,47,48,48,49,49,50,50,51,51,52,52,53,53,54,54,55,55,56,56,57,57,58,58,59,59,60,60,61,61,62,62,63,63,64,64,65,65,66,66,67,67,68,68,69,69,70,70,71,71,72,72,73,73,74,74,75,75,76,76,77,77,78,78,79,79,80,80,81,81,82,82,83,83,84,84,85,85,86,86,87,87,88,88,89,89,90,90,91,91,92,92,93,93,94,94),
CR=c(NA,NA,1.41,0.85,1.13,0.65,NA,NA,2.13,1.61,7.31,3.8,1.65,2.32,1.13,2.3,0.99,1.5,1.32,3.95,7.2,2.97,0.83,1.55,NA,6.5,0.89,1.2,1.52,7,8.68,7.41,NA,0.86,NA,1.92,NA,1.31,7.8,1.78,NA,1.67,NA,NA,NA,NA,NA,2.25,0.98,0.82,3.94,1.14,12,2.58,2.42,2.59,NA,NA,NA,NA,6.6,3.22,NA,2.02,2.43,1.96,0.82,1.64,1.81,1.53,1.01,5.21,8.33,1.14,1.49,6,5.6,2,3.33,4.08,NA,NA,1.14,1.25,0.85,5.42,0.85,0.65,1.02,1.33,1.1,1.12,NA,NA,1.53,1.76,2,0.85,2.9,5,4.09,2.68,0.98,1.48,0.66,0.57,5.72,2.34,0.93,2.39,1.39,1.44,4.77,2.39,1.79,1.2,0.81,1.25,4.69,1.22,1.92,1.48,2.46,NA,NA,2.53,1.12,1.74,3.45,1.22,1.27,2.61,1.75,0.82,NA,1.4,NA,5.1,1.24,1.5,1.94,1.24,1.04,1.24,NA,2.39,NA,2.07,2.19,1.6,6,6.38,1.17,1.2,5.62,6.39,1.82,1.31,NA,1.18,3.71,2.03,5.4,2.17,NA,1.94,1.57,1.44,1.35,1.63,1.24,1.54,1.5,NA,NA,NA,NA,1.44,NA,2.19,7.98,2.15,1.71,1.45,NA,0.98,2.37,1.58), N = 188)
I know that the error is because of "NA" in variable "CR" but i don't know how to solve it. i'll appreciate any help.

You have 2 options:
1) Remove the missing CR and the corresponding elements of n, dgf, and subject (and reduce N accordingly).
2) Define a stochastic relation for CR within your model, so that the model estimates the missing CR and uses these estimates in the logistic regression. Something like:
for(i in 1:N){
CR[i] ~ dnorm(CR_mu, CR_tau)
}
CR_mu ~ dnorm(0, 10^-6)
CR_tau ~ drama(0.001, 0.001)
CR_mu and CR_tau are probably not of interest but can be monitored if you want.
Note that both approaches assume that CR are missing at random (and not e.g. censored) - if CR are missing not at random this will give you biased results.
Matt

How do I sum the coefficients of a polynomial in Maxima?

I came up with this nice thing, which I am calling 'partition function for symmetric groups'
Z[0]:1;
Z[n]:=expand(sum((n-1)!/i!*z[n-i]*Z[i], i, 0, n-1));
Z[4];
6*z[4]+8*z[1]*z[3]+3*z[2]^2+6*z[1]^2*z[2]+z[1]^4
The sum of the coefficients for Z[4] is 6+8+3+6+1 = 24 = 4!
which I am hoping corresponds to the fact that the group S4 has 6 elements like (abcd), 8 like (a)(bcd), 3 like (ab)(cd), 6 like (a)(b)(cd), and 1 like (a)(b)(c)(d)
So I thought to myself, the sum of the coefficients of Z[20] should be 20!
But life being somewhat on the short side, and fingers giving trouble, I was hoping to confirm this automatically. Can anyone help?
This sort of thing points a way:
Z[20],z[1]=1,z[2]=1,z[3]=1,z[4]=1,z[5]=1,z[6]=1,z[7]=1,z[8]=1;
But really...

I don't know a straightforward way to do that; coeff seems to handle only a single variable at a time. But here's a way to get the list you want. The basic idea is to extract the terms of Z[20] as a list, and then evaluate each term with z[1] = 1, z[2] = 1, ..., z[20] = 1.
(%i1) display2d : false $
(%i2) Z[0] : 1 $
(%i3) Z[n] := expand (sum ((n - 1)!/i!*z[n - i]*Z[i], i, 0, n-1)) $
(%i4) z1 : makelist (z[i] = 1, i, 1, 20);
(%o4) [z[1] = 1,z[2] = 1,z[3] = 1,z[4] = 1,z[5] = 1,z[6] = 1,z[7] = 1, ...]
(%i5) a : args (Z[20]);
(%o5) [121645100408832000*z[20],128047474114560000*z[1]*z[19],
67580611338240000*z[2]*z[18],67580611338240000*z[1]^2*z[18],
47703960944640000*z[3]*z[17],71555941416960000*z[1]*z[2]*z[17], ...]
(%i6) a1 : ev (a, z1);
(%o6) [121645100408832000,128047474114560000,67580611338240000, ...]
(%i7) apply ("+", a1);
(%o7) 2432902008176640000
(%i8) 20!;
(%o8) 2432902008176640000

Checking if lines intersect and if so return the coordinates

I've written some code below to check if two line segments intersect and if they do to tell me where. As input I have the (x,y) coordinates of both ends of each line. It appeared to be working correctly but now in the scenario where line A (532.87,787.79)(486.34,769.85) and line B (490.89,764.018)(478.98,783.129) it says they intersect at (770.136, 487.08) when the lines don't intersect at all.
Has anyone any idea what is incorrect in the below code?
double dy[2], dx[2], m[2], b[2];
double xint, yint, xi, yi;
WsqT_Location_Message *location_msg_ptr = OPC_NIL;
FIN (intersect (<args>));
dy[0] = y2 - y1;
dx[0] = x2 - x1;
dy[1] = y4 - y3;
dx[1] = x4 - x3;
m[0] = dy[0] / dx[0];
m[1] = dy[1] / dx[1];
b[0] = y1 - m[0] * x1;
b[1] = y3 - m[1] * x3;
if (m[0] != m[1])
{
//slopes not equal, compute intercept
xint = (b[0] - b[1]) / (m[1] - m[0]);
yint = m[1] * xint + b[1];
//is intercept in both line segments?
if ((xint <= max(x1, x2)) && (xint >= min(x1, x2)) &&
(yint <= max(y1, y2)) && (yint >= min(y1, y2)) &&
(xint <= max(x3, x4)) && (xint >= min(x3, x4)) &&
(yint <= max(y3, y4)) && (yint >= min(y3, y4)))
{
if (xi && yi)
{
xi = xint;
yi = yint;
location_msg_ptr = (WsqT_Location_Message*)op_prg_mem_alloc(sizeof(WsqT_Location_Message));
location_msg_ptr->current_latitude = xi;
location_msg_ptr->current_longitude = yi;
}
FRET(location_msg_ptr);
}
}
FRET(location_msg_ptr);
}

There is an absolutely great and simple theory about lines and their intersections that is based on adding an extra dimensions to your points and lines. In this theory a line can be created from two points with one line of code and the point of line intersection can be calculated with one line of code. Moreover, points at the Infinity and lines at the Infinity can be represented with real numbers.
You probably heard about homogeneous representation when a point [x, y] is represented as [x, y, 1] and the line ax+by+c=0 is represented as [a, b, c]?
The transitioning to Cartesian coordinates for a general homogeneous representation of a point [x, y, w] is [x/w, y/w]. This little trick makes all the difference including representation of lines at infinity (e.g. [1, 0, 0]) and making line representation look similar to point one. This introduces a GREAT symmetry into formulas for numerous line/point manipulation and is an
absolute MUST to use in programming. For example,
It is very easy to find line intersections through vector product
p = l1xl2
A line can be created from two points is a similar way:
l=p1xp2
In the code of OpenCV it it just:
line = p1.cross(p2);
p = line1.cross(line2);
Note that there are no marginal cases (such as division by zero or parallel lines) to be concerned with here. My point is, I suggest to rewrite your code to take advantage of this elegant theory about lines and points.
Finally, if you don't use openCV, you can use a 3D point class and create your own cross product function similar to this one:
template<typename _Tp> inline Point3_<_Tp> Point3_<_Tp>::cross(const Point3_<_Tp>& pt) const
{
return Point3_<_Tp>(y*pt.z - z*pt.y, z*pt.x - x*pt.z, x*pt.y - y*pt.x);
}