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POHow to code a multiparameter log-likelihood function in R
 

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  1. POHow to code a multiparameter log-likelihood function in R
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    <p>I would like to estimate power of the following problem. I am interested in comparing two groups that both follow Weibull distribution. So, group A has two parameters (shape par = a1,scale par = b1) and two parameters has group B (a2, b2). By simulating random variables from distribution of interest (for example assuming different scale and shape parameters, i.e. a1=1.5*a2, and b1=b2*0.5; or either differences between groups are just in either shape or scale parameters), apply log-likelihood ratio test to test if a1=a2 and b1=b2 (or e.g. a1=a1, when we know that b1=b2), and estimate power of the test.</p> <p>The questions would be what are log-likelihoods for the full models, and how to code it in R when a)having exact data, and b) for interval-censored data ?</p> <p>That is, for reduced model (when a1=a2,b1=b2) log-likelihoods for exact and interval-censored data are:</p> <pre><code>LL.reduced.exact &lt;- function(par,data){sum(log(dweibull(data,shape=par[1],scale=par[2])))}; LL.reduced.interval.censored&lt;-function(par, data.lower, data.upper) {sum(log((1-pweibull(data.lower, par[1], par[2])) – (1-pweibull(data.upper, par[1],par[2]))))} </code></pre> <p>What is it for the full model, when a1!=a2, b1!=b2, taking into account two different observational schemes, i.e. when 4 parameters have to be estimated (or, in case when interested in looking at diferences in shape parameters, 3 parameters have to be estimated)? </p> <p>Is it possible to estimate it buy building two log-likelihoods for separate groups and add it together (i.e.<strong>LL.full&lt;-LL.group1+LL.group2</strong>)? </p> <p>Regarding log-likelihood for interval-censored data, censoring is non-informative and all observations are interval-censored. Any better ideas how to perform this task will be appreciated.</p> <p>Please, find the R Code for exact data below to illustrate the problem. Thank you very much in advance.</p> <pre><code>R Code: # n (sample size) = 500 # sim (number of simulations) = 1000 # alpha = .05 # Parameters of Weibull distributions: #group 1: a1=1, b1=20 #group 2: a2=1*1.5 b2=b1 n=500 sim=1000 alpha=.05 a1=1 b1=20 a2=a1*1.5 b2=b1 #OR: a1=1, b1=20, a2=a1*1.5, b2=b1*0.5 # the main question is how to build this log-likelihood model, when a1!=a2, and b1=b2 # (or a1!=a2, and b1!=b2) LL.full&lt;-????? LL.reduced &lt;- function(par,data){sum(log(dweibull(data,shape=par[1],scale=par[2])))} LR.test&lt;-function(red,full,df) { lrt&lt;-(-2)*(red-full) pvalue&lt;-1-pchisq(lrt,df) return(data.frame(lrt,pvalue)) } rejections&lt;-NULL for (i in 1:sim) { RV1&lt;-rweibull (n, a1, b1) RV2&lt;-rweibull (n, a2, b2) RV.Total&lt;-c(RV1, RV2) par.start&lt;-c(1, 15) mle.full&lt;- ???????????? mle.reduced&lt;-optim(par.start, LL, data=RV.Total, control=list(fnscale=-1)) LL.full&lt;-????? LL.reduced&lt;-mle.reduced$value LRT&lt;-LR.test(LL.reduced, LL.full, 1) rejections1&lt;-ifelse(LRT$pvalue&lt;alpha,1,0) rejections&lt;-c(rejections, rejections1) } table(rejections) sum(table(rejections)[[2]])/sim # estimated power </code></pre>
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    1. COThis question appears to be off-topic because it is about how to derive a log-likelihood and thus not within the scope of Stack Overflow. It should be migrated to stats.stackexchange.com.
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    2. COThe question could be made on-topic by a minor rephrasing, such as 'how to code a multiparameter log-likelihood function in R'
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