Last updated: 2019-03-31

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Knit directory: fiveMinuteStats/analysis/

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  • Likelihood Ratio.


Suppose that we are considering whether to model some data \(X\) as normal or log-normal. In this case we’ll assume the truth is that the data are log normal, which we can simulate as follows:

X = exp(rnorm(1000,-5,2))

We will use \(Z\) to denote \(\log(X)\):

Z = log(X)

And let’s check by graphing which looks more normal:


Version Author Date
c3b365a John Blischak 2017-01-02

So it is pretty clear that the model ``\(M_2: \log(X)\) is normal" is better than the model “\(M_1: X\) is normal”.

Now consider computing a “log-likelihood” for each model.

To compute a log-likelihood under the model “X is normal” we need to also specify a mean and variance (or standard deviation). We use the sample mean and variance here:

sum(dnorm(X, mean=mean(X), sd=sd(X),log=TRUE))
[1] 43.45732

Doing the same for \(Z\) we obtain:

sum(dnorm(Z, mean=mean(Z), sd=sd(Z),log=TRUE))
[1] -2110.333

Done this way the log-likelihood for \(M_1\) appears much larger than the log-likelihood for \(M_2\), contradicting both the graphical evidence and the way the data were simulated.

The right way

The explanation here is that it does not make sense to compare a likelihood for \(Z\) with a likelihood for \(X\) because even though \(Z\) and \(X\) are 1-1 mappings of one another (\(Z\) is determined by \(X\), and vice versa), they are formally not the same data. That is, it does not make sense to compute \[\text{"LLR"} := \log(p(X|M_1)/p(Z|M_2))\].

However, we could compute a log-likelihood ratio for this problem as \[\text{LLR} := log(p(X|M_1)/p(X|M_2)).\] Here we are using the fact that the model \(M_2\) for \(Z\) actually implies a model for \(X\): \(Z\) is normal if and only if \(X\) is log-normal. So a sensible LLR would be given by:

sum(dnorm(X, mean=mean(X), sd=sd(X),log=TRUE)) - sum(dlnorm(X, meanlog=mean(Z), sdlog=sd(Z),log=TRUE))
[1] -2753.814

The fact that the LLR is very negative supports the graphical evidence that \(M_2\) is a much better fitting model (and indeed, as we know – since we simulated the data – \(M_2\) is the true model).

R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.1

Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib

[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] workflowr_1.2.0 Rcpp_1.0.0      digest_0.6.18   rprojroot_1.3-2
 [5] backports_1.1.3 git2r_0.24.0    magrittr_1.5    evaluate_0.12  
 [9] stringi_1.2.4   fs_1.2.6        whisker_0.3-2   rmarkdown_1.11 
[13] tools_3.5.2     stringr_1.3.1   glue_1.3.0      xfun_0.4       
[17] yaml_2.2.0      compiler_3.5.2  htmltools_0.3.6 knitr_1.21     

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