Last updated: 2019-03-31

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Knit directory: fiveMinuteStats/analysis/

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# Pre-requisites

• Likelihood Ratio.

# Example

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:

par(mfcol=c(2,2))
hist(X)
hist(Z)
qqnorm(X)
qqnorm(Z)

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).

sessionInfo()
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

locale:
[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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