Last updated: 2019-08-13

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Rmd f40b084 Jason Willwerscheid 2019-08-13 wflow_publish(“analysis/pvals.Rmd”)

Introduction

To assess goodness of fit, I’d like to be able to look at the distribution of \(p\)-values for the residuals from a flashier fit.

The problem is that the model that’s being fitted doesn’t correspond to a plausible (or even feasible) model for generating the data. In effect, we’re fitting \[ \log(Y_{ij} / \lambda_j + 1) = LF' + E,\ E_{ij} \sim N(0, \sigma_{ij}^2), \] so that \(Y_{ij}\) is modelled as having a shifted lognormal distribution \[ Y_{ij} \sim \text{lognormal}(\mu_{ij}, \sigma_{ij}^2) - \lambda_j, \] where \(\mu_{ij} = (LF')_{ij} + \log(\lambda_j)\).

This implies that \(Y_{ij}\) is supported on the interval \((-\lambda_j, \infty)\), whereas a feasible data-generating model would only be supported on the nonnegative integers.

To calculate \(p\)-values, I’d like to at least use a feasible model. To this end, I find a discrete distribution that is close to the shifted lognormal distribution fitted by the model and that can be found easily.

Here’s how I proceed: First, to guarantee that \(\mathbb{E}Y_{ij}\) is positive, I put all of the mass from \((-\lambda_j, 0)\) onto a point mass at zero, which gives a mixture of a point mass at zero and a shifted and truncated lognormal distribution with support on the positive real line: \[ Y_{ij} \sim \pi_0 \delta_0 + (1 - \pi_0) \left[ \text{TLN}(\mu_{ij}, \sigma_{ij}^2; \lambda_j, \infty) - \lambda_j \right]\]

Next, I match moments to find a Poisson or negative binomial distribution that approximates this mixture. If the variance of the mixture is less than the expectation, I choose the Poisson distribution with the same mean. Otherwise, I match the first and second moments to get a negative binomial distribution.

Finally, I use a randomization strategy to get a continuous range of \(p\)-values. For example, if \(Y_{ij}\) is approximately \(\text{Poisson}(\nu_{ij})\), then I draw \[ c_{ij} \sim \text{Unif}[0, 1] \] and set \[ p_{ij} = c_{ij} \cdot \text{ppois}(Y_{ij} - 1; \nu_{ij}) + (1 - c_{ij}) \cdot \text{ppois}(Y_{ij}; \nu_{ij}) \]

Mixture proportions

Calculating \(\pi_0\) is straightforward: \[ \begin{aligned} \mathbb{P}(\text{lognormal}(\mu, \sigma^2) < \lambda) &= \mathbb{P}(N(\mu, \sigma^2) < \log(\lambda)) \\ &= \Phi \left( \frac{\log(\lambda) - \mu}{\sigma} \right) \end{aligned} \] Similarly, \[ 1 - \pi_0 = \Phi \left( \frac{\mu - \log(\lambda)}{\sigma} \right) \]

Moments of the truncated lognormal

If \[ X \sim \text{truncated-normal}(\mu, \sigma^2; \log(\lambda), \infty), \] then \[ e^X \sim \text{TLN}(\mu, \sigma^2; \lambda, \infty). \] The MGF for the truncated normal gives \[ \mathbb{E}e^X = \text{exp}(\mu + \sigma^2 / 2) \left[ \frac{\Phi(\frac{\mu - \log(\lambda)}{\sigma} + \sigma)}{\Phi(\frac{\mu - \log(\lambda)}{\sigma})} \right] \] and \[ \mathbb{E}e^{2X} = \text{exp}(2\mu + 2\sigma^2) \left[ \frac{\Phi(\frac{\mu - \log(\lambda)}{\sigma} + 2\sigma)}{\Phi(\frac{\mu - \log(\lambda)}{\sigma})} \right]. \]

Thus \[ \begin{aligned} \mathbb{E}Y_{ij} &= (1 - \pi_0) \left[ \mathbb{E} (\text{TLN}(\mu_{ij}, \sigma_{ij}^2; \log(\lambda_j), \infty)) - \lambda_j \right] \\ &= (1 - \pi_0) \left[ \text{exp}((LF')_{ij} + \log(\lambda_j) + \sigma_{ij}^2 / 2) \left( \frac{\Phi(\frac{(LF')_{ij}}{\sigma_{ij}} + \sigma_{ij})}{\Phi(\frac{(LF')_{ij}}{\sigma_{ij}})} \right) - \lambda_j \right] \\ &= \lambda_j \left[ \text{exp}((LF')_{ij} + \sigma_{ij}^2 / 2) \Phi \left( \frac{(LF')_{ij}}{\sigma_{ij}} + \sigma_{ij} \right) - \Phi \left(\frac{(LF')_{ij}}{\sigma_{ij}} \right) \right] \end{aligned} \] and \[ \begin{aligned} \mathbb{E}Y_{ij}^2 &= (1 - \pi_0) \left[ \mathbb{E} (\text{TLN}(\mu_{ij}, \sigma_{ij}^2; \log(\lambda_j), \infty)^2) + \lambda_j^2 - 2 \lambda_j \mathbb{E} (\text{TLN}(\mu_{ij}, \sigma_{ij}^2; \log(\lambda_j), \infty)) \right] \\ &= (1 - \pi_0) \left[ \mathbb{E} (\text{TLN}(\mu_{ij}, \sigma_{ij}^2; \log(\lambda_j), \infty)^2) + \lambda_j^2 - 2 \lambda_j \left( \frac{\mathbb{E}Y_{ij}}{1 - \pi_0} + \lambda_j \right) \right] \\ &= (1 - \pi_0) \left[ \text{exp}(2(LF')_{ij} + 2 \log(\lambda_j) + 2\sigma_{ij}^2) \left( \frac{\Phi(\frac{(LF')_{ij}}{\sigma_{ij}} + 2 \sigma_{ij})}{\Phi(\frac{(LF')_{ij}}{\sigma_{ij}})} \right) - \lambda_j^2 \right] - 2 \lambda_j \mathbb{E} Y_{ij} \\ &= \lambda_j^2 \left[ \text{exp}(2(LF')_{ij} + 2\sigma_{ij}^2) \Phi \left( \frac{(LF')_{ij}}{\sigma_{ij}} + 2 \sigma_{ij} \right) - \Phi \left( \frac{(LF')_{ij}}{\sigma_{ij}} \right) \right] - 2 \lambda_j \mathbb{E} Y_{ij} \end{aligned} \]

Estimating LF’

Rather than trying to take expectations with respect to \(L\) and \(F\), I simply use the plug-in estimator \(\mathbb{E}(LF')\). One could try sampling from the posterior on \(LF'\) to see whether the \(p\)-value plots change, but I don’t think that it’s worth the trouble given that so many other approximations are already being made.

sessionInfo()
R version 3.5.3 (2019-03-11)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.6

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_1.0.1      digest_0.6.18   rprojroot_1.3-2
 [5] backports_1.1.3 git2r_0.25.2    magrittr_1.5    evaluate_0.13  
 [9] stringi_1.4.3   fs_1.2.7        whisker_0.3-2   rmarkdown_1.12 
[13] tools_3.5.3     stringr_1.4.0   glue_1.3.1      xfun_0.6       
[17] yaml_2.2.0      compiler_3.5.3  htmltools_0.3.6 knitr_1.22