Last updated: 2020-03-04

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Rmd 7cb2741 DongyueXie 2020-03-04 wflow_publish(“analysis/binomialthinning.Rmd”)

Introduction

Investigate why effects with larger se are bigger.

Assume we have \(n\) samples and a fraction \(p\) of them belong to group 1 and the rest belong to group 2. So \(x = (1,1,1,...,1,0,0,0,...,0)^T\in R^n\) and \(\sum_ix_i=np\). Under this setting, in simple linear regression \(y = a+\beta x + \epsilon\), \(\epsilon\sim N(0,\sigma^2)\), the variance of \(\hat\beta_j\) is \(\hat s^2 = \frac{n\sigma^2}{n\sum_ix_i^2-(\sum_ix_i)^2}=\frac{\sigma^2}{np-np^2}\). For fixed \(n\) and \(p\), if \(\hat s\) is large, then this means \(\hat\sigma^2\) is large hence \(\sigma^2\) is large.

We now need to figure out the relationship between \(\beta\) and \(\sigma^2\).

Let’s assume we have RNA-Seq count data \(z_i\sim Poisson(\lambda)\) for \(i=1,2,...,n\). In binomial thinning, \(\beta\) is the log2 fold change between groups. Now assume \(\beta>0\), according to Gerard and Stephens(2017), the new(thinned) data vector is \(w_i\sim Poisson(\mu_i)\), where \(\mu_i=2^{-\beta(1-x_i)}\lambda\). The response \(y\) in the simple linear regression is the log transformation of \(w\), \(y_i=\log(w_i)\), \(i=1,2,...n\).

The Taylor series expansion of \(\log w_i\) around \(\mu_i\) is \(\log(w_i)\approx \log(\mu_i)+\frac{w_i-\mu_i}{\mu_i}\). So the mean of \(\log(w_i)\) is \(\log(\mu_i) = \lambda - \beta(1-x_i)\) and variance \(\frac{1}{\mu_i} = \frac{1}{2^{-\beta(1-x_i)}\lambda}\). So if \(\beta\) is large, then \(Var(\log(w_i))\) is large if \(x_i=0\). This explains the why effects with larger se are bigger.