binomial thinning

Introduction

Last updated: 2020-03-04

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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 $\hat\beta$ 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.

binomial thinning

DongyueXie

2020-03-04

Introduction