Taylor expansions for single cell transforms

scRNA-seq count data can be understood as noisy observations of true gene expression (Sarkar and Stephens 2020) \( \DeclareMathOperator\E{E} \DeclareMathOperator\Poi{Poisson} \DeclareMathOperator\V{V} \newcommand\xiplus{x_{i+}} \)

\begin{equation} x_{ij} \mid \xiplus, \lambda_{ij} \sim \Poi(\xiplus \lambda_{ij}), \end{equation}

where \(x_{ij}\) denotes the number of molecules of gene \(j\) observed in cell \(i\), \(\xiplus \triangleq \sum_j x_{ij}\) denotes the total number of molecules observed in cell \(i\), and \(\lambda_{ij}\) denotes the true (relative) gene expression of gene \(j\) in cell \(i\). However, analyzing observations according to this measurement model can be difficult. A common strategy is to instead analyze log-transformed values, which can be understood as MLEs of (1) assuming a log link \(\theta_{ij} = \ln\lambda_{ij}\).

\begin{align} x_{ij} \mid \xiplus, \theta_{ij} &\sim \Poi(\xiplus \exp(\theta_{ij}))\\ \hat\theta_{ij} &= \ln\left(\frac{x_{ij}}{\xiplus}\right). \end{align}

Lun 2018 characterized a bias in using log-transformed values

\begin{equation} y_{ij} = \ln\left(\frac{x_{ij} + \epsilon}{\xiplus}\right), \end{equation}

where \(\epsilon\) is a pseudocount to handle the case \(x_{ij} = 0\), to estimate the mean of log true gene expression. Absorbing \(\epsilon\) into \(x\) and taking a Taylor expansion,

\begin{align} \ln x &\approx \ln \E[x] + \frac{x - \E[x]}{\E[x]} - \frac{(x - \E[x])^2}{2 \E[x]^2}\\ \E[\ln x] &\approx \ln \E[x] - \frac{\V[x]}{2 \E[x]^2}. \end{align}

Sarkar et al. 2019 characterized a bias in using log-transformed values to estimate the variance of log true gene expression. Taking Taylor expansions,

\begin{align} \E[\ln x]^2 &\approx (\ln \E[x])^2 - \ln \E[x] \frac{\V[x]}{\E[x]^2} + \frac{\V[x]^2}{4 \E[x]^4}\\ (\ln x)^2 &\approx (\ln \E[x])^2 + 2 \ln \E[x] (x - \E[x]) + \frac{(x - \E[x])^2}{\E[x]}\\ \E[(\ln x)^2] &\approx (\ln \E[x])^2 + \frac{\V[x]}{\E[x]}\\ \V[\ln x] &= \E[(\ln x)^2] - \E[\ln x]^2\\ &\approx \frac{\V[x]}{\E[x]} + \ln \E[x] \frac{\V[x]}{\E[x]^2} - \frac{\V[x]^2}{4 \E[x]^4}. \end{align}