How should I deduce the variance and expectation of the log of a variable?

Question

I read this paper "voom: precision weights unlock linear model analysis tools for RNA-seq read counts", in the methods, the "Delta rule for log-cpm" section:

The RNA-seq data consist of a matrix of read counts $r_{gi}$, for RNA samples $i=1$ to $n$, and genes $g=1$ to $G$. Write $R_i$ for the total number of mapped reads for sample $i$: $$R_i=\sum_{g=1}^{G}r_{gi}$$ They define the log-counts per million (log-cpm) value for each count as: $$y_{gi}=\log_2\left(\frac{r_{gi}+0.5}{R_i+1}\times 10^6\right)$$

Write $\lambda=E(r)$ for the expected value of a read count given the experimental conditions, and suppose that: $$\text{Var}(r)=\lambda+\phi\lambda^2$$ If $r$ is large, then the log-cpm value of the observation is: $$y\approx\log_2(r)-\log_2(R)+6\log_2(10)$$ where $R$ is the library size. The analysis is conditional on $R$, so $R$ is treated as a constant. It follows that $\text{Var}(y)\approx \text{Var}(\log_2 r)$. If $\lambda$ also is large, then: $$(\log_2 r)(\ln 2)\approx \ln r \approx \ln\lambda+\frac{r-\lambda}{\lambda}$$ so $$\text{Var}(y)(\ln 2)^2\approx\frac{\text{Var}(r)}{\lambda^2}=\frac{1}{\lambda}+\phi$$ How should I deduce the last 2 equations?

Answer 1

With the corrected version of the equations, this follows from a standard approximation to the variance of a natural log of a random variable, given the variance of the random variable. See this page, along with its warnings about when this approximation might get you into trouble.

Work first with the natural log of $r$. The Taylor expansion of $\ln r$ about the mean value $\lambda$ for $r$ is:

$$ \ln r \approx \ln \lambda + \frac{(r-\lambda)}{\lambda} - \frac{(r-\lambda)^2}{2\lambda^2}+\dots,$$

based on the first and second derivatives of the natural log function at $r=\lambda$. The terms after the first two are assumed to become small with large $\lambda$.

Then continue by taking the expectations of both sides. As $\lambda$ is the mean of $r$, the expectation of the second term is 0, leaving:

$$E[\ln r] \approx \ln E[r].$$

Now, working with the first two terms of the Taylor expansion, take the variance of both sides, the expected value of squared deviations from the mean:

$$\text{Var}(\ln r) \approx E\left[\left(\ln \lambda + \frac{(r-\lambda)}{\lambda} - \ln E[r]\right)^2\right]$$

$E[r]$ is $\lambda$, so the first and third terms within the parentheses on the right cancel, leaving:

$$\text{Var}(\ln r)\approx E\left[ \left(\frac{(r-\lambda)}{\lambda} \right)^2\right]=\frac{\text{Var }r}{\lambda^2}.$$

Given the assumed form for the variance of $r$, the rightmost part of the last display immediately follows. The $\ln 2$ factors translate between the natural log and the $\log_2$ scales.

Answer 2

Based on Taylor expansions for the moments of functions of random variables $$ \begin{align} \log X & = \log (\mu_X+X-\mu_X) \\ &\approx \log (\mu_X)+ \log' (\mu_X)(X-\mu_X)+\frac{1}{2}\log''(\mu_X)(X-\mu_X)^2 \\ &\approx\log (E(X))+\frac{X-E(X)}{E(X)} \end{align} $$ so take expectations and variances on both sides,

$$E(\log X)\approx\log (E(X))$$ $$\text{Var}(\log X)\approx E\left[\frac{X-E(X)}{E(X)}\right]^2=\frac{\text{Var}(X)}{(E(X))^2}$$ So

$$E[\log r]\approx\log (\lambda)+\frac{r-\lambda}{\lambda} $$

$$\text{Var}[\log r]\approx\frac{\text{Var}(r)}{\lambda^2} $$