# Standard error of fold changes

I have a conceptual problem to understand the standard error of the ratio of two random variables after error propagation.

Let $X$ and $Y$ be two random variables with means $\bar x$ and $\bar y$ and standard errors $se_x = \frac{\sigma_x}{\sqrt{m}}$ and $se_y = \frac{\sigma_y}{\sqrt{n}}$, where $m$ and $n$ are the sample sizes. I assume that $X$ and $Y$ are normally distributed. Their means do not necessarily have to differ significantly (which I analyse by a Welch test).

Currently I calculate the ration by $r = \frac{\bar x}{\bar y}$ and the according standard error by: $$se_r = r \cdot \sqrt{\left(\frac{se_x}{\bar x}\right)^2 + \left(\frac{se_y}{\bar y}\right)^2}$$ Usually my sample sizes are small (between 3 and 10) which may result in large individual errors $se_x$ and $se_y$. The choice if I calculate $r = \frac{\bar x}{\bar y}$ or $r = \frac{\bar y}{\bar x}$ is arbitrary (r is just a fold change). So what I don't understand is, that $se_r$ would be the same for $r$ and $1/r$, no matter if $\bar y$ and $\bar x$ differ by order of magnitudes. And this is why I am wondering if I calculate $se_r$ correctly and if the error interval $r \pm se_r$ shouldn't be asymmetrical.

Does it make in such cases only sense to calculate log-folds (i.e. $\log_{10}(r))$)? Would the standard error in this case be calculated as $\frac{se_r}{r\cdot\ln{10}}$?

Any clarification is highly appreciated.

• Are you aware of Fieller's theorem? The case of $log_{10}(r)$ is also discussed on the Wikipedia page. – COOLSerdash Jun 5 '13 at 10:48
• @COOLSerdash no I wasn't but your hint brought me over the last couple of hours to a lot of helpful ressources (+1). Especially stats.stackexchange.com/questions/16349/… – Beasterfield Jun 5 '13 at 13:39
• Sorry, I should have linked the post directly! I could have saved you hours of searching :) – COOLSerdash Jun 5 '13 at 13:48
• @COOLSerdash maybe one more question concerning the wikipedia page: en.wikipedia.org/wiki/Fieller's_theorem Would it be wrong to use always the approximation formula for case 2? Because even for cases where $\sqrt{Var(b)} \ll b$, $1 - g$ converges towards $1$? – Beasterfield Jun 6 '13 at 11:49