# Can I assume that the expected value of this log ratio is 0?

I am taking two measurements from subjects: measurement A and measurement B.

I have some strong theoretical reasons to think that my measurements of A and B should on average be the same as each other within any individual subject. However, there's considerable natural variation. Some people are high on A and low on B, and vice-versa.

I have been instructed to calculate $\log\frac{A}{B}$ for each person in the sample, and then average those $\log\frac{A}{B}$ values across the whole sample.

I know it's a mistake to assume the expected value of a ratio is the ratio of the expected values. So even if I assume that A and B are on average the same as each other, I can't assume that the expected value of $\frac{A}{B}$ is 1.

However, I thought that maybe I could get around this by taking the log. I thought that if I could assume A and B have the same expected value, then the expected value of $\log\frac{A}{B}$ really would be 0. Am I wrong about this?

If so, why? And is there any other transformation I could have done that would have created the desired effect of producing an expected value of 0 for the outcome variable?

I've repeated this process with several samples, and so far I've always ended up with a mean $\log\frac{A}{B}$ above 0.

• Perhaps a more radical rethinking would be useful: why do you need to estimate an expectation of $A/B$ in the first place? Is it really relevant to your work? Quite possibly the expectation of $\log(A/B)$ would be more meaningful. – whuber Jul 19 '16 at 14:00
• I strongly suspect that knowing the expectation of $\log\frac{A}{B}$ would be more meaningful for what I'm doing. Is there some easy way to work that out based on the values of A and B I observe in the data? – user1205901 - Reinstate Monica Jul 19 '16 at 14:04
• Yes: since $\log(A/B)=\log(A)-\log(B)$, linearity of expectation implies $\mathbb{E}(\log(A/B))=\mathbb{E}(\log(A)) - \mathbb{E}(\log(B)).$ You can estimate the latter expectations by averaging the logs of the data. – whuber Jul 19 '16 at 14:10

It is possible to construct $A$ and $B$ that have the same expected values but for which $\log(A/B)$ has expected value very far from 0.
Consider a simple example, where $B$ always takes value 1 (expected value 1) and $A$ takes value 0.5 with probability 0.5 and takes value 1.5 with probability 0.5 (expected value 1). $A/B$ takes value 0.5 with probability 0.5 and takes value 1.5 with probability 0.5, so $\log(A/B)$ takes value $-0.693\ldots$ with probability 0.5 and value $0.405\ldots$ with probability 0.5. Therefore the expected value of $\log(A/B)$ is non-zero (negative).
Indeed the expected value of $\log(A/B)$ can be very far from 0. Consider if $A$ had instead taken value 0 with probability 0.5 and 2 with probability 0.5 (still expected value 1). Now $\log(A/B)$ takes value $-\infty$ with probability 0.5 and value $0.693\ldots$ with probability 0.5, meaning its expected value is $-\infty$. By flipping $A$ and $B$ in this example, we end up with a case where $\log(A/B)$ has expected value $\infty$.
Noting that $\log(A/B) = \log(A)-\log(B)$, $\mathbb{E}[\log(A/B)] = \mathbb{E}[\log(A)] - \mathbb{E}[\log(B)]$. Therefore, $\log(A/B)$ has expected value of 0 only when you have selected $A$ and $B$ such that $\log(A)$ and $\log(B)$ have the same expected value. This is true of some $A$ and $B$ with the same expected value but not others (as seen above).