# How to combine standard errors for correlated variables

I was wondering what would be the formula for calculating the standard error of a quantity (A) that is the ratio of 2 quantities (A = B/C) if B and C are correlated?

According to page 2 of http://www.met.rdg.ac.uk/~swrhgnrj/combining_errors.pdf the formula for independent variables would be: However, how do I account for the covariance of B and C?

• Does this answer your question? Does the variance of a sum equal the sum of the variances? In particular, see the answer from @DouglasZare for the formula with covariances. Also, see the Wikipedia page. – EdM Jun 29 at 20:57
• @EdM - this question is about ratios – Henry Jun 29 at 21:06
• The formula you show is ultimately based on a log transformation of the initial ratio--$\log {b/c} = \log b - \log c$--plus the formula for the derivative of the log of a variable. Just as you would include the covariance term to adjust the formula in your linked reference for the sum or difference of 2 correlated variables, you include the covariance of the log-transformed ratio (subtracting twice the covariance in this case, as you have a difference between the logarithmic terms). – EdM Jun 29 at 21:17
• @EdM The log transformation is unnecessary and too limiting (what do you do when $B$ has appreciable chances of being negative??). – whuber Jun 29 at 21:51
• @whuber I agree. Love your answer, as usual. – EdM Jun 29 at 21:54

I find a little algebraic manipulation of the following nature to provide a congenial path to solving problems like this -- where you know the covariance matrix of variables $$(B,C)$$ and wish to estimate the variance of some function of them, such as $$B/C.$$ (This is often called the "Delta Method.")

Write

$$B = \beta + X,\ C = \gamma + Y$$

where $$\beta$$ is the expectation of $$B$$ and $$\gamma$$ that of $$C.$$ This makes $$(X,Y)$$ a zero-mean random variable with the same variances and covariance as $$(B,C).$$ Seemingly nothing is accomplished, but this decomposition is algebraically suggestive, as in

$$A = \frac{B}{C} = \frac{\beta+X}{\gamma+Y} = \left(\frac{\beta}{\gamma}\right) \frac{1 + X/\beta}{1+Y/\gamma}.$$

That is, $$A$$ is proportional to a ratio of two numbers that might both be close to unity. This is the circumstance that permits an approximate calculation of the variance of $$A$$ based only on the covariance matrix of $$(B,C).$$

Right away this division by $$\gamma$$ shows the futility of attempting a solution when $$\gamma \approx 0.$$ (See https://stats.stackexchange.com/a/299765/919 for illustrations of what goes wrong when dividing one random variable by another that has a good chance of coming very close to zero.)

Assuming $$\gamma$$ is reasonably far from $$0,$$ the foregoing expression also hints at the possibility of approximating the second fraction using the MacLaurin series for $$(1+Y/\gamma)^{-1},$$ which will be possible provided there is little change that $$|Y/\gamma|\ge 1$$ (outside the range of absolute convergence of this expansion). In other words, further suppose the distribution of $$C$$ is concentrated between $$0$$ and $$2\gamma.$$ In this case the series gives

\begin{aligned} \frac{1 + X/\beta}{1+Y/\gamma} &= \left(1 + X/\beta\right)\left(1 - (Y/\gamma) + O\left((Y/\gamma)^2\right)\right)\\&= 1 + X/\beta - Y/\gamma + O\left(\left(X/\beta\right)(Y/\gamma)^2\right).\end{aligned}

We may neglect the last term provided the chance that $$(X/\beta)(Y/\gamma)^2$$ being large is tiny. This is tantamount to supposing most of the probability of $$Y$$ is very close to $$\gamma$$ and that $$X$$ and $$Y^2$$ are not too strongly correlated. In this case

\begin{aligned} \operatorname{Var}(A) &\approx \left(\frac{\beta}{\gamma}\right)^2\operatorname{Var}(1 + X/\beta - Y/\gamma)\\ &= \left(\frac{\beta}{\gamma}\right)^2\left( \frac{1}{\beta^2}\operatorname{Var}(B) + \frac{1}{\gamma^2}\operatorname{Var}(C) - \frac{2}{\beta\gamma}\operatorname{Cov}(B,C)\right) \\ &= \frac{1}{\gamma^2} \operatorname{Var}(B) + \frac{\beta^2}{\gamma^4}\operatorname{Var}(C) - \frac{2\beta}{\gamma^3}\operatorname{Cov}(B,C). \end{aligned}

You might wonder why I fuss over the assumptions. They matter. One way to check them is to generate Normally distributed variates $$B$$ and $$C$$ in a simulation: it will provide a good estimate of the variance of $$A$$ and, to the extent $$A$$ appears approximately Normally distributed, will confirm the three bold assumptions needed to rely on this result do indeed hold.

For instance, with the covariance matrix $$\pmatrix{1&-0.9\\-0.9&1}$$ and means $$(\beta,\gamma)=(5, 10),$$ the approximation does OK (left panel): The variance of these 100,000 simulated values is $$0.0233,$$ close to the formula's value of $$0.0215.$$ But reducing $$\gamma$$ from $$10$$ to $$4,$$ which looks innocent enough ($$4$$ is still four standard deviations of $$C$$ away from $$0$$) has profound effects due to the strong correlation of $$B$$ and $$C,$$ as seen in the right hand histogram. Evidently $$C$$ has a small but appreciable chance of being nearly $$0,$$ creating large values of $$B/C$$ (both negative and positive). This is a case where we should not neglect the $$XY^2$$ term in the MacLaurin expansion. Now the variance of these 100,000 simulated values of $$A$$ is $$2.200$$ but the formula gives $$0.301,$$ far too small.

This is the R code that generated the first figure. A small change in the third line generates the second figure.

n <- 1e5   # Simulation size
beta <- 5
gamma <- 10
Sigma <- matrix(c(1, -0.9, -0.9, 1), 2)

library(MASS) #mvrnorm

bc <- mvrnorm(n, c(beta, gamma), Sigma)
A <- bc[, 1] / bc[, 2]
#
# Report the simulated and approximate variances.
#
signif(c(Var(A)=var(A),
Approx=(Sigma[1,1]/gamma^2 + beta^2*Sigma[2,2]/gamma^4 - 2*beta/gamma^3*Sigma[1,2])),
3)

hist(A, freq=FALSE, breaks=50, col="#f0f0f0")
curve(dnorm(x, mean(A), sd(A)), col="SkyBlue", lwd=2, add=TRUE)