What is 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 forindependent variableswould be:

However, how do I account for the covariance of B and C?

**Answer**

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=β+X, C=γ+Y

where β is the expectation of B and γ 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=BC=β+Xγ+Y=(βγ)1+X/β1+Y/γ.

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 γ shows the futility of attempting a solution when γ≈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 γ 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/γ)−1, which will be possible provided there is little change that |Y/γ|≥1 (outside the range of absolute convergence of this expansion). In other words, **further suppose the distribution of C is concentrated between 0 and 2γ.** In this case the series gives

1+X/β1+Y/γ=(1+X/β)(1−(Y/γ)+O((Y/γ)2))=1+X/β−Y/γ+O((X/β)(Y/γ)2).

**We may neglect the last term provided the chance that (X/β)(Y/γ)2 being large is tiny.** This is tantamount to supposing most of the probability of Y is *very* close to γ and that X and Y2 are not too strongly correlated. In this case

Var(A)≈(βγ)2Var(1+X/β−Y/γ)=(βγ)2(1β2Var(B)+1γ2Var(C)−2βγCov(B,C))=1γ2Var(B)+β2γ4Var(C)−2βγ3Cov(B,C).

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 (1−0.9−0.91) and means (β,γ)=(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 γ 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 XY2 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)
```

**Attribution***Source : Link , Question Author : Ralphael M. , Answer Author : whuber*