# Correlation can help approximate variance of a difference

Sunday December 19, 2021

Even if random variables $$\textbf{A}$$ and $$\textbf{δ}$$ are uncorrelated, realized data $$A$$ and $$\delta$$ will happen to have some covariance. If $$B = A + \delta$$, then $$\text{var}(B) = \text{var}(A) + \text{var}(\delta) + 2\text{cov}(A, \delta)$$, and estimating any one variance by the others will be off by the $$2\text{cov}(A, \delta)$$ term. For $$\text{var}(\delta)$$, however, the estimate $$\text{var}(B)(1 - \text{corr}(A, B)^2)$$ will usually be closer to correct, at the cost of being slightly biased. (But don't forget that doing $$\text{var}( \delta )$$ directly is a great first choice.)

### Three methods for $$\text{var}( \delta )$$, when $$B = A + \delta$$

The case of $$B = A + \delta$$ (so that $$\delta = B - A$$) is central to the paired t-test, for example. The variance of $$A$$ and $$B$$ could each be large, but the variance of $$\delta$$ can still be small, making it easier to reject the null for $$\textbf{δ}$$, for example.

Be careful: We're trying to estimate a variance, so we're interested in the variance of the estimate of the variance, which can be confusing. Hopefully the language is clear enough here.

#### 1) Variance of Differences

The obvious thing to do is to subtract and calculate the variance of the differences: $$\text{var}( \delta ) = \text{var}( B - A )$$. This is a good idea and what you should do. It's unbiased and has low variance (for the estimate of the true variance of $$\textbf{δ}$$).

Why not do it like this? Honestly I'm not sure. Maybe you don't have complete data, and you're looking for a $$\delta_1 - \delta_2$$ where the means of $$A_1$$ and $$A_2$$ are assumed equal so you're using $$B_1 - B_2$$ but you want the (smaller) variance of $$\delta_1 - \delta_2$$? So you'll estimate some things with complete cases even though the main effect is estimated with all $$B$$ data? Why not still use this method with the complete cases? Maybe you're not using simple subtraction, but some more complex regression with multiple variables? In that case, why not use variance of the residuals directly? Do you just want a more complicated method?

#### 2) Difference of Variances

By the variance sum law, $$\text{var}( \textbf{B} ) = \text{var}( \textbf{A} ) + \text{var}( \textbf{δ} )$$, if $$\textbf{A}$$ and $$\textbf{δ}$$ are uncorrelated, so $$\text{var}( \textbf{δ} ) = \text{var}( \textbf{B} ) - \text{var}( \textbf{A} )$$. Real data is not generally perfectly uncorrelated, however: $$\text{var}(\delta) = \text{var}(B) - \text{var}(A) - 2\text{cov}(A, \delta)$$. The covariance term is zero in expectation, so estimating $$\text{var}(\delta) = \text{var}(B) - \text{var}(A)$$ is unbiased. But the $$2\text{cov}(A, \delta)$$ is a kind of noise term, and adds variance to the estimate of $$\text{var}( \textbf{δ} )$$.

#### 3) Using Correlation

It isn't obvious, but using $$\text{var}(B)(1 - \text{corr}(A, B)^2)$$ to estimate $$\text{var}( \textbf{δ} )$$ is much like using the Difference of Variances, but with a generally smaller (but always positive) error coming from $$\text{cov}(A, \delta)$$. Proceeding in small steps:

$\text{var}(\delta) = \text{var}(B) - \text{var}(A) - 2\text{cov}(A, \delta)$

$\text{var}(\delta) = \text{var}(B) - \left( \text{var}(A) + 2\text{cov}(A, \delta) \right)$

$\text{var}(\delta) = \text{var}(B) \left( 1 - \frac{\text{var}(A) + 2\text{cov}(A, \delta)}{ \text{var}(B) } \right)$

$\text{var}(\delta) = \text{var}(B) \left( 1 - \frac{\text{var}(A)^2 + 2\text{var}(A)\text{cov}(A, \delta)}{ \text{var}(A) \text{var}(B) } \right)$

This has been algebraic. Now add $$\text{cov}(A, \delta)^2$$ to the fractional term's numerator. (This is an error in the estimate, to get to the destination.)

$\text{var}(\delta) = \text{var}(B) \left( 1 - \frac{\text{var}(A)^2 + 2\text{var}(A)\text{cov}(A, \delta) + \text{cov}(A, \delta)^2 }{ \text{var}(A) \text{var}(B) } \right)$

$\text{var}(\delta) = \text{var}(B) \left( 1 - \frac{ ( \text{var}(A) + \text{cov}(A, \delta))^2 }{ \text{var}(A) \text{var}(B) } \right)$

The bit squared in the fraction's numerator is $$\sum{(A_i - \bar{A} )^2} + \sum{(A_i - \bar{A} )(\delta_i - \bar{\delta})}$$, which is $$\sum{(A_i - \bar{A})(A_i - \bar{A} + \delta_i - \bar{\delta})}$$. Since $$B_i = A_i + \delta_i$$ and $$\bar{B} = \bar{A} + \bar{\delta}$$, that's $$\sum{(A_i - \bar{A})(B_i - \bar{B})}$$, which is $$\text{cov}(A, B)$$.

$\text{var}(\delta) = \text{var}(B) \left( 1 - \frac{ \text{cov}(A, B)^2 }{ \text{var}(A) \text{var}(B) } \right)$

So at last, by definition:

$\text{var}(\delta) = \text{var}(B) \left( 1 - \text{corr}(A, B)^2 \right)$

QED. The error introduced is $$\text{cov}(A, \delta)^2 / \text{var}(A)$$, which will tend to be considerably smaller in absolute value than the $$2\text{cov}(A, \delta)$$ error in the Difference of Variances method because $$0 \le |\text{cov}(A, \delta)| / \text{var}(A) \ll 2$$. This error is however positive in expectation, so the estimate is biased: it will be slightly too small.

### Computational demonstration

Generating 1,000 datasets where each of $$A$$ and $$\delta$$ have 100 values drawn at random from Gaussians with variance 9 and 16, respectively, the three methods above generate estimates of $$\text{var}( \textbf{δ} )$$ as follows:

| Method                  | Mean var(d) estimate | var(estimate) |
|-------------------------|----------------------|---------------|
| Variance of Differences |                15.98 |          5.10 |
| Difference of Variances |                15.95 |         10.69 |
| Using Correlation       |                15.82 |          5.12 |

As expected, the method Using Correlation comes out more below the true value of 16 on average, but its variance is comparable to that of the Variance of Differences.

The bias is already small with just 100 samples, and gets smaller still as samples get bigger and sample correlations tend to be smaller.

Here's clumsy R code to do the experiment:

trials = 1000
samples = 100

set.seed(0)
est1 = c()
est2 = c()
est3 = c()

for (i in 1:trials) {
A = rnorm(samples, sd=3)  # var=9
d = rnorm(samples, sd=4)  # var=16
B = A + d  # var=25, in population sense
est1 = c(est1, var(B - A))  # same as var(d)
est2 = c(est2, var(B) - var(A))
est3 = c(est3, var(B) * (1 - cor(A, B)^2))
}

mean(est1)
##  15.98799
mean(est2)
##  15.9451
mean(est3)
##  15.82234
var(est1)
##  5.09602
var(est2)
##  10.69393
var(est3)
##  5.117983