# Seemingly arbitrary upper bound on correlation between two binomials

I know I'm doing something wrong, but I cannot find what it is, although it's a rather simple deduction. The context is genotypes as binomial random variables, but I managed to express the problem in purely mathematical terms.

I define two sets of Bernoulli random variables, with different probabilities. Their variances are defined:

$$X_{1},\dots,X_{n}\overset{\text{iid}}{\sim}\text{Ber}(\pi) \Rightarrow\text{var}(X_{i})=\pi(1-\pi) \\ Y_{1},\dots,Y_{n}\overset{\text{iid}}{\sim}\text{Ber}(\theta) \Rightarrow\text{var}(Y_{j})=\theta(1-\theta)$$

I establish some constant correlation between any pair of variables where one is from the first group and the other one from the second group:

$$\rho_{X_{i},Y_{j}}=r\quad\forall\left(i,j\right)$$

And now I introduce two binomial variables, which are the sum of each group:

$$A =\sum_{i=1}^{n}X_{i}\sim\text{Bin}(n,\pi)\Rightarrow\text{var}(A)=n\pi(1-\pi) \\ B =\sum_{j=1}^{n}Y_{j}\sim\text{Bin}(n,\theta)\Rightarrow\text{var}(B)=n\theta(1-\theta)$$

After this, I want to find the correlation between A and B, given the known correlation between any pair of $$\left(X_{i},Y_{j}\right)$$, which seems straightforward:

\begin{aligned} \rho_{A,B} &=\frac{\text{cov}(A,B)}{\left[\text{var}(A)\text{var}(B)\right]^{1/2}} \\[1.4ex] &=\frac{\text{cov}(\sum_{i=1}^{n}X_{i},\sum_{j=1}^{n}Y_{j})}{\left[n\pi(1-\pi)n\theta(1-\theta)\right]^{1/2}} \\[1.4ex] &=\frac{\sum_{i=1}^{n}\sum_{j=1}^{n}\text{cov}(X_{i},Y_{j})}{n\left[\pi(1-\pi)\theta(1-\theta)\right]^{1/2}} \\[1.4ex] &=\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{\text{cov}(X_{i},Y_{j})}{\left[\text{var}(X_{i})\text{var}(Y_{j})\right]^{1/2}} \\[1.4ex] &=\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\rho_{X_{i},Y_{j}} \\[1.4ex] &=\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}r \\[1.4ex] &=\frac{1}{n}n^{2}r \\[1.4ex] &=nr \end{aligned}

Since $$r$$ and $$\rho_{A,B}$$ are both correlations, they must be in $$[0,1]$$. But then, a seemingly arbitrary constrain on $$r$$ appears, which depends on any arbitrary $$n$$ that I might choose to build the binomial variables:

$$\rho_{A,B}=nr\leq1\Longleftrightarrow r\leq\frac{1}{n}$$

So the correlation $$\rho_{X_i, Y_j}$$ is now bound by $$\frac{1}{n}$$, where $$n$$ can be any natural number. I must be missing some error along the way, but I can't find it after hours of struggling with these equations. Any ideas?

I see nothing "wrong" per-say with any of the steps in your derivation. Rather, I think you have established the following fact:

Suppose that $$\text{Cov}(X_i, Y_j) = r$$ for all $$i$$ and $$j$$, and this property holds for all $$n$$. Then it must be the case that $$r=0$$.

In other words, it is not possible to have constant (non-zero) correlation between every $$X_i$$ and every $$Y_j$$.

A weaker condition, that might be of interest to you, is:

For all $$i=1,2,\cdots n$$, we have $$\text{Cov}(X_i, Y_i) = r$$ and for all $$i\neq j$$, we have $$\text{Cov}(X_i, Y_j) = s.$$

In the limit, $$n\rightarrow\infty$$, we can show that $$s$$ must be $$0$$, and we can infer some convenient covariance/correlation relationships. If we repeat your calculations under this modified assumption we have \begin{align*} \text{Cor}(A, B) &= \cdots \\[1.2ex] &=\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \text{Cor}(X_i, Y_j) \\[1.2ex] &= \frac{1}{n}\left(\sum_{i=1}^n \text{Cor}(X_i, Y_i) + 2\sum_{1\leq i < j \leq n} \text{Cor}(X_i, Y_j)\right) \\[1.2ex] &= r + \frac{2}{n}\binom{n}{2} s \\ &= r + (n-1)s \end{align*}

Since $$\text{Cor}(A,B) \in [-1,1]$$ for all $$n$$, it must be the case that $$s=0$$.

Under the stated assumptions, we now have $$\text{Cor}(A, B) =\text{Cor}(X_i, Y_i)$$ and $$\text{Cov}(A, B) = n\text{Cov}(X_i, Y_i)$$

• Thanks! So, in my particular problem, $n = 2$, so that bound, which in my case is $r < \frac{1}{2}$, seems not to be an error. – jmberros Dec 3 '19 at 14:53
• Why do you have $n=2$? This is not stated in the problem, correct? That does seem to be an upper bound, but I'm not entirely sure how you would specify that in a probability model. It is easy to enforce $Cov(X_i, Y_i) = r$, but not so easy (at least I don't see how off the top of my head) to enforce $Cov(X_i, Y_j) = r$ for $i\neq j$. – knrumsey Dec 3 '19 at 16:19
• Yeah, I didn't clarify $n=2$ in the problem, I kept it general. But my use case was human genotypes, which have 2 alleles (excluding sexual chromosomes). – jmberros Jan 21 at 15:51