Perfectly Correlated Bernoulli Random Variables

probability

Here is a fun little exercise: show that any two perfectly correlated Bernoulli random variables $X$ and $Y$ must be equal.

Proof: The assumption can be stated as $$\frac{Cov(X, Y)}{\sigma_X \cdot \sigma_Y} = 1$$

Let $Z_X = (X - \mu_X) / \sigma_X$, $Z_Y = (Y - \mu_Y) / \sigma_Y$ be the standardizations so that $\mathbb{E}(Z_X) = 0$, $Var(Z_X) = 1$, and similarly for $Z_Y$. Then we have $$Cov(Z_X, Z_Y) = 1.$$

Now observe that $$Var(Z_X - Z_Y) = Var(Z_X) + Var(Z_Y) - 2 \cdot Cov(Z_X, Z_Y) = 2 - 2 \cdot 1 = 0$$

so that $Z_X - Z_Y = 0$ a.e. and therefore $Z_X = Z_Y$ a.e. But then $Y = aX + b$ for some constants $a$ and $b$, so either $Y = X$ or $Y = 1 - X$ since $X$ and $Y$ are Bernoulli. In the second case we have

$$Cov(X, 1 - X) = Cov(X, 1) + Cov(X, -X) = -Var(X)$$

and $\sigma_X \cdot \sigma_{1 - X} = \mu_X \cdot (1 - \mu_X) = Var(X)$ since $X$ is Bernoulli so the correlation in that case is $-1$. Therefore $X = Y$ as was to be shown. $\square$