On the convexity of variance^[1]

This past January, my dad shared a cute fact with me: the convexity of variance depends on how you define addition!

The most straightforward way to define the sum of random variables is pointwise: \((X+Y)(\omega) = X(\omega)+Y(\omega).\) In this case, \( \mathbb{V}(\alpha X+(1-\alpha)Y) = \alpha^2 \mathbb{V}(X) + (1-\alpha)^2 \mathbb{V}(Y) + 2\alpha(1-\alpha) \text{Cov}(X, Y)\leq \alpha^2 \mathbb{V}(X) + (1-\alpha)^2 \mathbb{V}(Y) \leq \alpha^2 \mathbb{V}(X) + (1-\alpha)^2 \mathbb{V}(Y) + 2\alpha(1-\alpha) \sqrt{\mathbb{V}(X)\mathbb{V}(Y)}\leq \alpha^2 \mathbb{V}(X) + (1-\alpha)^2 \mathbb{V}(Y) = (\alpha \text{sd}(X) + (1-\alpha)\text{sd}(Y))^2 \leq \alpha\mathbb{V}(X) + (1-\alpha)\mathbb{V}(Y)\), where in the last step we used the convexity of \(f(x) = x^2\). So we get convexity of variance here!

Now, another common way to define addition on random variables is with the mixture. Let's denote this by \(\oplus.\) Here, \(\mathbb{V}(\alpha X\oplus (1-\alpha)Y) = \alpha \mathbb{V}(X) + (1-\alpha)\mathbb{V}(Y) + \mathbb{V}[\cdot] \geq \alpha \mathbb{V}(X) + (1-\alpha)\mathbb{V}(Y)\), therefore in this case variance is concave.