# Is the variance concave?

Let $$X$$ be a discrete random variables with values in the set $$\{x_1,\ldots, x_n\}\subset\mathbb{R}$$. Denote by $$p_i$$ the probability that $$X=x_i$$. We can then regard the variance $$\mbox{Var}(X)$$ as a function of the vector $$p\in \Delta^{n-1}\subseteq\mathbb{R}^n$$.

Will it be a concave function of $$p$$? With $$n=2$$, we get

$$\mbox{Var} (X)=p_1p_2(x_1-x_2)^2=p_1(1-p_1)(x_1-x_2)^2$$

which is concave. I'm not sure how to generalize this beyond two dimensions though.

We have $\mathrm{var}_p\left(X\right)=\mathbb{E}_p\left[X^2\right]-\mathbb{E}_p\left[X\right]^2$. Furthermore, for $\lambda\in\left[0,1\right]$ we can write $$\mathbb{E}_{\lambda p+(1-\lambda)q}\left[X^2\right]=\lambda\mathbb{E}_p\left[X^2\right]+(1-\lambda)\mathbb{E}_q\left[X^2\right].$$ Convexity of $\left(\cdot\right)^2$ and Jensen's inequality then yield $$\lambda \mathbb{E}_p\left[X\right]^2+(1-\lambda) \mathbb{E}_q\left[X\right]^2\geq \left(\lambda\mathbb{E}_p\left[X\right]+(1-\lambda)\mathbb{E}_q\left[X\right]\right)^2=\mathbb{E}_{\lambda p+(1-\lambda)q}\left[X\right]^2.$$

Thus we deduce $$\mathrm{var}_{\lambda p+(1-\lambda)q}\left(X\right)=\mathbb{E}_{\lambda p+(1-\lambda)q}\left[X^2\right]-\mathbb{E}_{\lambda p+(1-\lambda)q}\left[X\right]^2\\\geq\lambda\mathrm{var}_p\left(X\right)+(1-\lambda)\mathrm{var}_q\left(X\right),$$ that means the variance is concave.

• Thank you, that was a very clean solution. Oct 8, 2012 at 20:03

We look at the Hessian and see it's negative semi-definite because

i) function itself

\begin{align*} f(p) &= \text{Var}(X)\\ &= E[X^2] - E[X]^2\\ &= p^Ty - (p^Tx)^2, \end{align*} where $$y = [x_1^2,\,\ldots,\,x_n^2]^T$$.

\begin{align*} \nabla f &= y - 2(p^Tx)x\\ \nabla f^2 &= -2xx^T. \end{align*} $$\nabla f^2$$ is negative semi-definite for all $$p$$ $$\Leftrightarrow$$ $$f$$ is concave.