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For example, we know that

$var(x) = \sum\limits_i^n (p_ia_i)^2 - (\sum\limits_i^n (p_ia_i))^2$

Then

$\dfrac{\partial Var(x)}{\partial p_k} = \sum\limits_i^n (2p_ia_i) a_k - 2\sum\limits_i^n (p_ia_i)a_k = 0$

But we know that Variance is a concave function. Did I do something wrong in taking the derivative?

Olórin
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1 Answers1

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Check that $$\begin{align}\frac{\partial\operatorname{Var}(x)}{\partial p_k} &=\sum_i\frac{\partial}{\partial p_k}(p_ia_i)^2-\frac{\partial}{\partial p_k}\left(\sum_ip_ia_i\right)^2 \\ &=\frac{\partial}{\partial p_k}(p_ka_k)^2-\frac{\partial}{\partial p_k}\left(\sum_ip_ia_i\right)^2 \\ &= 2a_k^2-2a_k\sum_ip_ia_i\end{align}$$

Hagen von Eitzen
  • 374,180
  • Hmm there is something wrong with your last line, it should be $2p_ka_k^2 - 2a_k \sum_ip_ia_i$, but if you take the second derivative of that the whole thing becomes 0 so it is inconclusive – Olórin Oct 06 '15 at 01:48