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In the answer to this question, the correction term is claimed to be nonnegative: $$p_A \mu_A^2 + p_B \mu_B^2 - p_A^2 \mu_A^2 - p_B^2 \mu_B^2 - 2 p_A p_B \mu_A \mu_B \geq 0$$ where $p_A + p_B = 1$, $0 < p_A < 1$, and $0 < p_B < 1$. How can this be shown?

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CCL
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    Hi, welcome to CrossValidated. It may be better to ask for a proof in a comment of the answer in the original question. – Marc Claesen May 20 '13 at 15:30
  • Hi Marc. Thanks for the response. I don't see a comment link near the answer of the original question. – CCL May 20 '13 at 15:35
  • Oh, right. Once your reputation increases you will be able to add comments on answers. I guess for now this new question may the best (only) solution. – Marc Claesen May 20 '13 at 15:36
  • Because this expression was obtained as the variance of a random variable, it is necessarily nonnegative, QED. (If you happened to find it can represent a negative value, that would only mean an algebraic mistake occurred in the calculation of this result.) The non-negativity also follows immediately by applying Jensen's Inequality to $x\to x^2$ upon noting this is a convex function. (That's easy to see with the original expression and much harder to see in the expanded-out expression given here.) – whuber May 20 '13 at 16:02
  • Thanks @whuber for the response. I agree that $\mu^{(2)} - \mu^2 = p_A (\mu_A^2 + \sigma_A^2) + p_B (\mu_B^2 + \sigma_B^2) - (p_A \mu_A + p_B \mu_B)^2 \geq 0$. But why must the correction term $p_A \mu_A^2 + p_B \mu_B^2 - (p_A \mu_A + p_B \mu_B)^2 \geq 0$? – CCL May 20 '13 at 16:54
  • @CCL Merely set $\sigma_A=\sigma_B=0$. – whuber May 20 '13 at 17:24
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    @whuber I think I understand. Since the above relation must hold for any choice of PDFs $f_A(x)$ and $f_B(x)$, choose $f_A$ and $f_B$ so that $\sigma_A=\sigma_B=0$. Then the relation must still hold, and we get the desired result. – CCL May 20 '13 at 17:44
  • That's exactly right. And in case you don't like to consider Normal distributions with zero variances, just notice that $\mu^{(2)}-\mu^2$ is a continuous function of $\sigma_A$ and $\sigma_B$ (for any Real values they might have, positive or negative), whence its non-negativity for all *strictly positive* $\sigma_A$ and $\sigma_B$ implies its non-negativity in the limits as $\sigma_A \to 0$ and $\sigma_B \to 0$. – whuber May 20 '13 at 17:49

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