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Given the Variance $\sum_{n=1}^{\infty}\frac{c\Delta\mu_n-\Delta(\sigma_{n}^{²}+\mu_{n}^2)}{P_{n}}$ and $ \sum_{n=1}^{\infty}P_{n} = \mu$. I would like to minimize Var by choosing the best $P_{n}$. Can somebody give me a hint how to get to the solution $P_{n}=const*\sqrt{c\Delta\mu_n-\Delta(\sigma_{n}^{²}+\mu_{n}^2)}$.

'const' stems from the constraint $ \sum_{n=1}^{\infty}P_{n} = \mu$.

I should probably use Lagrange and differentiate with respect to P. I tried it a couple of times but my Problem is I think to get the correct Lagrange function.

Thank you

Jordan
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1 Answers1

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Using aLagrangian mulitplier, you minimize$$\sum_n \frac{v_n}{P_n}+\lambda\left(\sum_n Pn-\mu\right).$$

Then cancelling the gradient, $$-\frac{v_n}{P_n^2}+\lambda=0$$

gives

$$P_n=\sqrt{\frac{v_n}{\lambda}}.$$

As you say,

$$c=\dfrac1{\sqrt\lambda}=\dfrac{\mu}{\displaystyle\sum_n\sqrt{v_n}}.$$

  • Thank you for your help! I have two additional questions regarding your solution.
    1. Why do you calculate $ +\lambda(...)$ and not $- \lambda(...)$. According to the lagrange formula it is '-'. Or is that because the term in brackets is $0$ so it doesnt make any difference?
    2. How do you get from $ \frac{1}{\sqrt{\lambda}} $ to $ \frac{\mu}{\sum_{n} \sqrt{v_{n}}} $ ?
     – Jordan Dec 10 '20 at 09:46
  • Dont mind the 2nd question. I got it. – Jordan Dec 10 '20 at 09:54
  • @Jordan: adding a positive (negative) is the same as subtracting a negative (positive), isn't it ? –  Dec 10 '20 at 09:58