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I have the following function I need to differentiate wrt. $C_i$ but cannot seem to re-arrange the form to match the correct solution

$$C=m^{1/{1-\theta}}\cdot\Big(\sum_{i=1}^m C_i^{{\theta-1}/\theta}\Big)^{\theta/{\theta-1}}$$

Differentiating $C$ wrt. $C_i$ I get:

$$m^{1/{1-\theta}}\cdot(\theta/{\theta-1}) \cdot \Big(\sum_{i=1}^mC_i^{{\theta-1}/\theta}\Big)^{(\theta/{\theta-1})-1}\cdot({\theta-1}/\theta)\cdot C_i^{({\theta-1}/\theta)-1} \Rightarrow$$

$$m^{1/{1-\theta}} \cdot \Big(\sum_{i=1}^mC_i^{{\theta-1}/\theta}\Big)^{1/{\theta-1}}\cdot C_i^{-1/\theta}$$

Now, one is supposed to reduce it to the following form:

$$\Big( \frac{C}{mC_i} \Big)^{\frac{1}{\theta}}$$

I can easily see how the $C_i$ appears in the expression, but have a hard time seeing

$$\big( \frac{C}{m}\big)^{\frac{1}{\theta}}=m^{1/{1-\theta}} \cdot \Big(\sum_{i=1}^mC_i^{{\theta-1}/\theta}\Big)^{1/{\theta-1}}$$

any help is highly appreciated :-) thanks

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  • $\begingroup$ curiosity question: Is this a likelihood function of some distribution ? $\endgroup$ Commented Sep 4, 2018 at 14:51
  • $\begingroup$ No, but I can see how the theta might allude to that. It is a CES (constant elasticity of substitution) function :-) $\endgroup$
    – user469216
    Commented Sep 4, 2018 at 14:52

1 Answer 1

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So you're asking why \begin{equation} \big( \frac{C}{m}\big)^{\frac{1}{\theta}}= m^{\frac{1}{1-\theta}} (\sum_{i=1}^mC_i^{{\theta-1}/\theta})^{\frac{1}{\theta-1}} \end{equation} Let \begin{equation} A = \sum_{i=1}^mC_i^{{\theta-1}/\theta} \end{equation} \begin{equation} \big( \frac{C}{m}\big)^{\frac{1}{\theta}} = \Big( \frac{m^{\frac{1}{1-\theta}} A^{\frac{\theta}{\theta-1}}}{m}\Big)^{\frac{1}{\theta}} = m^{(\frac{1}{1 - \theta } - 1 ) \frac{1}{\theta}} A^{\frac{\theta}{\theta-1}\frac{1}{\theta}} = m^{\frac{1}{1-\theta}} A^{\frac{1}{\theta-1}} = m^{\frac{1}{1-\theta}} (\sum_{i=1}^mC_i^{{\theta-1}/\theta})^{\frac{1}{\theta-1}} \end{equation}

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