0
$\begingroup$

I have the following set of equations

$$P_i=\frac{\gamma}{1-\gamma}\frac{M}{C}\Bigg(\frac{C}{m C_i}\Bigg)^{\frac{1}{\theta}} \tag{1}$$

$$P=\Bigg[\frac{1}{m} \sum_{i=1}^mP_i^{1-\theta} \Bigg]^{\frac{1}{1-\theta}} \tag{2}$$

$$C=m^{\frac{1}{1-\theta}}\Bigg[\sum_{i=1}^m C_i^{\frac{\theta-1}{\theta}}\Bigg]^{\frac{\theta}{\theta-1}} \tag{3}$$

By inserting $Pi_i$ into $P$ it should be possible to obtain:

$$P=\frac{\gamma}{1-\gamma}\frac{M}{C} \tag{4}$$

By inserting $P_i$ into $P$ I've obtained:

$$P=\frac{\gamma}{1-\gamma}\frac{M}{C}\cdot \Bigg[\frac{1}{m} \sum_{i=1}^m\Bigg(\Big(\frac{C}{m C_i}\Big)^{\frac{1}{\theta}} \Bigg)^{1-\theta} \Bigg]^{\frac{1}{1-\theta}} $$

I've tried dragging the $C/m$ inside the summation out to yield a constant in front of the summation as: $$\frac{1}{m}\cdot\Big(\frac{C}{m}\Big)^{\frac{1-\theta}{\theta}}=C^{\frac{1-\theta}{\theta}}\cdot\Big(\frac{1}{m}\Big)^{\frac{1-\theta}{\theta}+1}=C^{\frac{1-\theta}{\theta}}\cdot\Big(\frac{1}{m}\Big)^{\frac{1}{\theta}}$$

Then inside the summation you have: $$\Bigg(\sum_{i=1}^m \Big(\frac{1}{C_i}\Big)^{\frac{1}{\theta}}\Bigg)^{1-\theta} = \Bigg( \sum_{i=1}^m C_i ^\theta \Bigg) ^{1-\theta}$$

But I can't see it getting me anywhere.

Any help is highly appreciated :) Thanks!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.