I have the following Utility function: \begin{align} U = w^\prime\mu \end{align}

and Langrangian function subject to constraint: \begin{align} F (w, \lambda)= w^\prime\mu - \lambda(w^\prime i - 1) \end{align}

I would like to have an expression for w so that I can calculate the weights. Does anyone know how to solve the system and obtain the function for 'w'?

Thanks in advance!


Shouldn't the utility function be like this?

$$ L(w, \lambda) = w' \mu - \lambda\left( \sum_{i=1}^{n} w_i -1 \right) $$

In that case, let: $w'\mu $ be $\sum_{i=1}^{n} w_i \mu_i$, with the constraint that the weights must sum $1$.

$$ L(w, \lambda) = \sum_{i=1}^{n} w_i \mu_i -\lambda\left( \sum_{i=1}^{n} w_i -1 \right) $$

Then you just have to find each derivative for the problem, meaning:

$$ \frac{\partial L}{\partial w_i} = 0 \ , \forall i $$

$$ \frac{\partial L}{\partial \lambda} = 0 $$

If you had more constraints, you'd have to find also the derivative for the rest of constraints, meaning a vector $\lambda' $

  • 1
    $\begingroup$ Yes, I think so! thanks. Do you know how to solve it, because I am still stuck.. $\endgroup$ – user9891079 Oct 10 '18 at 10:28
  • $\begingroup$ But I still cannot figure out what the expressions for 'w' is.. I am trying to find the derivatives and set them equal to each other, but cannot find an expression for 'w' $\endgroup$ – user9891079 Oct 10 '18 at 12:40
  • $\begingroup$ There was a typo in my text. Look the derivatives, it will help now, I hope. But do you have a numerical example to which you want a solution? $\endgroup$ – YetAnotherUsr Oct 11 '18 at 11:19

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