0
$\begingroup$

I am thankful to any hints:

What I have:

Simple log-utility form: $u = \log c_1 + \beta \log c_2$

Budget constraints:

$c_1 + s \leq w$

$c_2 \leq R\; s$

Problem:

For utility maximization: $s = \frac{\beta}{1+\beta} \cdot w \; \; \; \; \;$ I am not getting this!

I have tried this:

Since the utility is monotonic, we use equality and then I substitue $c_1 \; and \; c_2$ so I get:

$ max \; \; \log (w-s) + \log Rs = 0$

Thus, $\frac{1}{w-s} + \frac{\beta}{s} = 0$

$ \implies s = - \beta w - \beta s \therefore$

$s = \frac{- \beta}{1 + \beta} w$ !!

Any ideas are appreciated.

$\endgroup$
1
$\begingroup$

The cost is an increasing function of $c_1$, so we can take $c_1 = w-s$.

If $\beta<0$, then we can choose $c_2$ close to zero to get arbitrarily large cost, so presumably we have $\beta \ge 0$. In this case, the cost is a non-decreasing function of $c_2$, so we should take $c_2 = Rs$. The utility now reduces to the unconstrained $f(s)= \log (w-s) + \beta \log (Rs)$.

Differentiating gives $-\frac{1}{w-s} + \beta \frac{1}{s} = 0$ (note the first minus sign!), which yields $s = \frac{\beta}{1+\beta} w$.

$\endgroup$
  • $\begingroup$ Oh God - I forgot to multiply by -1, thanks heaps Copper :) If I do such a mistake in real life, I am screwed :) $\endgroup$ – Ali NasserEddine Jun 24 '13 at 6:40
  • 1
    $\begingroup$ Everyone makes mistakess. $\endgroup$ – copper.hat Jun 24 '13 at 6:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.