0
$\begingroup$

I have to maximize following expected utility function using Kuhn tucker conditions -

enter image description here

Since expected utility function are increasing $C_{1,t}$ and $C_{2,t}$ so constraints (i) and (ii) will hold with equality. Thus, I substituted these constraints into the objective function.

Note: Here, only $q_t^e$ is the only variable and expected utility function is maximized subject to $q_t^e$

For simplification, I denoted M= (1-$q_t^e$)$w_t$ + $q_t^e$$w_t$$z_t$

This made $C_{1,t}$ = nM and $C_{2,t}$= (ARθ/$z_t$)M

I also broke down constraint iii into two constraints $q_t^e$≤1 and -$q_t^e$≤0

I set up the legrange:

L = (1-π)[${(nM)}^{-γ}$]/(-γ) + π[${((ARθ/z_t)M)}^{-γ}$]/(-γ) + $λ_1$(1-$q_t^e$) + $λ_2$(0+$q_t^e$)

Differentiating with respect to $q_t^e$

(1-π)[${(nM)}^{-γ-1}$][n($z_t$-1)] + π[${((ARθ/z_t)M)}^{-γ-1}$][(ARθ/$z_t$)($z_t$-1)] = $λ_1$ - $λ_2$

$λ_1$(1-$q_t^e$) =0; $λ_1$≥0

$λ_2$(0+$q_t^e$) = 0; $λ_2$≥0

Case 1: $λ_1$=0 and $λ_2$

I got (1-π)[${(nM)}^{-γ-1}$][n($z_t$-1)] + π[${((ARθ/z_t)M)}^{-γ-1}$][(ARθ/$z_t$)($z_t$-1)] =0

Which boils down to M=0

which gave me - - > $q_t^e$ = 1/(1-$z_t$)

But solution provided is of the following form -

enter image description here

My answer doesn't match the solution provided. Can someone please look at my solution and tell me what did I do wrong?

$\endgroup$
1
$\begingroup$

After substituting for $c_{1t}$ and $c_{2t}$, the problem can be rewritten as : \begin{eqnarray*} \max_{q_t^e} \ & \frac{(1- \pi) (n\omega_t)^{-\gamma} }{-\gamma} \left((1-q_t^e) + z_tq_t^e\right)^{-\gamma} + \frac{\pi (AR\theta\omega_t)^{-\gamma } }{-\gamma z_t^{ -\gamma }} \left((1-q_t^e) + z_tq_t^e\right)^{-\gamma} \\ \text{s.t. } & q_t^e \in [0, 1]\end{eqnarray*}

which is same as \begin{eqnarray*} \max_{q_t^e} \ & \alpha \left((1-q_t^e) + z_tq_t^e\right)^{-\gamma} \\ \text{s.t. } & q_t^e \in [0, 1]\end{eqnarray*} where $\alpha = \dfrac{(1- \pi) (n\omega_t)^{-\gamma} }{-\gamma} + \dfrac{\pi (AR\theta\omega_t)^{-\gamma } }{-\gamma z_t^{ -\gamma }} $

Observe that this is equivalent to solving : \begin{eqnarray*} \max_{q_t^e} \ & (1-q_t^e) + z_tq_t^e \\ \text{s.t. } & q_t^e \in [0, 1]\end{eqnarray*}

This is an objective that is linear in choice variable $q_t^e$, which is increasing in $q_t^e$ when $z_t > 1$, decreasing in $q_t^e$ when $z_t < 1$, and is a constant for $z_t = 1$. Consequently, the solution is \begin{eqnarray*} q_t^e \in \begin{cases} \{1\} & \text{if } z_t > 1 \\ \{0\} & \text{if } z_t < 1 \\ [0, 1] & \text{if } z_t = 1 \end{cases} \end{eqnarray*}

$\endgroup$
  • $\begingroup$ Thank you so much Sir for a clear answer! $\endgroup$ – Elina Gilbert Nov 13 at 23:58

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.