Are there interior solutions to this constrained optimization problem?

Question

I am getting some confusing results solving this problem:

$$\max_{c_0\geq 0, c_1\geq 0} \ R(1-c_0) [p t_1 + (1-p) c_1^{-\lambda} t_2] \\ s.t. \ c_0+c_1 \leq 1$$

where $\lambda>1$, $R>0$, $p$ is the probability of $t_1>0$ and $(1-p)$ is the probability of $t_2<0$. I need to derive what parameter values give an interior solution and which give a corner solution. Currently, I have been getting contradictions, so I am unsure if the problem itself has a solution.

Any help or pointers would be much appreciated. Thanks in advance

Answer 1

Suppose $c_1 = 0$, we have $R(1-c_0)pt_1$, then we obtain corner optimal point if $t_1>0$ w.p. 1. If $c_1 = 1$, we have $R(1-c_0)[pt_1+(1-p)t_2]$. We obtain a corner optimal point iff $pt_1 + (1-p)t_2 >0$. Maybe we can interpret the latter case as: if $t$ is random variable (taking some positive or negative value), then it takes a positive value with more probability than a negative value (hence a positive mean).
So I think $p$ is the only parameter where we can derive some condition. Did you try this way?

Answer 2

Since $R>0$ we can ignore $R$ in finding the maximizer. We cannot have $c_1=0$, so the only possibilities for solutions on the boundary are with $c_0+c_1=1$ or $c_0=0$.

Suppose $c_0+c_1=1$. Then the objective function is equal to

$$ pt_1c_1+(1-p)c_1^{1-\lambda}t_2$$

Each term is nondecreasing and at least one term is strictly increasing $c_1$, so the maximum occurs at $c_1=1$. Hence $(c_0,c_1)=(0,1)$ is a candidate for a maximizer. The value of the objective function at this candidate is $pt_1+(1-p)t_2$.

Suppose $c_0=0$. Then the objective function is equal to

$$pt_1+(1-p)c_1^{-\lambda}t_2$$

The second term is nondecreasing in $c_1$. If $p<1$ then it is strictly increasing and the maximum occurs at $c_1=1$ (the candidate solution found above). If $p=1$, then any $c_1>0$ is a candidate for a solution and the value of the objective function at the candidate is $t_1$.

Finally, suppose $c_0,c_1>0$ and $c_0+c_1<1$. Then $(1-c_0)>0$ and so the objective function is nondecreasing in $c_1$. Thus we get the same candidates as the previous case.

So, if $p<1$, the maximizer is $(c_0,c_1)=(0,1)$. If $p=1$, then any $(c_0,c_1)$ with $c_0=0$ and $c_1>0$ is a maximizer.