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I would like to maximize a recursive functional equation but I am struggling with setting up the problem correctly. The equation I am interested in captures the nature of a decision process over time and has the following form:

$$f(l,h,d)=\max_{l\leqslant m\leqslant h}[(1-d)m(\frac{h-m}{h-l})+d^2f(l,m,d)]$$

The first term of the expression to be maximized is the current value of the situation I am trying to represent while the second term is its future value. The code below and the additional information that follows is intended to convey the idea of what I would like to do in some more detail:

f0[l_,h0_,d_] = Maximize[{(1-d)*m1*(h0-m1)/(h0-l)+d^2*f1[l,m1,d], 0<l<=m1<=h0 && 0<d<1}, {m1}]
f1[l_,m1_,d_] = Maximize[{(1-d)*m2*(m1-m2)/(m1-l)+d^2*f2[l,m2,d], 0<l<=m2<=m1 && 0<d<1}, {m2}]
f2[l_,m2_,d_] = Maximize[{(1-d)*m3*(m2-m3)/(m2-l)+d^2*f3[l,m3,d], 0<l<=m3<=m2 && 0<d<1}, {m3}]
f3[l_,m3_,d_] = ...

The function $f_0$ is the original function at time 0, the expression $(h-m)/(h-l)$ is the $cdf$ of a uniform distribution with support $[l,h]$ at time 0, $m_i$ is a choice variable with respect to which the maximization should be done at each iteration step, and $d$ is a discount factor. The function $f_1$ is the next iteration of the process which feeds into the RHS of $f_0$, and so on.

Basically, what should happen during the maximization is that at time $t$ the value for $m_i$ that maximizes the current iteration serves as the upper bound on the distribution at time $t+1$. So, m2 = m1 /. Maximize[{(1-d)*m1*(h0-m1)/(h0-l)+d^2*f1[l,m1,d], 0<l<=m1<=h0 && 0<d<1}, {m1}][[2]]. Over time, therefore, the distribution is increasingly truncated from above.

For sufficiently large intervals, the maximizer $m_i$ will lie in the interior of the range $[l,m_{i+1}]$. As the truncation process continues, however, the relative weight of the discount factor increases until it entirely eats up the benefits of delay. This means that $m_i$ is continuously pushed towards $l$ and that $m_n=l$ for some finite $n$. If we define $f_n(l,l,d)\equiv l/(1+d)$, the equation should have a unique solution and the sequence of optimizations should converge in a finite number of iterations.

What I have not managed is to figure out is how to interlink the individual functions f0, f1, ... in order to actually implement this process.

I am grateful for any help!

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    $\begingroup$ Can you give us a list of sample parameters to play with? e.g. l, d, h0, etc. $\endgroup$ – march Feb 2 '16 at 16:42
  • $\begingroup$ I usually set l=1, d=0.7 or so and h0 from somewhere in the single digits up to somewhere around 50. $\endgroup$ – m.user Feb 2 '16 at 20:55
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Just as a quick an dirty start, I would try this:

$maxStep = 3; (* or what you believe necessary *)
    $eps = 0.001; (* some limit to tell that l and m are now sufficiently close *)

ClearAll[f];
f[l_, m_, d_, $maxStep ] = l / (l + d); (* end by maxStep *)

f[l_, leps_, d_, _ ] /; l <= leps < l + $eps = l / (l + d) (* end by l approx equal m *)

f[l_, h_, d_, n_] /; l < h && 0 < d < 1 && 1 <= n < $maxStep := f[l, h, d, n] = Module[

   {m},

   MaxValue[
     {
       (1 - d) * m * (h - m)/(h - l) + d^2 * f[l, m, d, n + 1],

       (* s.t. *) 
       0 < l <= m <= h
     }
     ,
     m
   ]
]

f[ 4., 6., 0.1, 1]

3.63636

| improve this answer | |
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  • $\begingroup$ I used MaxValue (and also NMaxValue) but Maximize might work also. Hope that helps. $\endgroup$ – gwr Feb 2 '16 at 17:10
  • $\begingroup$ gwr, this works perfectly, thanks for your help! I greatly appreciate it. $\endgroup$ – m.user Feb 3 '16 at 21:07

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