The marginal value theorem holds true under three fairly mild conditions:

1. The fixed cost $T$ is larger than zero.
2. The reward function $E(t)$ increases with $t$.
3. The slope $dE(t)/dt$ of the reward function decreases with $t$ (i.e. $E(t)$ is a diminishing returns function).

We wish to prove that the instantaneous reward rate $r(t)$ equals the average reward rate $R(t)$ when $R(t)$ is maximal. To achieve this, we need to find the value of $r(t)$ when $R(t)$ is maximal. To find the maximal average reward rate, we make use of the fact that its slope is zero at a maximum. The average reward rate is defined as \begin{eqnarray} R(t) & = & \frac{E(t)} {T+t}, \end{eqnarray} and its derivative is \begin{eqnarray} \frac{dR}{dt} & = & \frac{1}{T+t} \frac{dE}{dt} + E(t) \frac{d(T+t)^{-1}} {dt}, \label{eqaa} \end{eqnarray} where (by definition) \begin{eqnarray} \frac{dE} {dt} & = & r(t), \label{eqa} \end{eqnarray} is the instantaneous reward rate, and where \begin{eqnarray} \frac{d((T+t)^{-1}) } { dt} & = & \frac{-1} {(T+t)^{2}}. \label{eqb} \end{eqnarray} Substituting Equations \ref{eqa} and \ref{eqb} into Equation \ref{eqaa}, \begin{eqnarray} \frac{dR}{dt} & = & r(t) \frac{1}{T+t} - \frac{E(t)} {(T+t)^{2}} , \label{eqaab} \end{eqnarray} where \begin{eqnarray} \frac{ E(t) } {T+t} & = & R(t), \end{eqnarray} is the average reward rate, so that Equation \ref{eqaab} becomes \begin{eqnarray} \frac{dR}{dt} & = & \frac{r(t) }{T+t} - \frac{R(t)} {T+t}. \end{eqnarray} At a maximum, this is equal to zero, \begin{eqnarray} \frac{r(t) }{T+t} - \frac{R(t)} {T+t} & = & 0. \end{eqnarray} Finally, multiplying both sides by $(T+t)$, and re-arranging yields \begin{eqnarray} r(t) & = & R(t). \end{eqnarray} This proves that the average reward rate is maximal when the instantaneous reward rate equals the average reward rate.

Back to the main article