Optimal training loads

Suppose I’m training for an upcoming race. I want to choose the training load that maximises my expected performance on race day. The harder I train, the better my performance will be but the more likely I am to injure myself. How should I balance this trade-off between better performance and greater risk of injury?

We can model this choice problem as follows. Let $t\in [0,1]$ represent my training load and $a\in R$ my natural ability.¹ My performance on race day is some function $f(t,a)$ of $t$ and $a$. I assume that this function is increasing and concave in $t$ (so that there are positive but diminishing returns to training), and increasing in $a$.

I can’t compete if I get injured, which occurs with some probability $p(t,r)$ that depends on my training load and my natural resistance to injury $r\in R$. I assume that $p$ is increasing and convex in $t$ (so that training increases my likelihood of injury at an increasing rate), and decreasing in $r$.

My objective is to choose the training load $t^{\ast }$ that maximises my expected performance² $$\psi \left(t\right)=(1-p(t,r\left)\right)f(t,a).$$ My assumptions on the shapes of $f$ and $p$ imply that $\psi$ is concave in $t$. Therefore, the unique optimal training load $t^{\ast }$ satisfies the first-order condition (FOC) $$\begin{array}{cc}0& ={\psi }^{\prime }\left(t^{\ast }\right)\\ & =-p_t(t^{\ast },r)f(t^{\ast },a)+(1-p(t^{\ast },r\left)\right)f_t(t^{\ast },a),\end{array}$$ where ${\psi }^{\prime }$ denotes the derivative of $\psi$ with respect to $t$, and where $p_t$ and $f_t$ denote the partial derivatives of $p$ and $f$ with respect to $t$. The FOC can be rewritten as $$(1-p(t^{\ast },r\left)\right)f_t(t^{\ast },a)=p_t(t^{\ast },r)f(t^{\ast },a),$$ which shows that I should keep training until the marginal benefit of improved performance (the left-hand side) equals the marginal cost of injury becoming more probable (the right-hand side).

I can’t determine the value of $t^{\ast }$ without further assumptions on $f$ and $p$. However, I can determine the relationship between $t^{\ast }$ and the parameters $a$ and $r$. Since ${\psi }^{\prime \prime }\left(t\right)<0$ for all feasible $t$, the implicit function theorem (IFT) implies that $$sign\frac{\partial t^{\ast }}{\partial \theta }=sign\frac{\partial {\psi }^{\prime }\left(t^{\ast }\right)}{\partial \theta }$$ for each element $\theta$ of the symbol set $\{a,r\}$. Now $$\frac{\partial {\psi }^{\prime }\left(t^{\ast }\right)}{\partial a}=-p_t(t^{\ast },r)f_a(t^{\ast },a)+(1-p(t^{\ast },r\left)\right)f_{ta}(t^{\ast },a),$$ where $f_a$ and $f_{ta}$ denote the partial derivatives of $f$ and $f_t$ with respect to $a$, and $$\frac{\partial {\psi }^{\prime }\left(t^{\ast }\right)}{\partial r}=-p_{tr}(t^{\ast },r)f(t^{\ast },a)-p_r(t^{\ast },r)f_t(t^{\ast },a),$$ where $p_{tr}$ and $p_r$ denote the partial derivatives of $p_t$ and $p$ with respect to $r$. By Young’s theorem, the mixed partials $f_{ta}$ and $p_{tr}$ satisfy $$f_{ta}(t,a)=\frac{\partial }{\partial t}\left(\frac{\partial f(t,a)}{\partial a}\right)$$ and $$p_{tr}(t,r)=\frac{\partial }{\partial t}\left(\frac{\partial p(t,r)}{\partial r}\right)$$ for all feasible $t$, $a$ and $r$. Thus, it seems reasonable to assume that $f_{ta}(t,a)\le 0$ and $p_{tr}(t,r)\le 0$, which mean that training washes out the benefits of natural ability and resistance to injury. These assumptions, together with the IFT, imply that $t^{\ast }$ is decreasing in $a$ and increasing in $r$—that is, I should train harder if I become less naturally able or more resistant to injury.