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 $$a\in\mathbb{R}$$ 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 $$r\in\mathbb{R}$$. 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 $$t^*$$ that maximises my expected performance2 $$\psi(t)=(1-p(t,r))\,f(t,a).$$ My assumptions on the shapes of $$f$$ and $$p$$ imply that $$\psi$$ is concave in $$t$$. Therefore, the unique optimal training load $$t^*$$ satisfies the first-order condition (FOC) \begin{align} 0 &= \psi'(t^*) \\ &= -p_t(t^*,r)\,f(t^*,a)+(1-p(t^*,r))\,f_t(t^*,a), \end{align} where $$\psi'$$ 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 $$(1-p(t^*,r))\,f_t(t^*,a)=p_t(t^*,r)f(t^*,a),$$ 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 $$t^*$$ without further assumptions on $$f$$ and $$p$$. However, I can determine the relationship between $$t^*$$ and the parameters $$a$$ and $$r$$. Since $$\psi''(t)<0$$ for all feasible $$t$$, the implicit function theorem (IFT) implies that $$\mathrm{sign}\frac{\partial t^*}{\partial \theta}=\mathrm{sign}\frac{\partial \psi'(t^*)}{\partial \theta}$$ for each element $$\theta$$ of the symbol set $$\{a,r\}$$. Now $$\frac{\partial \psi'(t^*)}{\partial a}=-p_t(t^*,r)\,f_a(t^*,a)+(1-p(t^*,r))\,f_{ta}(t^*,a),$$ where $$f_a$$ and $$f_{ta}$$ denote the partial derivatives of $$f$$ and $$f_t$$ with respect to $$a$$, and $$\frac{\partial \psi'(t^*)}{\partial r}=-p_{tr}(t^*,r)\,f(t^*,a)-p_r(t^*,r)\,f_t(t^*,a),$$ 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 $$f_{ta}(t,a)\le0$$ and $$p_{tr}(t,r)\le0$$, which mean that training washes out the benefits of natural ability and resistance to injury. These assumptions, together with the IFT, imply that $$t^*$$ 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.

1. For example, $$t$$ could represent the proportion of time before the race that I spend training. ↩︎

2. I assume that $$f$$ and $$p$$ are twice continuously differentiable so that $$\psi$$ is too. ↩︎