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.