The impact of firms’ adjustments on the indirect cost of illness


Illness-related absenteeism reduces firms’ output, an effect referred to as indirect cost (IC) and often included in cost-of-illness or cost-effectiveness (of health technologies) studies. The companies may foresee this effect and modify hiring or contracting policies. We present a model allowing the estimation of IC with such adjustments. We show that the risk of illness does not change the general shape and properties of the (expected) marginal productivity function. We apply our model to several illustrative examples and show that firm’s adjustments impact IC in an ambiguous way, depending on detailed company/market characteristics: in some cases the company reduces the employment (further increasing IC), in another—the opposite happens. Contrary to previous findings, teamwork and shortfall penalties may reduce IC in some settings. Our analysis highlights that IC should be split into the result of companies preparing for and actually experiencing sick leaves.

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  1. 1.

    For example, in Austria (Walter and Zehetmayr 2006), Sweden (Pharmaceutical Benefits Board 2003), the Netherlands (College Voor Zorgverzekeringen 2006), Australia (Pharmaceutical Benefits Advisory Committee 2008), or Italy (Capri et al. 2001).

  2. 2.

    This assumption approximately holds, e.g., in Poland, where the employer pays the salary for the first 33 days of the illness, and then the public insurer takes over (the employee gets 80% of a regular wage). Similar model is used in Germany, where the employer pays full salary for at least first 6 weeks of illness, then the insurer pays 70% of salary. In Sweden, employer covers sick-leaves for the first 2 weeks of illness.


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This paper was partially prepared when M. Jakubczyk was at The University of Iowa, Tippie College of Business, which was possible thanks to the Fulbright Senior Award 2015/2016. We would also like to thank R. Amir for useful comments.

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Correspondence to Michał Jakubczyk.

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Proof of Proposition 1

We first show how \(\text {MP}^E({\cdot })\) can be presented in a different form, more convenient for the present proof. We start with rewriting Eq. 4 and then modify it (notice we write it for \((L+1)\)-th worker here and change how the summation range is denoted)

$$\begin{aligned} \begin{aligned}&\frac{\text {MP}^E(L+1)}{1-s}=\sum _{i=0}^L {L \atopwithdelims ()i} s^{L-i}(1-s)^i \text {MP}(i+1)\\&\quad =\sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-i}(1-s)^i\text {MP}(i+1) + \sum _{i=1}^L {L-1 \atopwithdelims ()i-1} s^{L-i}(1-s)^i\text {MP}(i+1)\\&\quad =s \sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-1-i}(1-s)^i\text {MP}(i+1)\\&\qquad + (1-s) \sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-1-i}(1-s)^i\text {MP}(i+2)\\&\quad =s\times \frac{\text {MP}^E(L)}{1-s} + (1-s) \sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-1-i}(1-s)^i\text {MP}(i+2), \end{aligned} \end{aligned}$$

where we use the fact that for \(i\in {\mathbf {N}}, 0<i<L, {L \atopwithdelims ()i}={L-1 \atopwithdelims ()i} + {L-1 \atopwithdelims ()i-1}\). Hence, \(\text {MP}^E(L+1)\) is a weighted average (with weights s and \(1-s\)) of \(\text {MP}^E(L)\) and a similarly calculated expression in which we take \(\text {MP}\)s moved one worker to the right. As \(\text {MP}({\cdot })\) is decreasing it must be that

$$\begin{aligned} \sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-1-i}(1-s)^i\text {MP}(i+1)\ge \sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i} s^{L-1-i}(1-s)^i\text {MP}(i+2), \end{aligned}$$

as \(\ge \) holds for each term of the summation. \(\square \)

Proof of Proposition 2

The proof takes several steps. First, notice that \(\text {MP}^E(1)=(1-s)\text {MP}(1)\), and \(\text {MP}^E(2)=(1-s)\left( s\times \text {MP}(1)+(1-s)\times \text {MP}(2)\right) \). Then, if \(\text {MP}(2)\ge \text {MP}(1)\), then \(\text {MP}^E(2)\ge \text {MP}^E(1)\) for any s. Thus, if \(\text {MP}({\cdot })\) is increasing initially, so is \(\text {MP}^E({\cdot })\).

Notice, that if \(\text {MP}({\cdot })\) is decreasing after some point, then \(\text {MP}^E\) must also decrease at some point. Say, we calculate \(\text {MP}^E({\cdot })\) for some \(0\ll L_1\ll L_2\), where already \(L_1\) is much greater than the point where \(\text {MP}({\cdot })\) started to be decreasing. Increasing \(L_2\) results in \(\text {MP}^E(L_2\)) being calculated (as a weighted sum in Eq. 4) averaging lower values of \(\text {MP}({\cdot })\) than when calculating \(\text {MP}^E(L_1)\) (the weights assigned to the values of \(\text {MP}({\cdot })\) prevailing in both sums can be made arbitrarily small by increasing \(L_2\)).

We now show that if \(\text {MP}^E(L+1)<\text {MP}^E(L)\) then \(\text {MP}^E({\cdot })\) must be strictly decreasing further on. This is the main part of the whole proof, and again it is done in several steps. First, \(\text {MP}^E({\cdot })\) is calculated as an expected value of \(\text {MP}({\cdot })\) function calculated for a variable distributed according to a binomial distribution. It is then useful to see how the probabilities change in this distribution. Denote by B(knp) the probability of k successes in n independent experiments, where a single success comes with probability p. We are interested in

$$\begin{aligned} \frac{B(k,n,p)}{B(k+1,n,p)}=\frac{{n \atopwithdelims ()k}p^k(1-p)^{n-k}}{{n\atopwithdelims ()k+1}p^{k+1}(1-p)^{n-k-1}}=\frac{k+1}{n-k}\times \frac{1-p}{p}, \end{aligned}$$

and this is increasing in k.

As noticed in the proof of Proposition 1, \(\text {MP}^E(L+1)\) is calculated as a weighted average of \(\text {MP}^E(L)\) (calculated based on \(\text {MP}({\cdot })\) for workers from 1 to L) and a weighted average of \(\text {MP}\) calculated for workers from 2 to \((L+1)\) with weights given by a binomial distribution \(B({\cdot },L-1,1-s)\). It will be useful to show that if

$$\begin{aligned} \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+1)>\sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+2), \end{aligned}$$

and so \(\text {MP}^E(L+1)<\text {MP}^E(L)\), then also

$$\begin{aligned} \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+2)>\sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+3). \end{aligned}$$

Let \({\varDelta }(i)\) denote \(\text {MP}(i)-\text {MP}(i-1)\), defined for \(i\ge 2\). \({\varDelta }({\cdot })\) is positive for the part of \(\text {MP}({\cdot })\) with increasing marginal productivity. Our assumptions guarantee that \({\varDelta }({\cdot })\) changes from positive to negative. Rewriting inequality 10

$$\begin{aligned} \sum _{i=0}^{L-1} B(i,L-1,1-s){\varDelta }(i+2)<0, \end{aligned}$$

and, analogously, in order to prove inequality 11 we need to show that

$$\begin{aligned} \sum _{i=0}^{L-1} B(i,L-1,1-s){\varDelta }(i+3)<0. \end{aligned}$$

Obviously some \({\varDelta }({\cdot })\) values in inequality 12 must be negative. Assume non-trivially that also some (for \(i\le i^*\)) are positive. Then

$$\begin{aligned} \begin{aligned}&\sum _{i=0}^{L-1} B(i,L-1,1-s){\varDelta }(i+2) \ge \sum _{i=1}^{L-1} B(i,L-1,1-s){\varDelta }(i+2)\\&\quad =\underbrace{\sum _{i=1}^{i^*} B(i,L-1,1-s){\varDelta }(i+2)}_{\text {(I)}} + \underbrace{\sum _{i=i^*+1}^{L-1} B(i,L-1,1-s){\varDelta }(i+2)}_{\text {(II)}}, \end{aligned} \end{aligned}$$

where (I) is positive, and (II) is negative. We have \(\text {(I)}+\text {(II)}<0\), and so

$$\begin{aligned} \sum _{i=1}^{i^*} B(i-1,L-1,1-s){\varDelta }(i+2) + \sum _{i=i^*+1}^{L-1} B(i-1,L-1,1-s){\varDelta }(i+2)<0, \end{aligned}$$

as Eq. 9 is increasing in k, and so the negative elements in (II) are inflated more than the positive elements in (I). Rewriting the last inequality we get

$$\begin{aligned} \begin{aligned}&\sum _{i=1}^{L-1} B(i-1,L-1,1-s){\varDelta }(i+2)= \sum _{i=0}^{L-2} B(i,L-1,1-s){\varDelta }(i+3)\\&\quad \ge \sum _{i=0}^{L-1} B(i,L-1,1-s){\varDelta }(i+3), \end{aligned} \end{aligned}$$

which proves inequality 13. As the final step, notice that just as we decomposed \(\text {MP}^E(L+1)\), we can decompose \(\text {MP}^E(L+2)\):

$$\begin{aligned} \begin{aligned} \frac{\text {MP}^E(L+2)}{1-s}&=s^2 \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+1)\\&\quad +2s(1-s) \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+2)\\&\quad +(1-s)^2 \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+3)\\&=s^2 \sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+1)\\&\quad +2s(1-s) \sum _{i=0}^{L-1} B(i,L-1,1-s) (\text {MP}(i+1)+{\varDelta }(i+2))\\&\quad +(1-s)^2 \sum _{i=0}^{L-1} B(i,L-1,1-s) (\text {MP}(i+1)+{\varDelta }(i+2)+{\varDelta }(i+3))\\&=\sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+1) +(1-s^2)\sum _{i=0}^{L-1} B(i,L-1,1-s) {\varDelta }(i+2)\\&\quad +(1-s)^2\sum _{i=0}^{L-1} B(i,L-1,1-s) {\varDelta }(i+3)<\sum _{i=0}^{L-1} B(i,L-1,1-s) \text {MP}(i+1) \\&\quad +(1-s)\sum _{i=0}^{L-1} B(i,L-1,1-s) {\varDelta }(i+2)=\frac{\text {MP}^E(L+1)}{1-s}, \end{aligned} \end{aligned}$$

where the inequality results from omitting a negative term and \(1-s^2\ge 1-s\). This finishes the proof of \(\text {MP}^E({\cdot })\) being decreasing ever since the first decrease.

Finally, to show that the first decrease will only happen for some L greater than the point at which \(\text {MP}({\cdot })\) attains its maximum, it’s enough to revert the argument presented in the proof of proposition 1. \(\square \)

Proof of Proposition 3

We will show that if for some L we have \(\text {MP}^E(L) \le \text {MP}^E(L+1) \ge \text {MP}^E(L+2)\), then the difference \(\text {MP}^E(L+2)-\text {MP}^E(L+1)\) is strictly increasing in s. Based on the decomposition presented in the proof of Proposition 1, this difference amounts to

$$\begin{aligned} s\times \text {MP}^E(L+1)+(1-s)^2\sum _{i=0}^L {L \atopwithdelims ()i}(1-s)^i s^{L-i}\text {MP}(i+2)-\text {MP}^E(L+1), \end{aligned}$$

and so to

$$\begin{aligned} (1-s)^2\sum _{i=0}^L {L \atopwithdelims ()i}(1-s)^i s^{L-i}\text {MP}(i+2)-(1-s)^2\sum _{i=0}^L {L \atopwithdelims ()i}(1-s)^i s^{L-i}\text {MP}(i+1) \end{aligned}$$

or, more concisely, to \((1-s)^2\sum _{i=0}^L {L \atopwithdelims ()i}(1-s)^i s^{L-i}{\varDelta }(i+2)\).

The derivative of this expression with respect to s is given as \(\text {(I)}+\text {(II)}+\text {(III)}+\text {(IV)}+\text {(V)}\), where

$$\begin{aligned} \text {(I)}&=-2(1-s)\sum _{i=0}^L {L \atopwithdelims ()i}(1-s)^i s^{L-i}{\varDelta }(i+2),\\ \text {(II)}&=(1-s)^2{L \atopwithdelims ()0}Ls^{L-1}{\varDelta }(2),\\ \text {(III)}&=-(1-s)^2\sum _{i=1}^{L-1}\frac{L!}{(i-1)!\,(L-i)!}(1-s)^{i-1}s^{L-i}{\varDelta }(i+2),\\ \text {(IV)}&=(1-s)^2\sum _{i=1}^{L-1}\frac{L!}{i!\,(L-i-1)!}(1-s)^i s^{L-i-1}{\varDelta }(i+2),\\ \text {(V)}&=-(1-s)^2{L \atopwithdelims ()L}L(1-s)^{L-1}{\varDelta }(L+2). \end{aligned}$$

Now, \(\text {(II)}+\text {(IV)}=(1-s)^2 L\sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i}(1-s)^i s^{L-1-i}{\varDelta }(i+2)\), while \(\text {(III)}+\text {(V)}=-(1-s)^2 L\sum _{i=0}^{L-1} {L-1 \atopwithdelims ()i}(1-s)^i s^{L-1-i}{\varDelta }(i+3)\).

Notice (cf. Eq. 8 in the proof of Proposition 1) that

$$\begin{aligned} \text {MP}^E(L+1)-\text {MP}^E(L)=(1-s)^2\sum _{i=0}^{L-1}{L-1 \atopwithdelims ()i}s^{L-1-i}(1-s)^i{\varDelta }(i+2). \end{aligned}$$

Accounting for our assumptions, that directly proves that \(\text {(II)}+\text {(IV)}\ge 0\) and \(\text {(I)}\ge 0\) (applying the formula to \(\text {MP}^E(L+2)-\text {MP}^E(L+1)\)).

We can also decompose \(\text {MP}^E(L+2)\) into three elements as shown in first three lines of Eq. 14 in the proof or Proposition 2. Again, accounting for the fact that \(\text {MP}^E(L) \le \text {MP}^E(L+1) \ge \text {MP}^E(L+2)\) it must be that \(\text {(III)}+\text {(V)}\ge 0\).

If either \(\text {MP}^E(L) < \text {MP}^E(L+1)\) or \(\text {MP}^E(L+1) > \text {MP}^E(L+2)\), then the derivative is strictly positive (for \(s<1\)). \(\square \)

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Jakubczyk, M., Koń, B. The impact of firms’ adjustments on the indirect cost of illness. Int J Health Econ Manag. 17, 377–394 (2017).

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  • Absenteeism
  • Indirect cost
  • Teamwork
  • Output shortfall
  • Friction cost method
  • Societal perspective

JEL Classification

  • D21
  • J24
  • J21
  • L23