On the joint production of research and training

Abstract

Universities and research institutions have the responsibility to produce science and to provide training to new generations of researchers. In this paper, we propose a model to analyze the determinants of a senior scientist’s decisions about allocating time between these tasks. The results of this decision depend upon the characteristics of the research project, the senior scientist’s concern for training and the expected innate ability of the junior scientist involved. We analyze the role that a regulator can play in defining both the value of scientific projects and the future population of independent scientists.

Appendix :

1.1 Proof of Lemma 1

The program can be rewritten as

$$\begin{array}{@{}rcl@{}} \underset{e_{R},e_{G}}{Max}\ \alpha e_{R}+a\delta \left( t-e_{R}\right) e_{R}+\beta a\left( t-e_{R}\right) \\ s.t.\qquad e_{R}\geq 0 \qquad \text{and}\qquad e_{R}\leq t \end{array} $$

The Lagrangian of this program is $\mathcal {L}=\alpha e_{R}+a\delta \left (t-e_{R}\right ) e_{R}+\beta a\left (t-e_{R}\right ) +\lambda e_{R}+\mu \left (t-e_{R}\right )$, which FOC is:

$$\alpha +a\delta \left( t-2e_{R}\right) -\beta a+\lambda -\mu =0. $$

Then cases to consider are (1) λ=0, μ=0 and $e_{R}=\frac {a\delta t+\alpha -\beta a}{ 2a\delta }$, which is a candidate when $\frac {a\delta t+\alpha -\beta a}{ 2a\delta }\geq 0$ and $\frac {a\delta t+\alpha -\beta a}{2a\delta }\leq t$, or equivalent, when $\alpha +a\left (\delta t-\beta \right ) \geq 0$ (which is always the case since δ t≥β≥0) and $\alpha -a\left (\delta t+\beta \right ) \leq 0$. (2) λ>0, μ=0 and e _R =0, which is a candidate when $\alpha +a\left (\delta t-\beta \right ) \leq 0$ and $\alpha -a\left (\delta t+\beta \right ) \leq 0$. This case never occurs since δ t≥β≥0. (3) λ=0, μ>0 and e _R =t, which asks for $\alpha -a\left (\delta t+\beta \right ) \geq 0$. Finally, the combination of these cases gives the results presented in the Lemma.

1.2 Proof of Lemma 3

We proceed for simplicity by regions of parameters:

Case 1: α−a β>0

In this case:

$$\begin{array}{@{}rcl@{}} u(t) &=&\frac{\left( a\delta t+\alpha \right)^{2}-\left( \beta a\right)^{2} }{4a\delta }+\frac{2\beta a\left( a\delta t-\alpha +\beta a\right) }{ 4a\delta }\ \text{if}\ t\geq \frac{\alpha -a\beta }{a\delta } \end{array} $$

(3)

$$\begin{array}{@{}rcl@{}} u(t) &=&\alpha t\ \text{if}\ t\leq \frac{\alpha -a\beta }{a\delta } \end{array} $$

(4)

Step (i) Let’s first consider the function u(t)=α t. The maximization problem is

$$\begin{array}{@{}rcl@{}} \underset{t}{Max}\ \left\{ \alpha t-\frac{c}{2}t^{2}\right\} \\ s.t.\qquad 0\leq t\leq \frac{\alpha -a\beta }{a\delta } \end{array} $$

The Lagrangian function is $\mathcal {L}=\alpha t-\frac {c}{2}t^{2}+\lambda (\frac {\alpha -a\beta }{a\delta }-t)+\mu t$. The FOC is

$$\alpha -ct-\lambda +\mu =0. $$

Then the possibilities are (i.1) λ=0 and μ=0, then $t=\frac { \alpha }{c}$, which is a candidate when $\frac {\alpha }{c}\leq \frac {\alpha -a\beta }{a\delta }\Longleftrightarrow \frac {a\delta \alpha }{\alpha -a\beta }\leq c$. (ii.2) λ>0 and μ=0, then $t=\frac {\alpha -a\beta }{ a\delta }$, which is a candidate when $\alpha -c(\frac {\alpha -a\beta }{ a\delta })>0$, i.e., $\frac {\alpha }{c}>\frac {\alpha -a\beta }{a\delta } \Longleftrightarrow \frac {a\delta \alpha }{\alpha -a\beta }>c$. (ii.3) λ=0 and μ>0, then t=0, which is a candidate if α<0 which is never the case. Step (ii) Let’s now consider the function $u(t)=\frac {\left (\alpha +at\delta \right )^{2}-(a\beta )^{2}}{4a\delta }+\frac {2a\beta (a\delta t-\alpha +a\beta )}{4a\delta }$. The maximization problem is

$$\begin{array}{@{}rcl@{}} \underset{t}{Max}\ \left\{ \frac{\left( \alpha +at\delta \right) ^{2}-(a\beta )^{2}}{4a\delta }+\frac{2a\beta (a\delta t-\alpha +a\beta )}{ 4a\delta }-\frac{c}{2}t^{2}\right\} \\ s.t.\qquad t\geq \frac{\alpha -a\beta }{a\delta } \end{array} $$

The Lagrangian is $\mathcal {L}=\frac {\left (\alpha +at\delta \right ) ^{2}-(a\beta )^{2}}{4a\delta }+\frac {2a\beta (a\delta t-\alpha +a\beta )}{ 4a\delta }-\frac {c}{2}t^{2}+\lambda (t-\frac {\alpha -a\beta }{a\delta })$. The FOC is

$$\frac{\alpha +at\delta +a\beta }{2}-ct+\lambda =0. $$

Then, the possibilities are: (ii.1) λ=0 and $t=\frac {\alpha +a\beta }{2c-a\delta }$, which is a candidate when $\frac {\alpha +a\beta }{ 2c-a\delta }>\frac {\alpha -a\beta }{a\delta }$ or, equivalently, when $\frac { a\delta \alpha }{\alpha -a\beta }\geq c$. (ii.2) λ>0 and $t=\frac { \alpha -a\beta }{a\delta }$, which asks for $\frac{a\delta \alpha }{\alpha -a\beta }<c.$, or equivalently $\frac {a\delta \alpha }{\alpha -a\beta }<c$. Step (iii) Comparing the results from the previews steps, and realizing that the candidate for solution in one of cases is a possible solution in the other case we obtain the result presented in the Lemma.

Case 2: α−a β<0

Here we have:

$$\begin{array}{@{}rcl@{}} u(t) &=&\frac{\left( a\delta t+\alpha \right)^{2}-\left( \beta a\right)^{2} }{4a\delta }+\frac{2\beta a\left( a\delta t-\alpha +\beta a\right) }{ 4a\delta }\ \text{if}\ t\geq \frac{a\beta -\alpha }{a\delta } \end{array} $$

(5)

$$\begin{array}{@{}rcl@{}} u(t) &=&a\beta t\ \text{if}\ t\leq \frac{a\beta -\alpha }{a\delta } \end{array} $$

(6)

Step (i) Let’s consider the function u(t)=a β t. The maximization problem is

$$\begin{array}{@{}rcl@{}} \underset{e_{R},e_{G}}{Max}\ \left\{ a\beta t-\frac{c}{2}t^{2}\right\} \\ s.t.\qquad 0\leq t\leq \frac{a\beta -\alpha }{a\delta } \end{array} $$

The Lagrangian function is $\mathcal {L}=a\beta t-\frac {c}{2}t^{2}+\lambda (\frac {a\beta -\alpha }{a\delta }-t)+\mu t$. The FOC is

$$a\beta -ct-\lambda +\mu =0. $$

Then the possibilities are (i.1) λ=0 and μ=0 which gives $t= \frac {a\beta }{c}$, which is a candidate when $\frac {a\beta }{c}\leq \frac { a\beta -\alpha }{a\delta }\Longleftrightarrow \frac {\beta a^{2}\delta }{ a\beta -\alpha }\leq c$. (ii.2) λ>0 and μ=0, and $t=\frac { a\beta -\alpha }{a\delta }$, which is a candidate when $\alpha -c\left (\frac {a\beta -\alpha }{a\delta }\right ) >0$, i.e., $\frac {\alpha }{c}>\frac { a\beta -\alpha }{a\delta }\Longleftrightarrow \frac {\beta a^{2}\delta }{ a\beta -\alpha }>c$. (ii.2) λ=0 and μ>0, and t=0, which is a candidate when a β<0, which is never the case. Step (ii) Let’s now consider the function $u(t)=\frac {\left (\alpha +at\delta \right )^{2}-(a\beta )^{2}}{4a\delta }+\frac {2a\beta (a\delta t-\alpha +a\beta )}{4a\delta }$. The maximization problem is

$$\begin{array}{@{}rcl@{}} \underset{t}{Max}\ \left\{ \frac{\left( \alpha +at\delta \right) ^{2}-(a\beta )^{2}}{4a\delta }+\frac{2a\beta (a\delta t-\alpha +a\beta )}{ 4a\delta }-\frac{c}{2}t^{2}\right\} \\ s.t.\qquad t\geq \frac{a\beta -\alpha }{a\delta } \end{array} $$

The Lagrangian function is $\mathcal {L}=\frac {\left (\alpha +at\delta \right )^{2}-(a\beta )^{2}}{4a\delta }+\frac {2a\beta (a\delta t-\alpha +a\beta )}{4a\delta }-\frac {c}{2}t^{2}+\lambda (t-\frac {a\beta -\alpha }{ a\delta })$. The FOC is

$$\frac{\alpha +at\delta +a\beta }{2}-ct+\lambda =0. $$

Then, the possibilities are: (ii.1) λ=0 and $t=\frac {\alpha +a\beta }{2c-a\delta }$, which is a candidate when $\frac {\alpha +a\beta }{ 2c-a\delta }>\frac {a\beta -\alpha }{a\delta }$ or, equivalently, when $\frac { \beta a^{2}\delta }{a\beta -\alpha }>c$. Note that $\frac {\beta a^{2}\delta }{a\beta -\alpha }<\frac {a\delta }{2}$ (ii.2) λ>0 and $t=\frac { a\beta -\alpha }{a\delta }$, which asks for $\frac {\alpha +a\delta (-\frac { \alpha -a\beta }{a\delta })+a\beta }{2}-c(\frac {a\beta -\alpha }{a\delta } )<0$, or equivalently $\frac {\beta a^{2}\delta }{a\beta -\alpha }<c$. Step (iii). Comparing the results from the previews steps, and realizing that the candidate for solution in one of cases is a possible solution in the other case we obtain the result presented in the Lemma.

Case 3: α−a β=0

In this case:

$$u(t)=t\frac{2\left( \alpha +\beta a\right) +a\delta t}{4}\text{if }t\geq 0 $$

Then the maximization problem of the senior is:

$$\begin{array}{@{}rcl@{}} \underset{t}{Max}\ \left\{ \frac{2t\left( \alpha +\beta a\right) +a\delta t^{2}}{4}-\frac{c}{2}t^{2}\right\} \\ s.t. \qquad t\geq 0 \end{array} $$

The Lagrangian function is $\mathcal {L}=\frac {2t\left (\alpha +\beta a\right ) +a\delta t^{2}}{4}-\frac {c}{2}t^{2}+\lambda t$. The FOC is

$$\frac{\alpha +a\beta +at\delta }{2}-ct+\lambda =0 $$

Then, the possibilities are: (i) λ=0 and $t=\frac {\alpha +a\beta }{2c-a\delta }$, which is a candidate when $\frac {\alpha +a\beta }{ 2c-a\delta }\geq 0$ or, equivalently, when $c>\frac {a\delta }{2}$. (ii) λ>0 and t=0, which asks for $\frac {\alpha +a\beta }{2}<0$, which is never the case.

1.3 Proof of Lemma 5

For the case β=0 and δ=1, we consider the candidates that satisfy a ²−(g T+A)a+α g≥0 and a ²−(g T+A)a+α g<0 sequentially and then we provide the solution as function of the parameters.

Step 1: a ²−(g T+A)a+α g≥0

The utility function to be considered is $u^{\ast }(t)=\alpha \left (T-\frac { a-A}{g}\right )$.

Depending on the parameters, the values of a such that a ²−(g T+A)a+α g=0 (they only exist if (g T+A)²−4α g≥0) will lie, or not, in the interval where the junior’s ability is defined, $\left [ A,gT+A\right ]$. When a=A, a ²−(g T+A)a+α g has a positive value if A T≤α, and a negative value otherwise. When a=g T+A, the value for a ²−(g T+A)a+α g is always positive. Hence we can define 3 possible regions to attain a solution: a) when A T≤α and (g T+A)²−4α g≥0; b) when A T≤α and (g T+A)²−4α g<0; and c) when A T>α and (g T+A)²−4α g≥0. The One last case could be A T>α and (g T+A)²−4α g<0, however it is not possible since this would mean that the function never has negative values and, simultaneously, has a negative value when a=A.

Case a) A T≤α and (g T+A)²−4α g≥0

In this case the optimal ability lies in either one of the two following regions: $A\leq a\leq \frac {gT+A-\sqrt {(gT+A)^{2}-4\alpha g}}{2}$ or $\frac { gT+A+\sqrt [2]{(gT+A)^{2}-4\alpha g}}{2}\leq a\leq gT+A$. We first formalize the problem with respect to the first region:

$$\begin{array}{@{}rcl@{}} &&\underset{a}{Max}\ \left\{ \alpha \left( T-\frac{a-A}{g}\right) \right\} \\ s.t\qquad a &\geq &A \\ a &\leq &\frac{gT+A-\sqrt{(gT+A)^{2}-4\alpha g}}{2} \end{array} $$

The lagrangian is $\mathcal {L}=\alpha \left (T-\frac {a-A}{g}\right ) +\lambda (a-A)+\mu (\frac {gT+A-\sqrt {(gT+A)^{2}-4\alpha g}}{2}-a)$. The FOC is $- \frac {\alpha }{g}+\lambda -\mu =0$

There are two possible cases for the lagrangian multipliers:

1) λ>0, μ=0. a=A is a candidate.

2) λ>0, μ>0. This holds when $A=\frac {gT+A-\sqrt { (gT+A)^{2}-4\alpha g}}{2}\Leftrightarrow AT=\alpha$ which is a particular case of 1). Hence a=A is again a candidate.

Formalizing the problem with respect to the second region, the FOC is:

$$-\frac{\alpha }{g}-\lambda +\mu =0 $$

One possible case exists for the lagrange multiplier:

1) λ=0, μ>0. In this case, $a=\frac {gT+A+\sqrt { (gT+A)^{2}-4\alpha g}}{2}$ is a candidate.

Since the utility function is decreasing in a, a=A is a candidate for the optimal ability.

Case b) A T≤α and (g T+A)²−4α g<0

The region to work with is A≤a≤g T+A, since the function always has positive values in this case. Since the utility function of the senior is decreasing in a, a=A is a candidate for the optimal ability.

Case c) A T>α and (g T+A)²−4α g≥0

This is a particular case of case 1.a), where we only consider the second region, hence $a=\frac {gT+A+\sqrt {(gT+A)^{2}-4\alpha g}}{2}$ is a candidate for the optimal ability.

We now summarize the candidates for step 1:

$$\begin{array}{@{}rcl@{}} a &=&A\qquad \text{if}\qquad AT\leq \alpha \text{ } \\ a &=&\frac{gT+A+\sqrt{(gT+A)^{2}-4\alpha g}}{2}\qquad \text{if}\qquad AT>\alpha \text{ and }(gT+A)^{2}-4\alpha g\geq 0 \end{array} $$

Step 2: a ²−(g T+A)a+α g≤0

The utility function to be considered in this case is $u^{\ast }(a,t)=\frac { \left (at+\alpha \right )^{2}}{4a}$.

Following the same logic as in step 1, we have 2 subcases to solve this problem: 2.a) when A T≥α and (g T+A)²−4α g≥0; 2.b) when A T≤α and (g T+A)²−4α g≥0. The other two subcases are impossible, since (g T+A)²−4α g<0 means the function always has positive values. Hence, in this step we take as given that (g T+A)²−4α g≥0.

Case a) A T≥α

There is one region for a to work with: $A\leq a\leq \frac {gT+A+\sqrt [2]{ (gT+A)^{2}-4\alpha g}}{2}$. The maximization problem is:

$$\begin{array}{@{}rcl@{}} &&\underset{a,t}{Max}\ \left\{ \frac{1}{4a}\left( a(T-\frac{a-A}{g} )+\alpha \right)^{2}\right\} \\ \ s.t.\ \ a &\geq &A\qquad \text{and }\qquad a\leq \frac{gT+A+ \sqrt{(gT+A)^{2}-4\alpha g}}{2} \end{array} $$

The FOC is:

$$\frac{1}{a^{2}}\left( a(T-\frac{a-A}{g})+\alpha \right) \left( a(T-\frac{3a-A }{g})-\alpha \right) +\lambda -\mu =0 $$

There are 4 possible cases for the lagrange multipliers:

1) λ>0, μ=0. Looking at the FOC, it must be that $a(T-\frac { 3a-A}{g})-\alpha <0$, that is, $a\in (\frac {gT+A-\sqrt {(gT+A)^{2}-12\alpha g} }{6},\frac {gT+A+\sqrt {(gT+A)^{2}-12\alpha g}}{6})$, which is the case when a=A.

2) λ=0, μ>0. In this case, $a=\frac {gT+A+\sqrt [2]{ (gT+A)^{2}-4\alpha g}}{2}$ is a candidate if

$a\in (\frac {gT+A-\sqrt {(gT+A)^{2}-12\alpha g}}{6},\frac {gT+A+\sqrt [2]{ (gT+A)^{2}-12\alpha g}}{6})$. We check that indeed a belongs to this interval. This holds only if (g T+A)²−12α g≥0.

3) λ=0, μ=0. $a=\frac {gT+A\pm \sqrt [2]{(gT+A)^{2}-12\alpha g}}{ 6}$ are candidates, provided (g T+A)²−12α g≥0.

4) λ>0, μ>0. This holds when $A=\frac {gT+A-\sqrt { (gT+A)^{2}-4\alpha g}}{2}$ $\Leftrightarrow AT=\alpha$. Hence a=A is a candidate again.

The utility function is decreasing from A until $\frac {gT+A-\sqrt { (gT+A)^{2}-12\alpha g}}{6}$, increasing from then on until $\frac {gT+A+\sqrt { (gT+A)^{2}-12\alpha g}}{6}$, and decreasing onwards. Hence, when (g T+A)²−12α g≥0$a=\frac {gT+A+\sqrt {(gT+A)^{2}-12\alpha g}}{6}$ is candidate for the optimal ability and when (g T+A)²−12α g<0, a=A .

Case b) A T≤α

There is one region for a to work with:$\frac {gT+A-\sqrt { (gT+A)^{2}-4\alpha g}}{2}\leq a\leq \frac {gT+A+\sqrt {(gT+A)^{2}-4\alpha g}}{2 }$. The FOC is the same as in case a), hence the possible cases for the lagrangean multipliers are:

1) λ>0,μ=0. $a=\frac {gT+A-\sqrt {(gT+A)^{2}-4\alpha g}}{2}$ and it must be that $a\in (\frac {gT+A-\sqrt {(gT+A)^{2}-12\alpha g}}{6}$,

$\frac {gT+A+\sqrt {(gT+A)^{2}-12\alpha g}}{6})$. Since $\frac {gT+A-\sqrt { (gT+A)^{2}-4\alpha g}}{2}<\frac {gT+A-\sqrt {(gT+A)^{2}-12\alpha g}}{6}$, indeed it is a candidate.

2) λ=0,μ>0. $a=\frac {gT+A+\sqrt {(gT+A)^{2}-4\alpha g}}{2}$ and it must be that $a\in (\frac {gT+A-\sqrt {(gT+A)^{2}-12\alpha g}}{6}$,

$\frac {gT+A+\sqrt {(gT+A)^{2}-12\alpha g}}{6})$. It is straightforward to check that a indeed belongs to this interval, so it is a candidate, provided that (g T+A)²−12α g≥0.

3) λ=0,μ=0. In this case, $a=\frac {gT+A\pm \sqrt { (gT+A)^{2}-12\alpha g}}{6}$ provided that (g T+A)²−12α g≥0.

4) λ>0,μ>0. In this case, $a=\frac {gT+A}{6}$, which happens when (g T+A)²−12α g=0, a particular case of 3).

Analyzing all the candidates and the behavior of the utility function, $a= \frac {gT+A+\sqrt {(gT+A)^{2}-12\alpha g}}{6}$ is a candidate for the optimal ability when (g T+A)²−12α g≥0 and $A=\frac {gT+A-\sqrt { (gT+A)^{2}-4\alpha g}}{2}$ when (g T+A)²−12α g<0.

We now summarize all candidates for case 2:

$$\begin{array}{@{}rcl@{}} a =\frac{gT+A+\sqrt{(gT+A)^{2}-12\alpha g}}{6}\qquad \text{if}\qquad (gT+A)^{2}-12\alpha g\geq 0 \\ a &=&A\qquad \text{if}\qquad AT\geq \alpha \text{ and }(gT+A)^{2}-12\alpha g<0 \\ a &=&\frac{gT+A-\sqrt{(gT+A)^{2}-4\alpha g}}{2}\qquad \text{if}\;\; AT<\alpha \text{ and }(gT+A)^{2}-12\alpha g<0 \end{array} $$

As function of the parameters, we evaluate and compare the utility of the solutions attained in step 1 and step 2. The final solutions for a ^∗ and t ^∗ (attained recursively) are:

• If (g T+A)²−12α g≥0, then $a^{\ast }=\frac {gT+A+\sqrt { (gT+A)^{2}-12\alpha g}}{6}$ and $t^{\ast }=\frac {5(gT+A)-\sqrt { (gT+A)^{2}-12\alpha g}}{6g}$

• If (g T+A)²−12α g<0, then a ^∗=A and t ^∗=T.