Is a bird in the hand worth two in the bush? Or, whether scientists should publish intermediate results

Thomas Boyer

doi:10.1007/s11229-012-0242-4

Synthese

January 2014, Volume 191, Issue 1, pp 17–35 | Cite as

Is a bird in the hand worth two in the bush? Or, whether scientists should publish intermediate results

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Thomas Boyer

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First Online: 23 February 2013

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Abstract

A part of the scientific literature consists of intermediate results within a longer project. Scientists often publish a first result in the course of their work, while aware that they should soon achieve a more advanced result from this preliminary result. Should they follow the proverb “a bird in the hand is worth two in the bush”, and publish any intermediate result they get? This is the normative question addressed in this paper. My aim is to clarify, to refine, and to assess informal arguments about the choice whether to publish intermediate results. To this end, I adopt a rational decision framework, supposing some utility or preferences, and I propose a formal model. The best publishing strategy turns out to depend on the research situation. In some simple circumstances, even selfish and short-minded scientists should publish their intermediate results, and should thus behave like their altruistic peers, i. e. like society would like them to behave. In other research situations, with inhomogeneous reward or difficulty profiles, the best strategy is opposite. These results suggest qualified philosophical morals.

Keywords

Social epistemology Intermediate results Publication Formal model

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Notes

Acknowledgments

Many thanks to Anouk Barberousse, Conor Mayo-Wilson, Jan Sprenger, Mikael Cozic, Philip Kitcher, Michael Strevens and an anonymous referee for their critical remarks and encouragements, which helped much improve this paper. Thanks also to participants in the EPSA conference in October 2011 in Athens (Greece), in the “Collective Dimension of Science” conference in December 2011 in Nancy (France), and in the “Progress of Science” conference in April 2012 in Tilburg (The Netherlands), for their remarks. The paper has been written while at the IHPST (UMR 8590, CNRS, Université de Paris 1 Panthéon-Sorbonne, ENS), at Columbia University (NY, USA), at the Archives Henri Poincaré (UMR 7117 CNRS, Université de Lorraine) and at Savoirs, Textes, Langage (UMR 8163 CNRS, Université Lille 1, Université Lille 3). A special financial support is acknowledged from the Université de Paris 1.

Appendix

Proof of Theorem 1 in Sect. 4.2

Theorem 1 invites to consider two general strategies:

strategy I: always immediately publish a passed step,
strategy II: never publish anything before the end of the game.

The theorem is proved by induction on the number of remaining temporal intervals, for scientist $A$. To start with, suppose there is just one temporal interval left. There are two cases: either $A$ is ahead of $B$, or the opposite (a similar position for $A$ and $B$ can be considered as a special case of the first or of the second case). Start with the first case, and define $a$ and $b$ such that: $B$ is $b$ steps ahead of the last published step (which might be step 0), and $A$ is $a$ steps ahead of $B$.

Firstly, suppose $B$ adopts strategy I. Let’s compare $A$’s reward according to the strategy she adopts. Suppose she adopts strategy I: she publishes $b+ a$ steps, while $B$ publishes $b$ steps, so $A$ immediately gets the reward $(a+b/2)v$. Then, $A$ and $B$ start on the same level for the last temporal interval. $A$’s final reward is given by Table 1, according to the research outcome of the final temporal interval.

Table 1

$A$’s reward during the last temporal interval, with strategy I

$A$ $\backslash $ $B$	Success	Failure
Success	$(a+b/2)v + v/2$	$(a+b/2)v+ v$
Failure	$(a+b/2)v$	$(a+b/2)v$

Suppose now $A$ adopts strategy II: she doesn’t publish now (while $B$ publishes her $b$ steps) and waits for the last time. $A$’s final reward is given by Table 2. For each cell, Table 1 is larger than Table 2. This shows that, when $A$ is in front of $B$ and $B$ adopts strategy I, then strategy I is better than strategy II for $A$.

Table 2

$A$’s reward during the last temporal interval, with strategy II

$A$ $\backslash $ $B$	Success	Failure
Success	$av + v/2 $	$(a+1)v$
Failure	$av-v/2$	$av$

Secondly, suppose $B$ adopts strategy II. Suppose $A$ adopts strategy I. She first publishes $b+a$ steps, while $B$ publishes nothing. So, her reward after the last temporal interval is given by $bv/2$ plus the cells of Table 1. Suppose now $A$ adopts strategy II. Then her reward after the last temporal interval is given by $bv/2$ plus the cells of Table 2. So, here again, strategy I is better than strategy II for $A$.

Consider now the second case: $B$ is ahead of $A$. Define $a$ and $b$ such that $A$ is $a$ steps ahead of the last published step, and $B$ is $b$ steps ahead of $A$. First, suppose that $B$ adopts strategy I. If $A$ adopts strategy I too, her final reward is given by Table 3. If she adopts strategy II, it is given by Table 3 minus $av/2$. So, strategy I is better for $A$.

Table 3

Expected rewards for $A$ and $B$

$A$ $\backslash $ $B$	Success	Failure
Success	$av/2 + v/2$	$av/2+ v$
Failure	$av/2$	$av/2$

Suppose now that $B$ adopts strategy II. If $A$ adopts strategy I, her final reward is given by Table 4. If she adopts strategy II, it is given by Table 3 minus $av/2$. So, strategy I is better for $A$ in this case too. So in any case, strategy I is better than strategy II for $A$, and this establishes the first case of the induction proof.

Table 4

$A$’s reward, according to the outcome of the first temporal interval

$A$ $\backslash $ $B$	Success	Failure
Success	$av + v/2$	$av+ v/2$
Failure	$av$	$av$

Suppose now it has been proved for some $m<n$ that, when $m$ temporal intervals remain, $A$ should publish every step she has passed. Consider $A$ when there remains $m+1$ temporal intervals. By hypothesis, $A$ should publish at the next temporal interval (when there remains only $m$ temporal intervals). If $B$ adopts strategy I, then she will publish at the next temporal interval, so $A$ is in the case already proved²⁴, where $m=1$. If $B$ adopts strategy II, $B$ doesn’t publish before the end, so nothing is changed whether $A$ publishes right now or just at the next temporal interval. Hence, strategy I is still better than strategy II for $A$, and Theorem 1 has been proved by induction.²⁵

Note that the conclusion that $A$ should always publish has not only been shown on average, but also for every possible outcome of research — for every comparison of the corresponding cells in the various tables. So, the theorem doesn’t depend on the hypothesis that the scientist is risk-neutral.

Proof of Theorem 2 in Sect. 4.2

It is useful to distinguish between two kinds of steps: the steps both $A$ and $B$ have passed but not yet published (call them the “common steps”), and the steps $A$ is the only one to have passed, if any (her “solitary steps”). I am going to show that solitary steps should be published. Then, taking into account the possibility of common steps cannot make publication a worse strategy than it is if there are only solitary steps (because nothing can be gained in not publishing them, in particular because the competitor’s strategy doesn’t depend on what is actually published, and a potential reward can be lost). For a result aiming at proving the superiority of publication, this is just fine. Strategies I and II are defined in the same way as in the proof of Theorem 1. Again, proving Theorem 2 amounts to showing that a scientist should prefer strategy I to strategy II.

As the chain is composed of $l$ steps, a total reward $lv$ will finally be distributed to the scientists. Because of their symmetric role, both scientists can expect a reward $lv/2$ if they adopt the same strategy. Table 5 shows the expected rewards for scientists I will argue for. Going from this Table 5 to Theorem 2 — $A$ should adopt strategy I — will not be difficult. Whatever strategy $B$ adopts, $A$ can expect a better reward if she adopts strategy I. And the same can be said for $B$. So, both should choose it, and it is a Nash equilibrium.

Table 5

Expected rewards for $A$ and $B$ (respectively), according to the strategies they adopt

$A \backslash B$	Strategy I	Strategy II
Strategy I	$lv/2$ ; $lv/2$	$>lv/2$ ; $<lv/2$
Strategy II	$<lv/2$ ; $>lv/2$	$lv/2$ ; $lv/2$

Table 5 could clearly be derived from the following proposition $P(k)$, if it was true for any $k$:

Proposition $P(k)$: If $B$ has adopted strategy I, if $A$ has some solitary steps, and if there remain $k$ steps before the end of the chain (i. e. $A$ is at the position $l-k$ ), then $A$ ’s expected reward is greater if she publishes her solitary steps.

So, let’s prove $P(k)$ by induction. In the initial case, $k=1$, and $A$ is at step $l-1$. She has some (say $a$) solitary steps. Suppose $A$ publishes her $a$ steps now. She immediately gets an $av$ reward, and both scientists are at step $l-1$. Given the symmetry between $A$ and $B, A$’s expected reward for the last step is $v/2$. Summing, her expected reward in this case is

$$\begin{aligned} E = av + \frac{v}{2}. \end{aligned}$$

(1)

Suppose now $A$ doesn’t publish her $a$ steps now. I am going to build a table (Table 6) for $A$’s expected reward in the long run (called $E_{6}$), according to the results after this next temporal interval. Note that it is $A$’s expected reward from this specific situation until the end of the game, and not only for the next temporal interval. Here is how table 6 is filled in. If $A$ succeeds, the game is over, and the rewards can be easily computed. If $A$ doesn’t succeed, the game is not over, so computing a reward isn’t straightforward. If both $A$ and $B$ fail, then they still are at the same position. So, $A$’s expected reward in this case is $E_{6}$ itself. In case $A$ fails and $B$ succeeds, $A$ is in position $l-1$, and $B$ in $l-a$, one step higher. Let’s call $E_{6}^{\prime }$ the expected reward in this situation. In any case $A$ cannot expect to publish more than $a$ steps, and perhaps she will have to share some of these rewards, so $E^{\prime }_{6}< av$ on average. This gives Table 6. To get the expected reward $E_{6} $, this table is weighted with the probability of the cells outcome. Some easy computation shows that $E_{6} < E$. In words: if $A$ has some solitary steps and if she is at the position $l-1$, her expected reward is larger if she publishes them now, than if she doesn’t. This proves $P(1)$.

Table 6

Expected reward $E_{6}$ for $A$ after one temporal interval, with $A$ starting $a$ positions ahead of $B$, who will publish after this temporal interval

$A \backslash $ $B$	Success	Failure
Success	$(a + 1/2)v$	$(a + 1)v$
Failure	$E_{6}^{\prime }$	$E_{6}$

Suppose now that $P(k)$ has been proved for some $k$. In the case $k+1, A$ is at step $l-k-1$. By $P(k)$, she knows that she will publish her solitary steps, if she has some, when she is on $l-k$, i. e. on the next step. Both $A$ and $B$ would publish if they reached step $l-k$ ($B$ would because she always publishes): we are tempted to say that this step plays the role of the end of a shorter chain, that we are in a situation where $P(1)$ applies, in order to argue that $A$ should publish now at $l-k$. This reasoning implicitly assumes that $A$ should maximize her reward when someone arrives at $l-k$ (if it is to play the role of the end of chain). Actually, the real goal for $A$ is to maximize her reward when either her or $B$ arrives at step $l$, the actual end of the chain. Hopefully, the only way to maximize the reward at $l$ is to maximize it at $l-k$ first: someone will publish at $l-k$, the two scientists will be together at this step, so there is no way for $A$ to keep any sort of better non-published lead for the real end of the chain. So, the argument of $P(1)$ can be used, and $P(k+1)$ is proved. As was argued before, this proves Table 5, and Theorem 2. ²⁶

Proof of Theorem 3 in Sect. 4.3.1

As before, it is clear that $A$ should publish a common step, and also that she cannot know in practice which steps are solitary or common. Strategies I and II are defined in the same way as in the proof of Theorem 1.

Theorem 3 is equivalent to stating that strategy II gives a larger average reward than strategy I. As the total reward distributed among scientists is constant at $v_1 + v_2$, it is equivalent to stating Table 7.

Table 7

$A$’s expected reward, in case $A$ is at step 1 and $B$ at step 0

$A{\backslash }B$	Success	Failure
Success	$v_1/2 + v_2$	$v_1+v_2$
Failure	$v_2/2$	$E_{6}$

Let’s consider the case of the bottom left cell. Let’s call $E_{4}$ the average reward that $A$ can expect after the first temporal interval. Table 8 is filled in: if $A$ and $B$ succeed, only $B$ publishes because $A$ follows strategy II; then, they are in a symmetrical position for the last step: $A$’s expected reward is $v_2/2$. If $A$ fails and $B$ succeeds, only $B$ publishes too, and $A$’s expected reward is $v_2/2$. If both scientists fail, they remain in the same position, and the reward is still $E_{4}$. If $A$ succeeds and $B$ fails, $A$’s reward is called $E_{6}$ and is computed below.

$E_{4}$ is computed by weighting Table 8 with Table 9:

$$\begin{aligned} E_{4} = p^2 \frac{v_2}{2} + p(1-p)\frac{v_2}{2} +p(1-p) E_{6} + (1-p)^2 E_{4} \end{aligned}$$

(2)

The evaluation of $E_{6}$ is made by filling in a similar Table 10.

Table 8

Expected rewards for $A$ and $B$, according to their strategies

$A{\backslash }B$	Strategy I	Strategy II
Strategy I	$v$ ; $v$	$<v$ ; $>v$
Strategy II	$>v$ ; $<v$	$v$ ; $v$

Table 9

$A$’s expected reward, according to the outcome of the first temporal interval

$A{\backslash }B$	Success	Failure
Success	$v/2$	$E_{9}$
Failure	$v/2$	$E_{8}$

Table 10

Expected reward $E_{9}$ for $A$, in case $A$ is at step 1 and $B$ at step 0

$A{\backslash } B$	Success	Failure
Success	$v/2+ v$	$2v$
Failure	$v/2$	$E_{9}$

Weighting Table 10 with Table 9, one computes

$$\begin{aligned} E_{6} = p^2 \Big ( \frac{v_1}{2} + v_2 \Big ) + p(1-p) \Big (v_1 + v_2 + \frac{v_2}{2} \Big ) + (1-p)^2 E_{6}. \end{aligned}$$

(3)

$E_{6}$ is extracted from it and injected in Eq. 2:

$$\begin{aligned} E_{4} = \frac{v_2}{2(2-p)} + \frac{1-p}{2(2-p)^2}\big [ (2-p) v_1 + (3-p) v_2 \big ] \end{aligned}$$

(4)

The condition expressed in the bottom left cell of Table 7 is that $E_{4} > \frac{v_1+v_2}{2}$. With Eq. 4, it is equivalent to

$$\begin{aligned} v_2 > \frac{2-p}{1-p} v_1. \end{aligned}$$

(5)

Proof of Theorem 4 in Sect. 4.3.2

The proof of Theorem 4 is very similar to the proof in Sect. , and I only give the key differences here. The total distributed reward is $2v$, so the expected rewards should be Tables 8 and 9.

$E_{8}$ is computed by weighting Table 9 with Table 6:

$$\begin{aligned} E_{8} = p_1 \frac{v}{2} + p_1(1-p_1)E_{9} + (1-p_1)^2 E_{8} \end{aligned}$$

(6)

$E_{9}$ is computed from Table 10, weighted with Table 11:

$$\begin{aligned} E_{9} = \frac{ p_1/2 + 2 p_2 - p_1 p_2}{p_1 + p_2 - p_1p_2} v. \end{aligned}$$

(7)

Table 11

Probability of $A$ and $B$’s outcomes of research, when $A$ is at step 1 and $B$ at step $0$

$A{\backslash } B$	Success	Failure
Success	$p_1 p_2$	$p_2(1-p_1)$
Failure	$p_1(1-p_2)$	$(1-p_1)(1-p_2)$

From Eqs. 7 and 6, one gets

$$\begin{aligned} E_{8} = \frac{v}{2(2-p_1)} + \frac{1-p_1}{2-p_1} \times \frac{p_1/2+2p_2-p_1p_2}{p_1 + p_2 - p_1 p_2}v \end{aligned}$$

(8)

The condition expressed in the bottom left cell of Table 8 ($E_{8}>v$) leads to Theorem 4.

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Thomas Boyer
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1.Savoirs, Textes, Langage (UMR 8163 CNRS, Université Lille 1, Université Lille 3) Villeneuve d’AscqFrance
2.Archives Henri Poincaré (UMR 7117 CNRS, Université de Lorraine) NancyFrance

\(A\) \(\backslash \) \(B\)	Success	Failure
Success	\((a+b/2)v + v/2\)	\((a+b/2)v+ v\)
Failure	\((a+b/2)v\)	\((a+b/2)v\)

\(A\) \(\backslash \) \(B\)	Success	Failure
Success	\(av + v/2 \)	\((a+1)v\)
Failure	\(av-v/2\)	\(av\)

\(A\) \(\backslash \) \(B\)	Success	Failure
Success	\(av/2 + v/2\)	\(av/2+ v\)
Failure	\(av/2\)	\(av/2\)

\(A\) \(\backslash \) \(B\)	Success	Failure
Success	\(av + v/2\)	\(av+ v/2\)
Failure	\(av\)	\(av\)

\(A \backslash B\)	Strategy I	Strategy II
Strategy I	\(lv/2\) ; \(lv/2\)	\(>lv/2\) ; \(<lv/2\)
Strategy II	\(<lv/2\) ; \(>lv/2\)	\(lv/2\) ; \(lv/2\)

Is a bird in the hand worth two in the bush? Or, whether scientists should publish intermediate results

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\(A \backslash \) \(B\)	Success	Failure
Success	\((a + 1/2)v\)	\((a + 1)v\)
Failure	\(E_{6}^{\prime }\)	\(E_{6}\)

\(A{\backslash }B\)	Strategy I	Strategy II
Strategy I	\(v\) ; \(v\)	\(<v\) ; \(>v\)
Strategy II	\(>v\) ; \(<v\)	\(v\) ; \(v\)

\(A{\backslash } B\)	Success	Failure
Success	\(p_1 p_2\)	\(p_2(1-p_1)\)
Failure	\(p_1(1-p_2)\)	\((1-p_1)(1-p_2)\)