The optimal amount of information to provide in an academic manuscript

Abstract

Authors may believe that having more information available about the research can help reviewers make better recommendations. However, too much information in a manuscript may create problems to the reviewers and may lead them to poorer recommendations. An information overload on the part of the reviewer might be a state in which she faces an amount of information comprising the accumulation of manuscript informational cues that inhibit the reviewer’s ability to optimally determine the best possible recommendation about the acceptance or rejection of the manuscript. Therefore the author wants to determine the amount of manuscript attributes to provide to reviewers. With this goal in mind we show that there is an intermediate number of manuscript attributes that maximizes the probability of acceptance. If too much research information is provided, some of it is not as useful for recommending acceptance, the average informativeness per research attribute evaluation is too low, and reviewers end up recommending rejection. If too little information is provided about the research, reviewers may end up not having sufficient details to recommend its acceptance. We also show that authors should provide more information to reviewers with more favorable initial valuation toward the research. For those reviewers with a less favorable prior attitude, the author should provide only the most important manuscript attributes. Given that expert reviewers face less load than potential readers, it follows that with respect to the target audience the optimal author’s strategy is also to trade off the amount of research information provided in the manuscript with the average informativeness of these items by selecting an intermediate number of attributes.

Appendices

Appendix A: Acceptance and rejection thresholds ${\overline{U}}$ and ${\underline{U}}$

Following Branco et al. (2016), by a Taylor expansion valid to terms of the first order and less in dt, the reviewer’s value function of the reviewer, V(u, t), becomes

$$\begin{aligned} V(u,t)= & {} - c \ dt + \frac{dt}{T} \max [ 0, u] \nonumber \\&+ \left( 1- \frac{dt}{T} \right) \left[ V (u, t) + V_u E(du) + V_t dt + \frac{1}{2} V_{uu} E[(du)^2] + V_{ut} E(du)dt \right] \end{aligned}$$

(5)

with $V_t$ and $V_u$ being the partial derivatives with respect to t and u respectively, $V_{uu}$ being the second derivative with respect to u, and $V_{ut}$ being the cross derivative with respect to u and t.

By definition we have that $du = {\overline{\sigma }}_T d w$ where u is the expected utility of recommending acceptance, w is a standardized Brownian motion, and ${\overline{\sigma }}_T$ is the instantaneous standard deviation of the Brownian motion.

Besides, the attribute importance $\sigma _i$ is continuous in i, with $\lim _{i \rightarrow \infty } \sigma _i =0$, and the average informativeness ${\overline{\sigma }}^2_T$ is given by the average of $\sigma ^2_i$ for manuscript attributes i in [0, T], ${\overline{\sigma }}^2_T = \frac{1}{T} \int _0^T \sigma ^2_i di.$ Therefore, the average importance of manuscript attributes ${\overline{\sigma }}^2_T$ decreases with the amount T of research information made available by the author. Hence, $E(du) =0$ and $E[(du)^2] ={\overline{\sigma }}^2_T dt$. We suppose that the reviewer’s problem does not change with the number of manuscript attributes already evaluated, i.e., $V_t = 0$, and therefore the value function of the reviewer, V(u, t) is only a function of u: $V(u,t) = V(u)$.

Then, by dividing in Eq. (5) by dt we get the differential equation that represents the value function of the reviewer V(u):

$$\begin{aligned} - c T + \max [ 0, u] - V + \frac{T{\overline{\sigma }}^2_T }{2} V_{uu} = 0 \end{aligned}$$

(6)

Now, solving this differential equation for $u >0$ we get:

$$\begin{aligned} V(u) = A_1 e^{\sqrt{\frac{2}{T{\overline{\sigma }}^2_T }}u} + A_2 e^{-\sqrt{\frac{2}{T{\overline{\sigma }}^2_T }}u} + u - c T \end{aligned}$$

(7)

with $A_1$ and $A_2$ being constants to be determined with the conditions for V(u) for when the reviewer decides to stop the evaluation. For $u<0$, solving Eq. (6), we get:

$$\begin{aligned} V(u) = A_3 e^{\sqrt{\frac{2}{T{\overline{\sigma }}^2_T }}u} + A_4 e^{-\sqrt{\frac{2}{T{\overline{\sigma }}^2_T }}u} - c T \end{aligned}$$

(8)

with $A_3$ and $A_4$ being constants to be determined with the boundary conditions for V(u), i.e., the constraints imposed on the constants for when the reviewer decides to stop the evaluation.

For example, by the smoothness of the valuation function V at $u=0$, we have that $V(0^+) = V(0^-)$. That is, the constraint $V(0^+) = V(0^-)$ follows from Eq. (1) for $u = 0^+$ or $u=0^-$. And this boundary condition $V(0^+) = V(0^-)$ leads to

$$\begin{aligned} A_1 + A_2 = A_3 + A_4 \end{aligned}$$

Also, by the same smoothness of the valuation function V at $u=0$, we have that $V_u(0^+) = V_u(0^-)$ which follows from the Taylor expansion in Eq. (5) for $u = 0^+$ or $u=0^-$ and noting that du has a symmetric distribution, and that

$$\begin{aligned} E[V(0+du)] = \frac{1}{2} \left[ V(0) + V_u(0) E(du | du>0 ) \right] + \frac{1}{2} \left[ V(0) - V_u(0) E(du | du >0 ) \right] \end{aligned}$$

The boundary condition $V_u(0^+) = V_u(0^-)$ leads to

$$\begin{aligned} A_1 - A_2 + \sqrt{\frac{T{\overline{\sigma }}^2_T }{2}} = A_3 - A_4 \end{aligned}$$

When the reviewer’s expected utility u is sufficiently large such that the reviewer is indifferent between continuing the evaluation and stopping the review recommending acceptance, the expected utility reaches $u = {\overline{U}}$ –with ${\overline{U}}$ being the acceptance threshold– and the reviewer recommends to accept the manuscript and gets expected utility ${\overline{U}}$, $V({\overline{U}}) = {\overline{U}}$ and $V_u ({\overline{U}}) =1$. The boundary condition $V({\overline{U}}) = {\overline{U}}$ leads to

$$\begin{aligned} A_1 e^{\sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\overline{U}}} + A_2 e^{-\sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\overline{U}}} = c T \end{aligned}$$

The condition $V_u ({\overline{U}}) =1$ is implied by the fact that the reviewer maximizes V(u, t) for all (u, t). This constraint states that when the reviewer’s utility u walks away from the acceptance threshold ${\overline{U}}$, in order for her to be indifferent between continuing to evaluate the manuscript and recommending acceptance right away, the marginal decrease in V(u) when u walks to the left would have to equal the marginal increase in the value function V(u) when u walks to the right (Branco et al. 2016). The boundary condition $V_u ({\overline{U}}) =1$ leads to

$$\begin{aligned} A_2 = A_1 e^{ 2 \sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\overline{U}}} \end{aligned}$$

If the reviewer’s expected utility u is sufficiently small such that the expected utility reaches $u = {\underline{U}}$ –with ${\underline{U}}$ being the rejection threshold–, then the reviewer recommends to reject the manuscript and gets 0, and we have the conditions $V({\underline{U}}) = 0$ and $V_u ({\underline{U}}) =0$. The boundary condition $V({\underline{U}}) = 0$ states that at $u= {\underline{U}}$, the reviewer expects zero utility from continuing to evaluate the manuscript. On the other hand, the constraint $V_u ({\underline{U}}) =0$ states that if utility u goes up as the reviewer continues to evaluate, the change of value function V(u) will be slow.

The condition $V({\underline{U}}) = 0$ leads to

$$\begin{aligned} A_3 e^{\sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\underline{U}}} + A_4 e^{-\sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\underline{U}}} = c T \end{aligned}$$

And the boundary condition $V_u ({\underline{U}}) =0$ leads to

$$\begin{aligned} A_4 = A_3 e^{ 2 \sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\underline{U}}} \end{aligned}$$

From the equations above, we get

$$\begin{aligned} A_1= A_4 = \frac{cT}{2} e^{- \sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\overline{U}}} \end{aligned}$$

and

$$\begin{aligned} A_2= A_3 = \frac{cT}{2} e^{ \sqrt{\frac{2}{T{\overline{\sigma }}^2_T } }{\overline{U}}} \end{aligned}$$

Therefore, it follows that the acceptance threshold ${\overline{U}}$ is

$$\begin{aligned} {\overline{U}} = \sqrt{ \frac{T {\overline{\sigma }}^2_T}{2}} \log \left[ \sqrt{ \frac{ {\overline{\sigma }}^2_T}{8 c^2 T}} + \sqrt{ 1 + \frac{ {\overline{\sigma }}^2_T}{8 c^2 T}} \ \right] \end{aligned}$$

and the rejection threshold is ${\underline{U}} = - {\overline{U}}$.

Appendix B: Reviewer’s probability of recommending acceptance P(u, T)

Following Branco et al. (2016), when the expected manuscript valuation u is negative and such that $u \in [- {\overline{U}},0]$, the probability of recommending acceptance P(u, T) is

$$\begin{aligned} P(u,T) = 0 \cdot dt/T + E \left[ P (u +du, T) \cdot \left( 1- dt/T \right) \right] \end{aligned}$$

(9)

where the first term in Eq. (9) represents the situation such that the review process runs out of research attributes to evaluate (with probability $\frac{dt}{T}$) and the expert stops the review process and makes a recommendation to reject the manuscript because u is negative (i.e., there exists a zero probability of recommending acceptance). The second term in Eq. (9) represents the situation such that the review process does not run out of research attributes to evaluate (with probability $1- \frac{dt}{T}$) and the expected value of the probability of recommending acceptance is $E \left[ P (u +du, T) \cdot \left( 1- dt/T \right) \right]$.

By a Taylor expansion valid to terms of the second order and less in du, we have

$$\begin{aligned} P(u,T) = \left( 1- dt/T \right) \cdot E [ P (u, T) + P_u (u, T) \cdot du + (1/2) \cdot P_{uu} (u, T) \cdot (du)^2 ] \end{aligned}$$

(10)

with $P_u$ being the partial derivative with respect to u, and $P_{uu}$ being the second derivative with respect to u.

By taking expectations we get:

$$\begin{aligned} P(u,T)= & {} \left( 1- dt/T \right) \cdot \left[ P (u, T) + P_u (u, T) \cdot E(du) + (1/2) \cdot P_{uu} (u, T) \cdot E[(du)^2] \right] \nonumber \\= & {} \left( 1- dt/T \right) \cdot \left[ P (u, T) + (1/2) \cdot P_{uu} (u, T) \cdot {\overline{\sigma }}^2_T dt \right] \end{aligned}$$

(11)

since $E(du) =0$ and $E[(du)^2] ={\overline{\sigma }}^2_T dt$.

From this equation, solving in P(u, T), we get the differential equation that represents the probability of recommending acceptance P(u, T)

$$\begin{aligned} P(u,T) = \frac{1}{2} \cdot P_{uu} (u, T) \cdot {\overline{\sigma }}^2_T \cdot T \end{aligned}$$

(12)

Now, solving this differential equation we get:

$$\begin{aligned} P(u,T) = B_1 e^{ \alpha u} + B_2 e^{ - \alpha u} \end{aligned}$$

(13)

where $\alpha = \sqrt{ \frac{2}{ T {\overline{\sigma }}^2_T } }$, and with $B_1$ and $B_2$ being constants to be determined with the boundary conditions for P.

Therefore, given that the probability of recommending acceptance P(u, T) is 0 if $u = - {\overline{U}}$, and that by symmetry $P(0,T) = 1/2$, we get

$$\begin{aligned} B_1 = \frac{e^{2 \alpha {\overline{U}}}}{2 \left( e^{2 \alpha {\overline{U}}} -1 \right) } \end{aligned}$$

and

$$\begin{aligned} B_2 = - \frac{1}{2 \left( e^{2 \alpha {\overline{U}}} -1 \right) } \end{aligned}$$

Appendix C: Proof of Proposition 1

Following Branco et al. (2016), first we have to show that the acceptance threshold ${\overline{U}}$ converges to 0 when the amount of information about the research to provide, T, goes to zero.

To this aim, using the L’Hôpital rule, it follows that

$$\begin{aligned} \lim _{T \rightarrow 0} {\overline{U}} = \sqrt{ \frac{\sigma _0^2}{2}} \lim _{x \rightarrow 0} \sqrt{x} \log \left[ 2 \sqrt{ \frac{ \sigma ^2_0}{8 c^2 x}} \ \right] =0 \end{aligned}$$

Second, we have to show that the acceptance threshold ${\overline{U}}$ also converges to 0 when the amount of information about the research to provide, T, goes to $\infty$. By the Lebesgue’s dominated convergence theorem it follows that

$$\begin{aligned} \lim _{T \rightarrow \infty } {\overline{\sigma }}^2_T = \int _0^1 \lim _{T \rightarrow \infty } \sigma ^2_{T_y} dy =0 \end{aligned}$$

Hence, taking limits when T converges to $\infty$ we get

$$\begin{aligned} \lim _{T \rightarrow \infty } {\overline{U}} = \lim _{T \rightarrow \infty } \frac{ {\overline{\sigma }}^2_T}{2} \lim _{T \rightarrow \infty } \frac{ T^{-\frac{3}{2}} \sqrt{ \frac{ {\overline{\sigma }}^2_T}{8 c^2 }} + T^{-2} \frac{ {\overline{\sigma }}^2_T}{8 c^2 } \left( 1 + \frac{ {\overline{\sigma }}^2_T}{8 c^2 T} \right) ^{-\frac{1}{2}} }{T^{-\frac{3}{2}} \left[ \sqrt{ \frac{ {\overline{\sigma }}^2_T}{8 c^2 T}} + \sqrt{ 1 + \frac{ {\overline{\sigma }}^2_T}{8 c^2 T}} \ \right] } =0 \end{aligned}$$

From previous results we have that when the acceptance threshold is ${\overline{U}} =0$, the rejection threshold is ${\underline{U}} = - {\overline{U}} =0$, and the reviewer does not do any evaluation, and she would recommend rejection of the manuscript as the initial valuation v is negative, i.e., the acceptance probability P(v, T) is zero. Then, given that this probability P(v, T) is continuous in the amount of information about the research to provide, T, it follows that an author who seeks to maximize the reviewer’s probability of recommending acceptance P(v, T) would not provide too much or too little research information in the manuscript.

Appendix D: Proof of Proposition 2

We have to show that when the reviewer’s prior valuation of the manuscript, v, becomes less negative, the author should provide more details for evaluation, and hence provide a higher quantity of manuscript attributes, T.

Consider any negative initial valuation of the manuscript $v_0$ and assume that $T_0 = T^*(v_0)$ is the optimal amount of research information for $v_0$ in order to maximize the reviewer’s probability of acceptance $P(v_0, T)$. Let ${\overline{U}}(T_0)$ be the optimal acceptance threshold given in Eq. (2) for $T_0$:

$$\begin{aligned} {\overline{U}} (T_0) = \sqrt{ \frac{T_0 {\overline{\sigma }}^2_{T_0}}{2}} \log \left[ \sqrt{ \frac{ {\overline{\sigma }}^2_{T_0}}{8 c^2 T_0}} + \sqrt{ 1 + \frac{ {\overline{\sigma }}^2_{T_0}}{8 c^2 T_0}} \ \right] \end{aligned}$$

We have to prove that the amount of manuscript information to provide $T^*(v)$ is increasing in the initial valuation v. To this aim, we have to show that there exists some positive $\delta$ such that for all $v \in (v_0- \delta , v_0)$ it follows that $T^*(v) \le T^*(v_0)$ with $T^*(v)$ being the optimal amount of research information for v.

If $v_0 \le - {\overline{U}} (T_0)$ as given above, then the reviewer’s acceptance probability $P(v_0,T_0)$ is zero. But this contradicts the assumption that there exists some $T'$ such that $P(v_0, T' )$ is strictly positive. Therefore, we have that $v_0 > - {\overline{U}} (T_0)$ and we can define $\delta = v_0 + {\overline{U}} (T_0)$.

Suppose that there exists some $v_1 \in (v_0 -\delta , v_0)$ such that $T_1 = T^*(v_1) > T_0$. For this case, in the following we can prove that $P(v_0, T_1) > P(v_0, T_0)$ which contradicts that $T_0 = T^*(v_0)$ is the optimal amount of research information for initial valuation $v_0$.

Let $u_s(v, \sigma ^2)$ represent the value of a Brownian motion at time s starting from initial valuation v with variance $\sigma ^2$. In short, Brownian motion is a stochastic process whose increments are independent, stationary and normal, and whose sample paths are continuous. Increments refer to the random variables of the form $u_{t+s} - u_s$. Stationary means that the distribution of this random variable is independent of s, while independent increments means that increments corresponding to time intervals that do not overlap are independent.

Following Branco et al. (2016), by the Markov Property of the Brownian Motion and the memory-less property of the constant hazard rate model, we get that the acceptance probability $P(v_0,T_0)$ evaluated at $v_0$ and $T_0$ can be represented as follows

$$\begin{aligned} P(v_0,T_0)= & {} P (0, T_0) \cdot \hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } 0 \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } v_1 \right] \\&+ P (v_1, T_0) \cdot \hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } v_1 \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } 0 \right] \end{aligned}$$

where $v_1 \in (v_0 -\delta , v_0)$, with $\delta = v_0 + {\overline{U}} (T_0)$, such that $T_1 = T^*(v_1) > T_0$. Also, $P(v_1, T_0) < P(v_1, T_1)$ by optimality of $T_1$ for initial manuscript valuation $v_1$, and by symmetry $P(0,T) = 1/2$ for all T.

Let $\gamma$ be a random variable distributed as an exponential with parameter one. Then, $T \gamma$ is distributed as an exponential with parameter T. Therefore, we get

$$\begin{aligned}&\hbox {Pr}\,\left[ u_s(v_0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } 0, \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } v_1 \right] \\&\quad =\hbox {Pr} \left[ u_s(0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } T_0 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \\&\quad =\hbox {Pr} \left[ u_{s/{\overline{\sigma }}^2_{T_0} }(0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } {\overline{\sigma }}^2_{T_0} T_0 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \end{aligned}$$

and by the property of Brownian motion that $\theta u_{s/\theta ^2} (0,1)$ is equal in distribution to $u_{s} (0,1)$ it follows that

$$\begin{aligned}&\hbox {Pr} \left[ u_{s/{\overline{\sigma }}^2_{T_0} }(0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } {\overline{\sigma }}^2_{T_0} T_0 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \\&\quad =\hbox {Pr} \left[ u_{s }(0, 1) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } {\overline{\sigma }}^2_{T_0} T_0 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \end{aligned}$$

Therefore, given that ${\overline{\sigma }}^2_{T} T$ is increasing as T increases, we get

$$\begin{aligned}&\hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } 0, \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } v_1 \right] \\&\quad =\hbox {Pr} \left[ u_{s }(0, 1) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } {\overline{\sigma }}^2_{T_0} T_0 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \\&\quad <\hbox {Pr} \left[ u_{s }(0, 1) \text{ reaches } -v_0, \text{ before } \text{ being } \text{ terminated } \text{ at } {\overline{\sigma }}^2_{T_1} T_1 \gamma \text{ or } \text{ reaching } v_1 -v_0 \right] \\&\quad =\hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_1}) \text{ reaches } 0, \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } v_1 \right] \end{aligned}$$

By other hand, using this same approach we can also demonstrate that

$$\begin{aligned}&\hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_0}) \text{ reaches } v_1, \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } 0 \right] \\&\quad <\hbox {Pr} \left[ u_s(v_0, {\overline{\sigma }}^2_{T_1}) \text{ reaches } v_1, \text{ before } \text{ being } \text{ terminated } \text{ or } \text{ reaching } 0 \right] \end{aligned}$$

Finally, putting all the equations above together, we have

$$\begin{aligned} P(v_0, T_1) > P(v_0, T_0) \end{aligned}$$

which contradicts that $T_0 = T^*(v_0)$ is the optimal amount of research information for initial valuation $v_0$.