The individual author’s publication–citation process: theory and practice

Abstract

The model proposed by Burrell (Information Processing and Management 28:637–645, 1992, Journal of Informetrics 1:16–25, 2007a) to describe the way that an individual author’s publication/citation career develops in time is investigated further, the aim being to describe in more detail the form of the citation distribution and the way it evolves over time. Both relative and actual frequency distributions are considered. Theoretical aspects are developed analytically and graphically and then illustrated using small empirical data sets relating to some well-known informetrics scholars. Perhaps surprisingly, it is found that the distribution may well be approximated in some cases by a simple geometric distribution.

Appendix

In the proofs we make free use of standard results regarding Poisson processes. The reader can refer to such standard references as Ross (1996) and Stirzaker (2005).

Proof of Proposition 2

The proof is constructed by conditioning on T, the time since publication, and Λ, the citation rate of a paper.

$$ G_{t} (z) = E\left[ {z^{{X_{t} }} } \right] = E_{T,\Uplambda } E\left[ {z^{{X_{t} }} \left| {T,\Uplambda } \right.} \right] $$

Given T and Λ, we have that citations following a Poisson distribution with mean ΛT so that the pgf is given by

$$ E\left[ {z^{{X_{t} }} \left| {T,\Uplambda } \right.} \right] = \exp \left[ {\Uplambda T(z - 1)} \right] $$

Now averaging over T, which has a uniform distribution on [0, t], we find

$$ \begin{aligned} E\left[ {z^{{X_{t} }} \left| \Uplambda \right.} \right] & = E_{T} E\left[ {z^{{X_{t} }} \left| {T,\Uplambda } \right.} \right] = \int {f_{T} (s)} e^{\Uplambda s(z - 1)} ds = \int\limits_{0}^{t} {\frac{1}{t}e^{\Uplambda s(z - 1)} } \\ & = \frac{1}{t\Uplambda (z - 1)}\left[ {e^{\Uplambda s(z - 1)} - 1} \right] \\ \end{aligned} $$

(A1)

Finally averaging over Λ, which has a gamma distribution as in (2), we get

$$ \begin{aligned} G_{t} (z) & = E\left[ {z^{{X_{t} }} } \right] = E_{\Uplambda } E\left[ {z^{{X_{t} }} \left| \Uplambda \right.} \right] = f_{\Uplambda } (\lambda )\frac{1}{t\lambda (z - 1)}\left[ {e^{\lambda t(z - 1)} - 1\,} \right]d\lambda \\ & = \frac{1}{t(z - 1)}\int\limits_{0}^{\infty } {\frac{{\alpha^{\nu } \lambda^{\nu - 1} }}{\Upgamma (\nu )}e^{ - \alpha \lambda } \frac{1}{\lambda }\left[ {e^{\lambda t(z - 1)} - 1} \right]} \,d\lambda \\ & = \frac{{\alpha^{\nu } }}{t(z - 1)\Upgamma (\nu )}\int\limits_{0}^{\infty } {\left[ {\lambda^{\nu - 2} e^{ - \lambda (\alpha - t(z - 1))} - \lambda^{\nu - 2} e^{ - \alpha \lambda } } \right]\,d\lambda } \\ & = \frac{{\alpha^{\nu } }}{t(z - 1)\Upgamma (\nu )}\left( {\frac{\Upgamma (\nu - 1)}{{(\alpha - t(z - 1))^{\nu - 1} }} - \frac{\Upgamma (\nu - 1)}{{\alpha^{\nu - 1} }}} \right) \\ \end{aligned} $$

(recognising each of the integrands as being proportional to a gamma pdf)

$$ \begin{aligned} & = \frac{\alpha }{(\nu - 1)t(z - 1)}\left[ {\frac{1}{{\left( {1 - (t/\alpha )(z - 1)} \right)^{\nu - 1} }} - 1} \right] \\ & = \frac{1}{(\nu - 1)s(z - 1)}\left[ {\frac{1}{{\left( {1 - s(z - 1)} \right)^{\nu - 1} }} - 1} \right] \\ \end{aligned} $$

(where s = t/α)

$$ \begin{aligned} & = \frac{1}{(\nu - 1)s(z - 1)}\left[ {\left( {\frac{1/(1 + s)}{1 - (s/(1 + s)z)}} \right)^{\nu - 1} - 1} \right] \\ & = \frac{1}{(\nu - 1)s(z - 1)}\left[ {\left( {\frac{p}{{\left( {1 - qz} \right)}}} \right)^{\nu - 1} - 1} \right] \\ \end{aligned} $$

(A2)

where, as before, $p = \frac{1}{1 + s} = \frac{\alpha }{\alpha + t} = 1 - q$

For the power series representation, this follows from expanding the first term inside the square brackets of (A2) using the general binomial expansion for negative powers and then simplifying. Thus

$$ \begin{aligned} \left( {\frac{p}{{\left( {1 - qz} \right)}}} \right)^{v - 1} - 1 & = \left( {\frac{1/(1 + s)}{1 - sz/(1 + s))}} \right)^{\nu - 1} - 1 = 1 - (s/(z - 1))^{ - (\nu - 1)} - 1 \\ & = \left( {1 + (\nu - 1)s(z - 1) + \frac{(\nu - 1)v}{2!}\left( {s(z - 1)^{2} } \right) + \cdots } \right) - 1 \\ & = (\nu - 1)s(z - 1) + \frac{(\nu - 1)v}{2!}\left( {s(z - 1)^{2} } \right) + \frac{(\nu - 1)\nu (\nu + 1)}{3!}\left( {s(z - 1)^{3} } \right) + \cdots \\ \end{aligned} $$

Now dividing through by $(\nu - 1)s(z - 1)$ gives the result.□

Remark

The power series representation could alternatively have been derived by expanding (A1) in standard power series form and then integrating term by term with respect to the gamma pdf of Λ.

Proof of Proposition 3

This is an application of a well-known result for probability generating functions. If the pgf G(z) of a random variable X is differentiated wrt z r times then

$$ G^{(r)} (z) = \frac{{d^{r} G(z)}}{{dz^{r} }} = E\left[ {X(X - 1) \ldots (X - r + 1)z^{X - r} } \right] $$

so that $G^{(r)} (1) = E\left[ {X(X - 1) \ldots (X - r + 1)} \right]$

Thus successively differentiating the power series expansion (7) and putting z = 1 each time gives the result.□