I show that if x and y are regarded as investments in R&amp;D and p is the probability of discovery and if p satis…es p(x+y)=p(x)+p(y)-p(x)p(y) then p is practically uniquely determined. 

Kultti, Klaus

Helsingin yliopiston digitaalinen arkisto

p(x+y)=p(x)+p(y)-p(x)p(y)Klaus Kultti¤Department of Economics, University of Helsinki,e-mail: klaus.kultti@helsinki.…Department of Economics, University of HelsinkiDiscussion Papers, No. 559:2003ISSN 1459-3696ISBN 952-10-0700-1April 2003AbstractI show that if x and y are regarded as investments in R&D and p isthe probability of discovery and if p satis…es p(x+y)=p(x)+p(y)-p(x)p(y)then p is practically uniquely determined.1 IntroductionIn modelling economic phenomena the following problem comes recurrently up.An agent invest in some activity which may be research and development, ed-ucation, job search or the agent may invest into keeping things secret. Thereturns to the investment are not deterministic but typically it is assumed thatthe higher the investment the higher the probability of success. Now there isan in…nite number of ways to pick up a probability measure that is increasingin the investment.In some cases one can expect more structure from the probability measure.For instance, in R&D, which is the example I use for the rest of the note, onemay think that R&D activity consist of choosing di¤erent paths of research, andeach of the paths either leads to a success or to a failure. Then the probabilityfunction p should satisfy the following equationp(x + y) = p(x) + (1 ¡ p(x)) p(y) (1)The interpretation is that it does not matter whether a …rm, say, invests amountx +y to R&D or whether it …rst invests x and then only in case this investment¤I thank Yrjö Vartia for directing me to consider functional equations.1results in a failure invest amount y more. Another way to look at (1) is thatit does not matter to the …rm whether it divides its R&D activity into manysmall sequential activities or puts up just one large activity. I show here thatthis structure already pins down the form of the probability function almostuniquely. The point here is to show how easily a form of a function may bededuced if one is able to formulate a functional equation of which (1) is anexample.Before going through the analysis I want to point out that here the focus isthe probability function not the objective function of a …rm. Dividing R&D intosmall pieces may still be preferable depending on the form of the cost function.For instance, ¤ the cost function were linear investing x +y to R&D would costx + y while …rst investing x and only in case of failure incurring the additionalinvestment y would result in expected costs x + (1 ¡ p(x)) y. Both investmentstrategies would lead to the same probability of success while the latter onewould result in lower costs. Thus, not dividing the investment into small piecesmust rely on some kind of non-convexities in the cost function.2 AnalysisA more convenient form to (1) isp(x + y) = p(x) + p(y) ¡ p(x)p(y) (2)Insert y = x into (2) to getp(2x) = 2p(x) ¡ (p(x))2 = ¡ (1 ¡ p(x))2 + 1 (3)In a similar wayp(3x) = p(2x + x) = p(2x) + p(x) ¡ p(2x)p(x) = 3p(x) ¡ 3 (p(x))2 + (p(x))3 =¡ (1 ¡ p(x))3 + 1 (4)where in the second equality (2) is used and in the third equality (3) is used.By induction it is straightforward to show thatp(nx) = ¡ (1 ¡ p(x))n + 1 (5)Notice that by (2) we getp(x) = p³2x2´= ¡³1 ¡ p³x2´´2+ 1 · 1 (6)Next let us determine the value of p at rational arguments. First we get from(5) inserting x = 1p(n) = ¡ (1 ¡ p(1))n + 1 (7)On the other hand inserting x = mn into (5) we getp(m) = ¡³1 ¡ p³ mn´´n+ 1 (8)2Evaluating (7) at n = m, equating the right hand sides of (7) and (8), andsolving for p¡ mn¢yieldsp³mn´= ¡ (1 ¡ p(1))m=n + 1 (9)By (6) p(x) is at most one. Let us assume that p(1) = 1: Then it followsfrom (2) thatp(x) = p(1 + (x ¡ 1) = p(1) + p(x ¡ 1) (1 ¡ p(1)) = 1 (10)for all x ¸ 1. Take x 2 [0; 1), and let n be an integer such that nx ¸ 1. Nowp(nx) = p (1 + (nx ¡ 1)) = p(1)+p(nx ¡1)¡p(1)p(nx ¡1) = ¡ (1 ¡ p(x))n +1where the second equality comes from (2) and the third equality comes from(5). By (10) this yields ¡ (1 ¡ p(x))n = 0, and thus p(x) = 1 for all x ¸ 0.If p(x) 6= 1 we can de…nec ´ log (1 ¡ p(1)) (11)Inserting this into (9) yieldsp³mn´= ¡ec mn + 1 (12)Next we assume that p is continuous, and as its value is known on a denseset or reals (rationals) we immediately get that for all x 2 R+ (12) implies thatp(x) = 1 ¡ ecx (13)In particular, (13) implies that p(0) = 0. Notice that in (11) c < 0, and de…ning° = ¡c (13) can be expressed in the formp(x) = 1 ¡ e¡°x (14)Thus, we get that the only continuous solutions to (1) for x 2 R+ arep(x) = 1 and p(x) = 1 ¡ e¡°x , ° > 0, where the latter one is the relevantsolution if we are after a non-trivial probability function. It is very pleasingthat the probability function is of this form as it is easy to apply, and behavesvery nicely; for instance, it is in…nitely many times continuously di¤erentiable.It should be noted that the same result can be derived using much weakerconditions, and that an almost identical derivation, as well as much more ma-terial can be found in Aczel (1966).ReferencesAczel J. 1966 Lectures on functional equations and their applications. Aca-demic Press: New York and London.3

p(x+y)=p(x)+p(y)-p(x)p(y)

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p(x+y)=p(x)+p(y)-p(x)p(y)

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