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I am trying to understand a paper of Maynard Smith (1974), that connects biology with game theory. I don't want to overwhelm you with useless stuff, but I have this definite integrals:

$$E(m)=\int_0^m (v-x)p(x) dx - \int_m^\infty mp(x) dx \tag{1}$$

We want to choose $p(x)$ such that $E(m)$ is the same constant $C$ for all $m$.

Now I'll copy exactly what he (Maynard Smith) says:

To find $p(x)$ we put $E(m) = E(m+\Delta m)$, so that

$$E(m)=\int_0^m (v-x)p(x) dx -\int_m^\infty mp(x) dx=\int_0^{m+\Delta m}(v-x)p(x)dx - \int_{m+\Delta m}^\infty( m+\Delta m) p(x) dx \tag{2}$$

After a little manipulation, remembering that $E(m)=\int_0^\infty p(x) dx =1$ this gives

$$ vp(m)= 1-\int_0^m p(x) dx \tag{3}$$

Equation $(3)$ is satisfied by the function $$ p(x) = (1/v ) e^{-x/v} \tag{4}$$

which is the equilibrium strategy we are seeking.

Since I am not a mathematician (but I have some knowledge of Calculus and Probability) I am having an hard time understanding this. I have no idea how we can go from $(2)$ to $(3)$ and from $(3)$ to $(4)$. Could anyone help? Thank you.

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Sorry for the mistakes. Now it should be fixed. – Vaaal Apr 13 '13 at 13:27
up vote 2 down vote accepted

The easiest thing to do is to apply the condition

$$\int_0^{\infty} dx \: p(x) = 1$$

at the beginning. This means rewriting $E(m)$ as

$$\begin{align}E(m) &= \int_0^m dx \: (v-x) p(x) - m + m \int_0^m dx \: p(x)\\ &= (v+m) \int_0^m dx \: p(x) - \int_0^m dx \: x p(x) - m\end{align}$$

Now write

$$\begin{align}\\ E(m+dm) &= (v+m+dm) \int_0^{m+dm} dx \:p(x) - \int_0^{m+dm} dx \: x p(x) - m-dm\\ &= (v+m+dm)\left[\int_0^m dx \: p(x) + \underbrace{\int_m^{m+dm} dx \: p(x)}_{p(m) dm} \right]\\ &- \left[\int_0^m dx \: x p(x) + \underbrace{\int_m^{m+dm} dx \: x p(x)}_{m p(m) dm} \right]-m-dm\\ &= (v+m) \int_0^m dx \: p(x) - \int_0^m dx \: x p(x) - m \\ &+\left [\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1 \right ] dm \\ &= E(m) + \left [\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1 \right ] dm\end{align}$$

The quantity in brackets is zero because $dm$ is arbitrary:

$$\int_0^m dx \: p(x) + (v+m) p(m) -m p(m)-1=0$$


$$v p(m) = 1-\int_0^m dx \: p(x)$$

as stated in step (3).

To get step (4), differentiate both sides with respect to $m$:

$$v p'(m) = -p(m) \implies p'(m) = -\frac{1}{v} p(m)$$

The general solution to this equation is

$$p(m) = A e^{-m/v}$$

$A=p(0)$ is determined by setting $m=0$ in the equation in step (3):

$$0=1-v p(0) \implies p(0)=\frac{1}{v}$$


$$p(m) = \frac{1}{v} e^{-m/v}$$

as stated in step (4).

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Thank you for your precise reply. I will mark your answer as right answer, but I have to be honest, I still do not understand some steps (most of the steps?). Maybe it's because me level in Calculus is quite basic. So I won't annoy you anymore with other questions and I'll try to figure it out by myself. Thank you very much! – Vaaal Apr 13 '13 at 14:16
@Vaaal: seriously, what steps do you not understand? I took pains to make the derivation comprehensible, but this sort of explanation is not a precise science. Feel free to ask some questions. – Ron Gordon Apr 13 '13 at 14:17
Hi Gordon, thank you for you kindness. My question can be quite trivial to an experienced mathematician, since I am a psychologist and I didn't have a formal training in calculus. Anyway, I miss the right path in the very beginning: when you rewrite E(m), where does the integral from m+dm to inf go? – Vaaal Apr 13 '13 at 14:24
As I said in the beginning, I replaced integral from $m$ to $\infty$ in the integral with $1-$ the integral from $o$ to $m$. That way I didn't have to mess around with that more messy limit. – Ron Gordon Apr 13 '13 at 14:50
In the second equation after "Now write" have you applied Taylor series (when you say that the integral from m to dm is equal to p(m)dm)? – Vaaal Apr 13 '13 at 16:47

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