Sex Allocation in Solitary Bees and Wasps


Jump to: navigation, search

The Objective Functions to Minimize

To get the equation that Frank cites on page 317, we can minimize either of the following functions:

F(\lambda)=\int_0^\lambda f(y)h(y)dy\int_\lambda^1 g(y)h(y)dy


E(\lambda)=\ln(F(\lambda))=\ln\left(\int_0^\lambda f(y)h(y)dy\right)+\ln\left(\int_\lambda^1 g(y)h(y)dy\right)

F is probably the more intuitive function: we can see that there's a trade-off point, λ, at which we switch from producing males to producing females. If we think of the two function f(x) and g(x) as "value" functions, with f(x) \ge g(x) over the interval x \in [0,1], and h(x) as the probability density of the resources distribution (taken as a beta density by Frank), then we're saying that we want to maximize the product of the value of the males along with the value of the females. If we take λ = 0 or λ = 1, F(0) = F(1) = 0; because F(x)\ge 0, we know that there is a maximum on the interior of the interval [0,1].

If we differentiate either expression, we will ultimately obtain the equation that Frank cites in his text:

\frac{f(\lambda)}{\int_0^\lambda f(y)h(y)dy}=\frac{g(\lambda)}{\int_\lambda^1 g(y)h(y)dy}

which we can then solve for λ, given the functions f, g, and h.

F(λ) can be normalized without changing the maximizing value of λ: hence we could use instead

F(\lambda)=\frac{\int_0^\lambda f(y)h(y)dy\int_\lambda^1 g(y)h(y)dy}{\int_0^1 f(y)h(y)dy\int_0^1 g(y)h(y)dy}=\left(\frac{\int_0^\lambda f(y)h(y)dy}{f_T}\right)\left(\frac{\int_\lambda^1 g(y)h(y)dy}{g_T}\right);

hence we can think of F(λ) as the product of the fraction of male value achieved by λ and the fraction of female value achieved from λ on out to the total resource allocation.

The Minimization with Specific f, g, and h

Assume that f(x) = xr, g(x) = x, and h(x) = κxa(1 - x)a.

We expand I=\int y^s y^a (1-y)^a dy = \int y^s y^a \left(\sum_{n=0}^a {n \choose k} (-1)^n y^{n}\right) dy


I=\sum_{n=0}^a {n \choose k} (-1)^n \int y^{n+a+s} dy =\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+s+1}}{n+a+s+1}|

Given our choices of f and g, we want to evaluate

I_f=\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+r+1}}{n+a+r+1}|_0^\lambda = \sum_{n=0}^a {n \choose k} (-1)^n \frac{\lambda^{n+a+r+1}}{n+a+r+1}


I_g=\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+2}}{n+a+2}|_\lambda^1=\sum_{n=0}^a {n \choose k} (-1)^n \frac{1-\lambda^{n+a+2}}{n+a+2}

Now the equation that we're to solve for λ can be written λrIg = λIf which we can reduce by dividing by λr (since λ = 0 is not a maximum); thus our equation becomes Ig - λ1 - rIf = 0, or

P(\lambda) = \sum_{n=0}^a {n \choose k} (-1)^n \left(\frac{1-\lambda^{n+a+2}}{n+a+2} - \lambda^{1-r}\frac{\lambda^{n+a+r+1}}{n+a+r+1} \right) = 0


P(\lambda) = \sum_{n=0}^a {n \choose k} (-1)^n \left(\frac{1-\lambda^{n+a+2}}{n+a+2} - \frac{\lambda^{n+a+2}}{n+a+r+1} \right) = 0


P(\lambda) = 
\sum_{n=0}^a {n \choose k} (-1)^n 
\frac{1}{n+a+2} - \lambda^{n+a+2}
\right)\right) = 0

Here's a little script that shows some of those functions P.

Personal tools