Demonstration that Eq.
20 is Always Positive and
Therefore
that 'Var' is Evolutionarily Stable Against Any Fix(x)
How does one show that eq. 20 always has a positive value (i.e., that E(var, fix(x)) is always greater than E(fix(x),fix(x))? The answer requires a bit of elementary calculus. There are two steps in this demonstration:
OK, here we go.
Step 1: Finding the cost where E(var, fix(x)) - E(fix(x),fix(x)) is minimal is where the calculus comes in. If we take the derivative of eq. 20 we get an equation for the instantaneous slopes (basically the slope from one point to the next) of eq. 20. If we then set this equal to a slope of zero and solve the differentiated equation 20 for cost, we get the cost where the difference between these two payoffs is minimal. Here are the steps:
| recall eq. 20: 2 * V * exp(-m/V) - 1.5 V + m |
Now we find the derivative of eq 20 and set it equal to zero:
d(2*V*exp(-m/V) - 1.5 *V + m) / dm = 0
To get the derivative, let's differentiate each term separately:
d(2*V*exp(-m/V)) / dm - d(1.5*V) / dm + m / dm = 0
and recalling that V = a constant for any game, then:
- 2*exp(-m/V) + 0 + 1 = 0
or:
- 2*exp(-m/V) + 1 = 0
solving for the cost m where the function is at a minimum:
- exp(-m/V) = 0.5
m/V = - ln(0.5)
m = 0.693 * V
| To see a graph of the instaneous slopes of eq. 20 with respect (i.e., the derivative of eq. 20) to cost , press here. |
Step #2 -- we go back to eq. 20:
E(var, fix(x)) - E(fix(x),fix(x)) = 2*V*exp(-m/V) - 1.5 V + m
and substitute in the value of cost (m) at the minimum (0.693 * V):
Min.{E(var,fix(x))-E(fix(x),fix(x))} = 2*V*exp(-0.693*V / V) - 1.5V + 0.693V
=2*V*(exp(-0.693)) - 1.5V +0.693V
now exp(-0.693) = 0.500. So:
=2*V*0.5 - 1.5V +0.693V
=V - 1.5V +0.693V
=0.193V
So, for any positive resource value V (and V could only have positive values or it would hardly be a resource!): E(var, fix(x)) > E(fix(x),fix(x)) so 'var' is evolutionarily stable. |