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I am looking at this paper, and they are modeling "cohesiveness" in a group by doing the following:

First, they define a "common action" of all individuals in a group by using the average:

$\bar{x}=\sum_{i=1}^{n} \bar{x}_{i} / n$

where the xi are generated from an arbitrary distribution.

They then define an individual action as:

$x_{i}=(1-c) \bar{x}_{i}+c \bar{x}$

...for an arbitrary value of the cohesiveness parameter c.

All of this plugs into the following survivor function:

$p(\mathbf{x})=100-\beta(\mathbf{x} \cdot \mathbf{x})=100-\beta \sum_{i=1}^{n} x_{i}^{2}$

...where x is an "action" (arbitrary number), n is the "group size", and beta is a "harshness" parameter from 0 to 1.

They then plot the following, using different values of c:

Plot of Survival Probability vs Group Cohesion

But if I try to create this plot, I get the same values for every c.

For example, if I use the following values (same chosen in their plot legend):

group_size = 5;
beta = 1;
cohesive_factors = [0, 0.25, 0.50, 0.75, 1];
action_value = 5;

...and I loop through the cohesive factor values like this (JavaScript):

cohesive_factors.forEach(function(c_factor) {
    const average_of_all_actions = (action_value*group_size)/group_size;
    const individual_action = ((1-c_factor)*action_value) + (c_factor*average_of_all_actions);
})

I get the same individual_action in each loop, which ends up leading to the same survival probability once entered into the full equation.

What am I missing here? I am taking the same action value, but they did say this was drawn from an arbitrary distribution, so should I be setting the action_value to a random number instead of hardcoding it to 5?

I do notice that in the paper they say “individual actions that are independent come from U[−a, a].”

So if a=5 in the above plot legend, does this mean I take a random number from [-5, 5]?

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  • $\begingroup$ $\forall i, \bar x_i = 5 \implies \bar x= 5 \implies \forall i, x_i = 5$. In resume, yes: the action value should vary, otherwise $p(\mathbb{x})$ is constant on $c$ (which is the weight in the convex combination between $\bar x$ and $\bar x_i$). $\endgroup$ Jun 14 at 14:28
  • $\begingroup$ By the way, you should replace the image equations by $\LaTeX$ equations. $\endgroup$ Jun 14 at 14:31
  • $\begingroup$ Thanks for your comment. I replaced the equations with latex. So should I be drawing a random number on the interval between -5 and +5? I see that makes much more sense, since the cohesion discounts the individual actions, which only makes sense if those actions are different than each other (diversity). $\endgroup$
    – Cybernetic
    Jun 14 at 15:22
  • $\begingroup$ Yes, you should be drawing from a non-degenerate distribution. Your distribution, as it is, is degenerate (concentrated in only one point). From what I read in the paper (skimming), you should sample from a uniform distribution between $-a$ and $a$. $\endgroup$ Jun 15 at 1:59
  • $\begingroup$ @econbernardo thanks!! $\endgroup$
    – Cybernetic
    Jun 15 at 5:50

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