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6h
comment scores of individuals and evaluation
No, each agent has the same number of heuristics $k$. But they all have a distinct heuristic set. It is the same calculation even if the terrain is fixed? The specific heuristic set should affect the score of agent $i$. Not all scores would be the same.
17h
comment scores of individuals and evaluation
So suppose an agent $i$ has heuristics $h_{i1},h_{i2},h_{i3}$. Then they begin at a marker $m$ and evaluate if the marker $h_{i1}$ steps to the right of $m$ is greater than the point value at $m$. If it is, they move to that marker and evaluate again using the second heuristic $h_{i2}$. If it is not, they stay at that marker and evaluate with the second heuristic $h_{i2}$. And so on. They do this until they find the maximum possible point value that they can attain with their heuristic set. This is $S_i$.
17h
comment scores of individuals and evaluation
Now that I think about it, the $S_i$ are all the same in the case of one heuristic I have presented. This is not what I was aiming for - I was simplifying the general problem, where each individual has $k$ heuristics (say $3$) and they cycle through them to find the maximum point that they can reach starting from each marker on the terrain.
18h
comment Adding and Subtracting Normal Distributions
You have found $\mu_E,\sigma_E$ correctly (the mean of distribution E is negative), I checked your numbers on that. Now, define $\bar{Z} = \frac{0-\mu_E}{\sigma_E}$. As I said in the above comment, you want to compute $$\mathbb{P}(Z > \bar{Z}) = 1 - \mathbb{P}(Z \leq \bar{Z}) = 1 - \Phi(\bar{Z}),$$ where $Z \sim N(0,1)$ and $\Phi$ is the standard normal cumulative distribution function. You will need to look up $\Phi(\bar{Z})$ in a table or use your calculator.
19h
comment Adding and Subtracting Normal Distributions
If $X$ has distribution E, then you want $$\mathbb{P}(X>0) = \mathbb{P}\left(Z >\frac{0 - \mu_E}{\sigma_E}\right),$$ where $Z \sim N(0,1)$.
19h
comment Adding and Subtracting Normal Distributions
This is correct, at least up until the distribution of E. Then you want to find the probability that a random variable with distribution E is greater than $0$.
20h
comment Chain rule and implicit dfferentiation
@user2850514 $\frac{\text{d}z}{\text{d}y} \neq z'$
20h
comment Chain rule and implicit dfferentiation
$z' = y'/y$ because of the chain rule - $y$ is a function of $x$.
20h
comment Chain rule and implicit dfferentiation
you can't use the chain rule on operators, it only works for derivatives of functions - look up the theorem and you'll see there is nothing about the differentiation operator @user2850514, i.e. $$\frac{\text{d}}{\text{d}x} \neq \frac{\text{d}}{\text{d}y}\frac{\text{d}y}{\text{d}x}$$
20h
revised power series of function
wrote things more nicely
21h
suggested approved edit on power series of function
21h
comment scores of individuals and evaluation
This would mean that every individual has the same score ($S_i$ is the same for each agent $i$) since the terrain and hence the maximum $p_m$ on the terrain is fixed. This doesn't make sense.
22h
comment scores of individuals and evaluation
Actually I guess what you said makes sense, but it just seems too simple for me to accept. I'm lost as to why the heuristic doesn't matter here.
22h
comment scores of individuals and evaluation
But $X_i^m$ is different for every individual $i$ because each individual has a different heuristic, so that $p_m'$ is different for every individual when they start at $p_m$.
1d
comment scores of individuals and evaluation
Or will they eventually evaluate at every point on the terrain? This is unclear to me.
1d
comment scores of individuals and evaluation
I don't think your answer is correct for the distribution of $S_i$. Indeed, not every agent $i$ can evaluate themselves at every point on the terrain, so the second equality that you have is incorrect. They each have different heuristics $h_i$ and can only evaluate themselves on points that are their heuristic steps away from each starting point $m$, or am I missing something?
Dec
15
revised scores of individuals and evaluation
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Dec
15
revised scores of individuals and evaluation
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Dec
14
revised scores of individuals and evaluation
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Dec
14
revised Prove or disprove the statement: if all the eigenvalues of a matrix are 0, then the matrix must be the zero matrix?
added 121 characters in body