SARSA_8states_v2.m

[Chuck-Chae]

function [loglik] = model_RL_8states_v2(parameters, subj)

num_state = 8; num_action = 2;

nd_alpha = parameters(1); % normally-distributed alpha alpha = 1/(1+exp(-nd_alpha)); % alpha (transformed to be between zero and one)

nd_beta = parameters(2); beta = exp(nd_beta);

% unpack data actions = subj.actions + 1; % adding 1 such that actions are {1, 2} outcome = subj.Outcome; color = subj.Color; orientation = subj.Orientation; mouth_dir = subj.Mouth_dir;

% counter over all trials ctr = 1;

% calculating numTrails U = unique(subj.block()); numTrails = zeros(1, length(U)); for i = 1:length(U) ind = find(subj.block() == U(i)); numTrails(i) = length(ind); end

% to save probability of choice. Currently NaNs, will be filled below p = nan(1, sum(numTrails));

for t1 = 1:length(numTrails) % number of trials T = numTrails(t1);

% Q-value for each action
q = 0.5 * ones(num_state, num_action); % Q-value for both actions initialized at 0

for t = 1:T
    % determine state
    state = mouth_dir(ctr) * 2 * 2 + orientation(ctr) * 2 + color(ctr) + 1;

    % Ensure state is within bounds
    if state > num_state || state < 1
        error('State index out of bounds: %d', state);
    end

    % probability of action 1
    p1 = 1 / (1 + exp(-beta * (q(state, 1) - q(state, 2))));

    % probability of action 2
    p2 = 1 - p1;

    % read info for the current trial
    a = actions(ctr); % action on this trial
    o = outcome(ctr); % outcome on this trial

    % store probability of the chosen action
    if a == 1
        p(ctr) = p1;
    elseif a == 2
        p(ctr) = p2;
    end

    % SARSA
    delta = o - q(state, a); % prediction error
    q(state, a) = q(state, a) + (alpha * delta);

    ctr = ctr + 1;
end

end

% log-likelihood is defined as the sum of log-probability of choice data % (given the parameters). loglik = sum(log(p + eps)); % Note that eps is a very small number in Matlab (type eps in the command % window to see how small it is), which does not have any effect in practice, % but it overcomes the problem of underflow when p is very very small % (effectively 0). end