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If you consider every action independent from the others, I assume the equation might be right. Consider the case with two actions. $\pi(a_1,a_2)$ is the joint probability of the two actions which becomes $\pi(a_1)\cdot \pi(a_2)$ in case $a_1$ and $a_2$ are independent. Then, $\log(\pi(a_1,a_2))$ will become $\log(\pi(a_1))+\log(\pi(a_2))$. I suppose this ...

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