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I don't understand the last equality.

Here is my derivation $r(s,a,s')=\sum_{r\in R} r p(r|s,a,s')=\sum_{r\in R} r \frac{p(s,a,s'|r)p(r)}{p(s,a,s')}$

Could you give me the correct steps to derive them?

Thank you

tesio
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1 Answers1

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I found out that

$\sum_{r\in R} r p(r|s,a,s') = \sum_{r\in R} r \frac{p(s,a,s',r)}{p(s,a,s')} = \sum_{r\in R} r \frac{p(s',r|s,a)}{p(s'|s,a)} $ by cancelling $p(s,a)$.

tesio
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