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I don't remember how to do math

Prove that $\sum_{j=1}^{k} r_{j}{\log p_{j}} -\sum_{j=1}^{k} r_{j}{\log r_{j}} <=0$

Can I just put all expressions in one sum symbol, use log probability $\log(p_{j})-log(r_{j})=log(p_{j}/r_{j})$. Then we use Jensen Inequality to get ${\log(}\sum_{j=1}^{k} r_{j}*{\ p_{j}/r_{j}} )$ We simplify both $r_{j}$ and we get ${\log(}\sum_{j=1}^{k} p_{j})=0$ (since I think $\sum_{j=1}^{k} p_{j})=1$

Is that the correct way to do things?

zzz247
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    related: https://math.stackexchange.com/questions/4097415/show-that-e-f-log-f-ge-e-f-log-g/4097420#4097420 – Fei Cao Apr 14 '21 at 19:07
  • @FeiCao Thank you very much. So it's good I guess, even though I need to justify more. – zzz247 Apr 14 '21 at 19:15

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