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Prove that $H_r(S) = H(s) / \log_2r$.

I'm not sure how to prove this. Any help would be greatly appreciated!

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  • $\begingroup$ Welcome to CS.SE! What have you tried? Where did you get stuck? Have you tried expanding the definition to see what the definition of each quantity is? We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. Also, see meta.cs.stackexchange.com/q/1284/755 $\endgroup$ – D.W. Nov 6 '16 at 18:33
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There's nothing to prove here, this follows by Definition. Assume a discrete r.v. $X$, then

$$H(X) \triangleq \mathbb{E}[-\log \Pr(X)]$$

and,

$$H_r (X) \triangleq \mathbb{E}[-\log_r \Pr(X)].$$

Therefore, the claim follows (using $E(aX)=aE(X)$ for a constant $a$). Note that in your question you use $H(X)$ as the binary entropy $H(X) \triangleq H_2(X) = \mathbb{E}[-\log_2 \Pr(X)]$, hence the multiplicative factor is $\log_2 r$.

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