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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Sign up to join this communityProve that $H_r(S) = H(s) / \log_2r$.
I'm not sure how to prove this. Any help would be greatly appreciated!
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 $\log_2 r$.