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Can someone explain the last two steps in the derivation given below?

This is the derivation of the entropy of a Gaussian random variable:

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Bobson Dugnutt
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user2233120
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  • As for the last step, $\frac{1}{2}=\frac{1}{2}\mathrm{ln}(e)$ and $\mathrm{ln}(a)+\mathrm{ln}(b)=\mathrm{ln}(ab)$. But I think you should show some more effort, other than just pasting a link. Do you have any ideas of your own? And if not, can you at least state the problem more elaboratly, because nowhere have you specified what $\phi$, $\mu$, $\sigma$ etc. are.... – Eric S. Mar 14 '16 at 07:52
  • The answer can be found here: https://math.stackexchange.com/questions/3906538/simple-algebra-clarification-needed-for-gaussian-entropy-formula – develarist Nov 24 '20 at 05:21

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Recall that the expectation of a constant is equal to the constant it self, and the expectation is linear operator so $$ -E[-\frac{(x-\mu)^2}{ 2\sigma^2}-\ln{\sqrt{2 \pi \sigma^2}}]=\frac{1}{ 2\sigma^2}E[(x-\mu)^2] + \frac{1}{2} \ln{{2 \pi \sigma^2}}$$ Recall also that $\ln(a^b)=b \ln(a)$ for $a$ non negative, $\ln(e)=1$, and $\ln(a)+\ln(b)=\ln(ab)$. Moreover, $E[(x-\mu)^2]=\sigma^2 $, then $$\frac{1}{ 2\sigma^2}E[(x-\mu)^2] + \frac{1}{2} \ln{{2 \pi \sigma^2}}= \frac{1}{ 2\sigma^2}\sigma^2 + \frac{1}{2} \ln{{2 \pi \sigma^2}}= \frac{1}{ 2}+\frac{1}{2} \ln{{2 \pi \sigma^2}}= \frac{1}{ 2}\ln(e)+\frac{1}{2} \ln{{2 \pi \sigma^2}} = \frac{1}{ 2}\ln{{2e \pi \sigma^2}}$$

Nizar
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