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Q. 9.15

Expert-verifiedFound in: Page 413

Book edition
9th

Author(s)
Sheldon M. Ross

Pages
432 pages

ISBN
9780321794772

A coin having probability p = 2 3 of coming up heads is flipped 6 times. Compute the entropy of the outcome of this experiment.

The entropy of the outcome of this experiment is$H\left(\mathbb{X}\right)=5.5098$.

We have given the probability of the coin $p=\frac{2}{3}$.

Considering random vector $\mathbb{X}=\left({X}_{1},\dots ,{X}_{6}\right)$ where ${X}_{i}$ are independent equally distributed variables with distribution which given as

${X}_{i}~\text{Bern}\left(\frac{2}{3}\right)$

Calculating the entropy of $\mathbb{X}$.

localid="1648135034976" $H\left(\mathbb{X}\right)=-\sum _{x\in {\left(0,1\right)}^{6}}p\left(x\right)\text{log}p\left(x\right)$

This sum contains ${2}^{6}$summands. Then,

localid="1648135254966" $\sum _{x\in {\left(0,1\right)}^{6}}=p\left(x\right)\text{log}p\left(x\right)=-\left(\begin{array}{c}6\\ 0\end{array}\right){\left(\frac{1}{3}\right)}^{6}\text{log}{\left(\frac{1}{3}\right)}^{6}\phantom{\rule{0ex}{0ex}}=-\left(\begin{array}{c}6\\ 1\end{array}\right){\left(\frac{1}{3}\right)}^{5}\frac{2}{3}\text{log}{\left(\frac{1}{3}\right)}^{5}\frac{1}{6}\phantom{\rule{0ex}{0ex}}=-\left(\begin{array}{c}6\\ 6\end{array}\right){\left(\frac{2}{3}\right)}^{6}\text{log}{\left(\frac{2}{3}\right)}^{6}\phantom{\rule{0ex}{0ex}}$

Hence,

$H\left(\mathbb{X}\right)=5.5098$

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