Aydin Jarvis

2022-10-13

Entropy of geometric random variable?

I am wondering how to derive the entropy of a geometric random variable? Or where I can find some proof/derivation? I tried to search online, but seems not much resources is available.

Here is the probability density function of geometric distribution: $(1-p{)}^{k-1}\phantom{\rule{thinmathspace}{0ex}}p$

Here is the entropy of a geometric distribution: $\frac{-(1-p){\mathrm{log}}_{2}(1-p)-p{\mathrm{log}}_{2}p}{p}$

Where p is the probability for the event to occur during each single experiment.

I am wondering how to derive the entropy of a geometric random variable? Or where I can find some proof/derivation? I tried to search online, but seems not much resources is available.

Here is the probability density function of geometric distribution: $(1-p{)}^{k-1}\phantom{\rule{thinmathspace}{0ex}}p$

Here is the entropy of a geometric distribution: $\frac{-(1-p){\mathrm{log}}_{2}(1-p)-p{\mathrm{log}}_{2}p}{p}$

Where p is the probability for the event to occur during each single experiment.

Dobricap

Beginner2022-10-14Added 14 answers

Step 1

Assume $P(X=k)=(1-p{)}^{k-1}p$, where $k\in {Z}^{+}$, then the entropy is

Step 2

$\begin{array}{rl}Entropy(X)& =\sum _{k=1}^{+\mathrm{\infty}}-(1-p{)}^{k-1}p\cdot {\mathrm{log}}_{2}((1-p{)}^{k-1}p)\\ & =-p\cdot {\mathrm{log}}_{2}(p)\sum _{k=1}^{+\mathrm{\infty}}(1-p{)}^{k-1}-p\cdot {\mathrm{log}}_{2}(1-p)\sum _{k=1}^{+\mathrm{\infty}}(k-1)(1-p{)}^{k-1}\\ & =-{\mathrm{log}}_{2}(p)-\frac{(1-p){\mathrm{log}}_{2}(1-p)}{p}\end{array}$

Assume $P(X=k)=(1-p{)}^{k-1}p$, where $k\in {Z}^{+}$, then the entropy is

Step 2

$\begin{array}{rl}Entropy(X)& =\sum _{k=1}^{+\mathrm{\infty}}-(1-p{)}^{k-1}p\cdot {\mathrm{log}}_{2}((1-p{)}^{k-1}p)\\ & =-p\cdot {\mathrm{log}}_{2}(p)\sum _{k=1}^{+\mathrm{\infty}}(1-p{)}^{k-1}-p\cdot {\mathrm{log}}_{2}(1-p)\sum _{k=1}^{+\mathrm{\infty}}(k-1)(1-p{)}^{k-1}\\ & =-{\mathrm{log}}_{2}(p)-\frac{(1-p){\mathrm{log}}_{2}(1-p)}{p}\end{array}$

Mattie Monroe

Beginner2022-10-15Added 3 answers

Step 1

It may be faster to derive the above entropy using the memoryless property of the Geometric distribution. Let Y be an auxiliary binary random variable such that $Y=0$ indicates $X>1$. Then $H(X)={H}_{B}(p)/p$ follows from the following:

$\begin{array}{rl}H(X,Y)& =H(X|Y)+H(Y)\\ & =H(Y|X)+H(X)\end{array}$

Step 2

Note that $\begin{array}{rl}H(X|Y)& =H(X|Y=1)P(Y=1)+H(X|Y=0)P(Y=0)=H(X|Y=0)(1-p)=(1-p)H(X)\\ H(Y|X)& =0\\ H(Y)& ={H}_{B}(p),\end{array}$

where the memoryless property indicates $H(X|Y=0)=H(X)$, and ${H}_{B}(p)$ is the binary entropy function.

It may be faster to derive the above entropy using the memoryless property of the Geometric distribution. Let Y be an auxiliary binary random variable such that $Y=0$ indicates $X>1$. Then $H(X)={H}_{B}(p)/p$ follows from the following:

$\begin{array}{rl}H(X,Y)& =H(X|Y)+H(Y)\\ & =H(Y|X)+H(X)\end{array}$

Step 2

Note that $\begin{array}{rl}H(X|Y)& =H(X|Y=1)P(Y=1)+H(X|Y=0)P(Y=0)=H(X|Y=0)(1-p)=(1-p)H(X)\\ H(Y|X)& =0\\ H(Y)& ={H}_{B}(p),\end{array}$

where the memoryless property indicates $H(X|Y=0)=H(X)$, and ${H}_{B}(p)$ is the binary entropy function.

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