161-A3.1

Activity: Mutual Information and Channel Capacity
Instructions: In this activity, you are tasked to
- Walk through the examples.
- Calculate the channel capacity of different channel models.
Should you have any questions, clarifications, or issues, please contact your instructor as soon as possible.

Example 1: Mutual Information

Given the following probabilities:

To get the entropies of $X$ and $Y$ , we need to calculate the marginal probabilities:

P_{X}=\{{\tfrac {1}{2}},{\tfrac {1}{4}},{\tfrac {1}{8}},{\tfrac {1}{8}}\}

(1)

P_{Y}=\{{\tfrac {1}{4}},{\tfrac {1}{4}},{\tfrac {1}{4}},{\tfrac {1}{4}}\}

(2)

And since:

H\left(A\right)=\sum _{i=1}^{n}P\left(a_{i}\right)\cdot \log _{2}\left({\frac {1}{P\left(a_{i}\right)}}\right)

(3)

We get:

H\left(X\right)={\frac {1}{2}}\log _{2}2+{\frac {1}{4}}\log _{2}4+{\frac {1}{8}}\log _{2}8+{\frac {1}{8}}\log _{2}8={\frac {7}{4}}\,\mathrm {bits} =1.75\,\mathrm {bits}

(4)

H\left(Y\right)={\frac {1}{4}}\log _{2}4+{\frac {1}{4}}\log _{2}4+{\frac {1}{4}}\log _{2}4+{\frac {1}{4}}\log _{2}4=2\,\mathrm {bits}

(5)

Calculating the conditional entropies using:

H\left(X\mid Y\right)=\sum _{i=1}^{4}\sum _{j=1}^{4}P\left(x_{i},y_{j}\right)\cdot \log _{2}\left({\frac {P\left(y_{j}\right)}{P\left(x_{i},y_{j}\right)}}\right)={\frac {11}{8}}\,\mathrm {bits} =1.375\,\mathrm {bits}

(6)

H\left(Y\mid X\right)=\sum _{i=1}^{4}\sum _{j=1}^{4}P\left(x_{i},y_{j}\right)\cdot \log _{2}\left({\frac {P\left(x_{i}\right)}{P\left(x_{i},y_{j}\right)}}\right)={\frac {13}{8}}\,\mathrm {bits} =1.625\,\mathrm {bits}

(7)

Note that $H\left(X\mid Y\right)\neq H\left(Y\mid X\right)$ . Calculating the mutual information, we get:

I\left(A;B\right)=H\left(X\right)-H\left(X\mid Y\right)={\frac {7}{4}}-{\frac {11}{8}}=0.375\,\mathrm {bits}

(8)

Or equivalently:

I\left(A;B\right)=H\left(Y\right)-H\left(Y\mid X\right)=2-{\frac {13}{8}}=0.375\,\mathrm {bits}

(9)

Let us try to understand what this means:

If we only consider $X$ , we have the a priori probabilities, $P_{X}$ for each blood type, and we can calculate the entropy, $H\left(X\right)$ , i.e. the expected value of the information we get when we observe the results of the blood test.
Will our expectations change if we do not have access to the blood test, but instead, we get to access (1) the person's susceptibility to skin cancer, and (2) the conditional probabilities in Table 1? Since we are given more information, we expect:
- The uncertainty to be equal to the original uncertainty if $X$ and $Y$ are independent, since $H\left(X\mid Y\right)=H\left(X\right)$ , thus $I\left(X;Y\right)=H\left(X\right)-H\left(X\mid Y\right)=0$ .
- A reduction in the uncertainty due to the additional correlated information given by $P\left(x_{i}\mid y_{j}\right)$ .