Probability of Error: Binary Communication Channel

bayes-theorembayesiandecision-theoryprobability

I have a binary communication channel shown below:

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with the following parameters:

  • $\pi_0 = 0.3$
  • $\pi_1 = 0.7$
  • $\epsilon_0 = 0.2$
  • $\epsilon_1 = 0.1$

Goal: compute the probability of error and probability of false alarm for the decision rule: always choose 0

Starting out:
$$P(Error) = P(A_0)P(B_1 | A_0) + P(A_1)P(B_0 | A_1)$$
and from the diagram,

  • $P(B_0|A_0) = 1 – .2 = 0.8$
  • $P(B_0|A_1) = 0.1$
  • $P(B_1 | A_0) = 0.2$
  • $P(B_1 | A_1) = 0.9$

so I can compute the following quantities:

  • $P(B_0) = P(B_0|A_0)P(A_0) + P(B_0|A_1)P(A_1) = (.8)(.3) + (.1)(.7) = .31$
  • $P(B_1) = P(B_1|A_1)P(A_1) + P(B_1|A_0)P(A_0) = (.9)(.7) + (.2)(.3) = .69$
  • $P(A_0|B_0) = \frac{P(B_0|A_0)P(A_0)}{P(B_0)} = \frac{(.8)(.3)}{.31} = .774$
  • $P(A_1|B_1) = \frac{P(B_1|A_1)P(A_1)}{P(B_1)} = \frac{(.9)(.7)}{.69} = .913$

Question:
Where does the decision rule: always choose $0$ come into play? Is the probability of error $.69$ since that's the value of $P(B_1)$? What if the decision rule was different: say keep the observation?

Thanks!

Best Answer

Obviously the optimum decision rule is: choose 1 if you receive 1, and choose 0 if you receive 0, because $P(A_1|B_1)>P(A_0|B_1)$, and $P(A_0|B_0)>P(A_1|B_0)$. The question is: what happens if you disregard what you have received and always choose 0. In that case, the probability of error would be $P(B_0|A_1)P(A_1)$, because the error only occurs when 1 is sent.

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