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A bit is transmitted through a noisy communication channel. It is uncertain whether the bit is 0 or 1, and it may also change values as it is being transmitted. Let \( X \) be a random variable representing the bit at the input end and \( Y \) be a random variable representing the bit at the output. We know that \( P(Y=0 \mid X=0) = 0.9 \) and \( P(Y=1 \mid X=1) = 0.8 \).

1. Suppose that \( P(X=0) = 0.4 \). Compute \( P(Y=0) \) and \( P(Y=1) \).
2. Compute \( P(X=Y) \) and \( P(X \neq Y) \) (remember that \( X \) and \( Y \) may each be either 0 or 1).
3. Suppose we observe \( Y \). Compute \( P(X=0 \mid Y=0) \) and \( P(X=1 \mid Y=1) \).
4. Now suppose that we take our bit and sequentially push it through multiple noisy channels identical to the original one. Write a short program to compute the probability that the output bit is 0 after 10, 50, and 100 transmissions, experimenting with different values of \( P(X=0) \). You should set up your program so that each probability can be computed with a simple change in the value of a variable. What do you notice about the results?

Answer :

Final answer:

The probabilities are as follows: P(Y=0) = 0.36, P(Y=1) = 0.48, P(X=Y) = 0.72, P(X≠Y) = 0.28, P(X=0∣Y=0) = 0.5, P(X=1∣Y=1) = 0.6667. By experimenting with different values of P(X=0) and the number of transmissions, we can observe the impact on the probability of the output bit being 0.

Explanation:

In a noisy communication channel, the probability of the output bit being 0 or 1 can be calculated based on the conditional probabilities P(Y=0∣X=0) and P(Y=1∣X=1). Given that P(Y=0∣X=0) = 0.9 and P(Y=1∣X=1) = 0.8, we can compute the probabilities as follows:

  1. P(Y=0) = P(Y=0∣X=0) * P(X=0) = 0.9 * 0.4 = 0.36
  2. P(Y=1) = P(Y=1∣X=1) * (1 - P(X=0)) = 0.8 * (1 - 0.4) = 0.48

To compute P(X=Y) and P(X≠Y), we need to consider all possible combinations of X and Y:

  • P(X=Y) = P(X=0, Y=0) + P(X=1, Y=1) = P(Y=0∣X=0) * P(X=0) + P(Y=1∣X=1) * (1 - P(X=0)) = 0.9 * 0.4 + 0.8 * (1 - 0.4) = 0.72
  • P(X≠Y) = P(X=0, Y=1) + P(X=1, Y=0) = P(Y=1∣X=0) * P(X=0) + P(Y=0∣X=1) * (1 - P(X=0)) = (1 - P(Y=0∣X=0)) * P(X=0) + (1 - P(Y=1∣X=1)) * (1 - P(X=0)) = (1 - 0.9) * 0.4 + (1 - 0.8) * (1 - 0.4) = 0.28

To compute P(X=0∣Y=0) and P(X=1∣Y=1), we can use Bayes' theorem:

  • P(X=0∣Y=0) = (P(Y=0∣X=0) * P(X=0)) / P(Y=0) = (0.9 * 0.4) / 0.36 = 0.5
  • P(X=1∣Y=1) = (P(Y=1∣X=1) * (1 - P(X=0))) / P(Y=1) = (0.8 * (1 - 0.4)) / 0.48 = 0.6667

To compute the probability of the output bit being 0 after multiple transmissions, we can write a short program that iteratively calculates the probability based on the number of transmissions and the value of P(X=0). By experimenting with different values of P(X=0) and the number of transmissions, we can observe the impact on the probability of the output bit being 0.

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