STA 247, Assignment 2, Part II(d) With P01=0.01 and P11=0.4, about 11% of the blocks have more than four errors, and therefore can't have their errors corrected. This estimate varies a bit from one run to the other, because 1000 blocks isn't enough to get a really precise answer. The average number of errors in a block is about 2. So the uncorrectable errors arise when the number of errors is greater than the average. The variance in the number of errors is about 4.5, so the standard deviation is a bit more than 2. So its not surprising that sometimes a block has more than 4 errors. With P01 and P11 both set to the steady state probability of an error, which is 1/61, the simulations show that only about 5% of the blocks have more than 4 errors, and hence can't have their errors corrected. The average number of errors is the same as before, but the variance is smaller, so the probability of getting more than four errors is reduced. (Note that if the average number of errors was GREATER than four, a bigger variance would make it MORE likely that the errors could be corrected, but as it is, the average number of errors is less than four, so a big variance makes things worse.) When P01=P11, the errors are actually independent, since it makes no difference whether an error occurred previously or not. The number of errors in a block with then have approximately a Poisson distribution, for which the mean and the variance are equal. This is consistent with what is seen in the simulations. One can calculate the average number of errors exactly. Using the fact that the steady state probability of an error is 1/61, the average number of errors in a block is 120/61 = 1.967213. This is consistent with the simulation results. Note that it makes no difference whether the errors are independent or not for this calculation. When P01=P11, we can calculate the probability of more than four errors nearly exactly by finding the probability that a Poisson random variable with mean 120/61 will have a value greater than 4. This can be done in R as follows: > 1-ppois(4,120/61) [1] 0.04974336 The result is consistent with the simulations.