Recall that in chapter 2 we dened a random variable

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: how to bound the probability of undetected error, assuming that each bit is in error with probability 0.001, independently of all other bits. Let Y denote the number of bits in error. As already discussed, there can be undetected bit errors only if Y ≥ 4. So the probability there exist undetected bit errors is less than or equal to P {Y ≥ 4}. We have Y = X1 + · · · + X64 , where Xi is one if the ith bit in the array is in error, and Xi = 0 otherwise. By assumption, the random variables X1 , . . . , X64 are independent Bernoulli random variables with parameter p = 0.001, so Y is a binomial random variable with parameters n = 64 and p = 0.001. The event {Y ≥ 4} can be expressed as the union of 644 events of the form {Xi = 1 for all i ∈ A}, indexed by the subsets A of {1, . . . , n} with |A| = 4. For such an A, P {Xi = 1 for all i ∈ A} = i∈A P {Xi = 1} = p4 . So the union bound with m = 644 yields P {Y ≥ 4} ≤ 644 p4 . Thus, 64 4 p = (953064)p4 = 0.953064 × 10−6 . 4 P (undetected errors) ≤ P {Y ≥ 4} ≤ A considerably tighter bound can be obtained...
View Full Document

This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

Ask a homework question - tutors are online