ABSTRACT
In what follows we make the simple assumption that positive and negative errors have the same probability of occurrence, namely, 12 . Now we can take expectations across the expression for x(n) and thereby determine the mean (i.e. average) accumulated measurement error:
E[x(n)] = [2E( j)− n]ε
where E(·) is the expectation operator. However, since positive and negative errors occur with equal probabilities, we should expect half of the n errors to be positive and the other half to be negative, in which case it follows that E( j)= 12n. Alternatively, we can use the fact that the number of positive measurement errors, j, is distributed as a binomial variate with a mean E( j) = np, where p is the probability of a positive error. Since, by assumption, p = 12 , it follows that the expected number of positive errors will be E( j) = np = 12n. Moreover, we can substitute this result into the expression for E[x(n)] given above, in which case the mean accumulated measurement error must be
E[x(n)] = [2E( j)− n]ε = (2× 12n− n)ε = 0
Hence, ‘on average’, the accumulatedmeasurement error will be zero – an unsurprising result given that the probabilities of a negative and positive measurement error are equal to 12 .
