Fundamentals of Probability and Statistics for Engineers

Parameter Estimation 281Estimates ^ 1 and ^ 2 of 1 ˆ m and 2 ˆ 2 based on the sample values givenby Table 8.1 are, following Equations (9.64) and (9.65),where x j , j ˆ 1, 2, . . . , 200, are sample values given in Table 8.1.Example 9.10. Problem: consider the binomial distributionEstimate parameter p based on a sample of size n.Answer: the method of moments suggests that we determine the estimator forp, ^P, by equating 1 to M 1 ˆ X. Sincewe have^ 1 ˆ 1 X 200x j 70;200jˆ1^ 2 ˆ 1200X 200jˆ1…x j^1 † 2 4;p X …k; p† ˆp k …1 p† 1 k ; k ˆ 0; 1: …9:66† 1 ˆ EfXg ˆp;The mean of^P is^P ˆ X:XEf ^Pg ˆ1nEfX j gˆp:njˆ1…9:67†…9:68†Hence it is an unbiased estimator. Its variance is given byvarf ^Pg ˆvarfXg ˆ2nIt is easy to derive the CRLB for this case and show that(9.67) is also efficient.p…1 p† ˆ : …9:69†n^P defined by EquationExample 9.11. Problem: a set of 214 observed gaps in traffic on a section ofArroyo Seco Freeway is given in Table 9.1. If the exponential density functionf …t; † ˆe t ; t 0; …9:70†is proposed for the gap, determine parameter from the data.TLFeBOOK