1.2 KiB
1.2 KiB
Ex 1
Given data: 4, 4, 7, 6, 6, 11, 1, 6, 3, 3 (n = 10 observations)
True parameter: \lambda = 5
Part 1: Estimate the expected value of X
Process: For a Poisson distribution, E[X] = λ. We use the sample mean as an unbiased estimator.
Solution:
\bar{x}_n = \frac{1}{n}\sum_{i=1}^{n} x_i = \frac{4+4+7+6+6+11+1+6+3+3}{10} = \frac{51}{10} = 5.1
Estimated $E[X] = 5.1$
Part 2: Estimate the variance using the unbiased estimator
Process: For Poisson distribution, \text{Var}(X) = \lambda. We use the corrected sample variance with (n-1) denominator.
Solution: First, calculate deviations from the mean:
(4-5.1)^2 = 1.21(4-5.1)^2 = 1.21(7-5.1)^2 = 3.61(6-5.1)^2 = 0.81(6-5.1)^2 = 0.81(11-5.1)^2 = 34.81(1-5.1)^2 = 16.81(6-5.1)^2 = 0.81(3-5.1)^2 = 4.41(3-5.1)^2 = 4.41
S_n^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \frac{1}{9} \times 68.9 = 7.66
Estimated $\text{Var}(X) = 7.66$
Theory behind this: S_n^2 = \frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X}_n)^2 is an unbiased estimator of \text{Var}(X), while the version with denominator n is biased. The (n-1) correction accounts for the loss of one degree of freedom when estimating the mean.