Let X denote the number of failures before the first success in a Bernoulli scheme with probabilityof success equal to θ, i.e. Pθ (X = k) = θ(1 − θ)k, for k = 0, 1, .... Find the method of momentsestimator for θ, based on the mean. What will be the precise value of the estimator, if in a sample ofn observations, the average number of failures is equal to 4? Find the method of moments estimatorfor θ, based on the variance. What will be the precise value of the estimator, if in a sample of nobservations, the variance of the number of failures is equal to 20?
Question
Let X denote the number of failures before the first success in a Bernoulli scheme with probabilityof success equal to θ, i.e. Pθ (X = k) = θ(1 − θ)k, for k = 0, 1, .... Find the method of momentsestimator for θ, based on the mean. What will be the precise value of the estimator, if in a sample ofn observations, the average number of failures is equal to 4? Find the method of moments estimatorfor θ, based on the variance. What will be the precise value of the estimator, if in a sample of nobservations, the variance of the number of failures is equal to 20?
Solution
The method of moments estimator for a parameter is found by setting the population moment equal to the sample moment and solving for the parameter.
- Based on the mean:
The mean of a geometric distribution (which is what this is) is (1-θ)/θ. Setting this equal to the sample mean (which we'll call m) gives us:
(1-θ)/θ = m
Solving for θ gives us:
θ = 1/(m+1)
So the method of moments estimator for θ based on the mean is 1/(m+1). If the average number of failures in a sample is 4, then the estimator for θ would be 1/(4+1) = 1/5 = 0.2.
- Based on the variance:
The variance of a geometric distribution is (1-θ)/θ^2. Setting this equal to the sample variance (which we'll call v) gives us:
(1-θ)/θ^2 = v
Solving for θ is a bit more complicated in this case, but with some algebra we can find that:
θ = 1/(v+1)
So the method of moments estimator for θ based on the variance is 1/(v+1). If the variance of the number of failures in a sample is 20, then the estimator for θ would be 1/(20+1) = 1/21 ≈ 0.0476.
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