Question

Argue that the proposed estimators 𝜆ˆ and 𝜆˜ below are both consistent and asymptotically normal. Then, give their asymptotic variances 𝑉(𝜆ˆ) and 𝑉(𝜆˜) , and decide if one of them is always bigger than the other.

Let 𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖯𝗈𝗂𝗌𝗌(𝜆) , for some 𝜆>0 . Let 𝜆̂ =𝑋⎯⎯⎯⎯⎯𝑛 and 𝜆̃ =−ln(𝑌⎯⎯⎯⎯𝑛) , where 𝑌𝑖=1{𝑋𝑖=0},𝑖=1,…,𝑛 .

𝑉(𝜆ˆ) =? and 𝑉(𝜆˜) =?

Answer 1

[AI-generated answer — Explain Bot]

To argue that the proposed estimators 𝜆̂ and 𝜆̃ are both consistent and asymptotically normal, we need to show that they satisfy the properties of consistency and asymptotic normality.

1. Consistency:
An estimator is consistent if it converges to the true parameter value as the sample size increases. In this case, we want to show that both 𝜆̂ and 𝜆̃ converge to 𝜆 as n approaches infinity.

a. 𝜆̂ = 𝑋⎯⎯⎯⎯⎯𝑛:
The sample mean 𝑋⎯⎯⎯⎯⎯𝑛 is a consistent estimator for the population mean due to the Law of Large Numbers. For an i.i.d. sequence of random variables 𝑋1, 𝑋2, ..., 𝑋𝑛 with a common distribution 𝖯𝗈𝗂𝗌𝗌(𝜆), the sample mean 𝑋⎯⎯⎯⎯⎯𝑛 converges to 𝜆 as n approaches infinity. Therefore, 𝜆̂ = 𝑋⎯⎯⎯⎯⎯𝑛 is a consistent estimator of 𝜆.

b. 𝜆̃ = −ln(𝑌⎯⎯⎯⎯𝑛):
In this case, 𝑌𝑖 = 1{𝑋𝑖 = 0} is an indicator variable that takes the value 1 if 𝑋𝑖 = 0, and 0 otherwise. So, 𝑌⎯⎯⎯⎯𝑛 represents the proportion of zeros in the sample. As n approaches infinity, the proportion of zeros in the sample will converge to the probability of getting a zero from the distribution 𝖯𝗈𝗂𝗌𝗌(𝜆). Taking the negative logarithm of this proportion, −ln(𝑌⎯⎯⎯⎯𝑛), will tend to 𝜆. Thus, 𝜆̃ = −ln(𝑌⎯⎯⎯⎯𝑛) is a consistent estimator of 𝜆.

2. Asymptotic Normality:
An estimator is asymptotically normal if, as the sample size increases, its distribution becomes approximately normal with mean 𝜆 and a finite variance. To determine the asymptotic variances 𝑉(𝜆̂) and 𝑉(𝜆̃), we need to calculate the variances of the estimators.

a. 𝑉(𝜆̂):
For the sample mean estimator 𝜆̂ = 𝑋⎯⎯⎯⎯⎯𝑛, the variance is given by Var(𝑋⎯⎯⎯⎯⎯𝑛) = Var(𝑋𝑖)/𝑛, where Var(𝑋𝑖) = 𝜆(1 - 𝜆) for 𝑖.𝑖.𝑑. random variables from the 𝖯𝗈𝗂𝗌𝗌 distribution. Thus, 𝑉(𝜆̂) = 𝜆(1 - 𝜆)/𝑛.

b. 𝑉(𝜆̃):
To find the variance of 𝜆̃ = −ln(𝑌⎯⎯⎯⎯𝑛), we need to calculate the variance of 𝑌⎯⎯⎯⎯𝑛. Since 𝑌⎯⎯⎯⎯𝑛 represents the proportion of zeros in the sample, it follows a binomial distribution with parameters 𝑛 and 𝜆. The variance of a binomial distribution with parameters 𝑛 and 𝜆 is given by Var(𝑌⎯⎯⎯⎯𝑛) = 𝑛𝜆(1 - 𝜆). Taking the negative logarithm does not change the variance, so 𝑉(𝜆̃) = 𝑛𝜆(1 - 𝜆).

Comparing the two asymptotic variances, we can see that 𝑉(𝜆̂) = 𝜆(1 - 𝜆)/𝑛, while 𝑉(𝜆̃) = 𝑛𝜆(1 - 𝜆). The ratio of the two variances is 𝑉(𝜆̃)/𝑉(𝜆̂) = (𝑛𝜆(1 - 𝜆))/ (𝜆(1 - 𝜆)/𝑛) = 𝑛². Thus, 𝑉(𝜆̃) is always greater than 𝑉(𝜆̂), with 𝑉(𝜆̃) growing quadratically with n.

In conclusion, both estimators 𝜆̂ and 𝜆̃ are consistent and asymptotically normal. However, the asymptotic variance 𝑉(𝜆̃) is always larger than the asymptotic variance 𝑉(𝜆̂).