Question

Estimators 𝑝ˆ,𝑝˜, are consistent and asymptotically normal.

Let 𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖦𝖾𝗈𝗆(𝑝) , for some 𝑝∈(0,1) .
That means that 𝐏(𝑋1=𝑘)=𝑝(1−𝑝)𝑘−1,for 𝑘=1,2,….
Let 𝑝ˆ = 1/𝑋𝑛,
and 𝑝˜ be the number of ones in the sample divided by 𝑛 .
Find asymptotic variances 𝑉(𝑝ˆ) and 𝑉(𝑝˜)

Answer 1

[AI-generated answer — Explain Bot]

To find the asymptotic variances of estimators 𝑝̂ and 𝑝̃, we need to use the properties of consistent and asymptotically normal estimators.

First, let's find the asymptotic variance of 𝑝̂. The estimator 𝑝̂ is defined as 𝑝̂ = 1/𝑋𝑛. We know that 𝑋₁, 𝑋₂, ..., 𝑋ₙ are independent and identically distributed (i.i.d) random variables with a Bernoulli distribution with parameter 𝑝.

From the given information, we know that 𝑋₁, 𝑋₂, ..., 𝑋ₙ follow a Bernoulli(p) distribution. Therefore, we can calculate the variance of 𝑋₁:

Var(𝑋₁) = 𝑝(1-𝑝)^(𝑘-1) = 𝑝(1-𝑝)

The variance of the sum of 𝑛 i.i.d Bernoulli random variables is equal to the sum of their variances, so:

Var(𝑋₁ + 𝑋₂ + ... + 𝑋ₙ) = 𝑛𝑝(1-𝑝)

Now, since 𝑝̂ = 1/𝑋ₙ, we can use the delta method to find the asymptotic variance. The delta method states that if 𝑌 = 𝑔(𝑋) is an estimator of 𝑡(𝑝), then the asymptotic variance of 𝑡̂(𝑝) is given by:

Var(𝑡̂(𝑝)) = 𝑔'²(𝑝) * Var(𝑋) / n

In this case, 𝑡(𝑝) = 1/𝑝 and 𝑋 = 𝑋ₙ. Therefore, 𝑔(𝑋) = 1/𝑋 and 𝑔'(𝑋) = -1/𝑋².

Plugging these values into the formula, we have:

Var(𝑝̂) = 𝑔'²(𝑋) * Var(𝑋) / n
= (-1/𝑋²)² * 𝑛𝑝(1-𝑝) / n
= 𝑝(1-𝑝) / (𝑛𝑋²)

Since 𝑋ₙ follows a Bernoulli distribution with parameter 𝑝, we can approximate 𝑋² by 𝑝(1-𝑝). Therefore:

Var(𝑝̂) ≈ 𝑝(1-𝑝) / (𝑛𝑝(1-𝑝))
= 1 / (𝑛𝑝(1-𝑝))

Next, let's find the asymptotic variance of 𝑝̃, which is the number of ones in the sample divided by 𝑛. The estimator 𝑝̃ is given by 𝑝̃ = number of ones in the sample / 𝑛.

Since the number of ones in the sample follows a binomial distribution with parameters 𝑛 and 𝑝, the variance of the number of ones in the sample is:

Var(number of ones in the sample) = 𝑛𝑝(1-𝑝)

Therefore, the variance of 𝑝̃ is:

Var(𝑝̃) = Var(number of ones in the sample) / 𝑛²
= 𝑛𝑝(1-𝑝) / 𝑛²
= 𝑝(1-𝑝) / 𝑛

So, the asymptotic variances of 𝑝̂ and 𝑝̃ are 𝑉(𝑝̂) = 1 / (𝑛𝑝(1-𝑝)) and 𝑉(𝑝̃) = 𝑝(1-𝑝) / 𝑛, respectively.