Question

Two estimators 𝜆ˆ and 𝜆˜ are consistent and asymptotically normal.

Let 𝑋1,…,𝑋𝑛∼𝑖.𝑖.𝑑.𝖤𝗑𝗉(𝜆) , for some 𝜆>0 .
Let 𝜆ˆ=1𝑋𝑛 and 𝜆̃ =−ln(𝑌𝑛) , where 𝑌𝑖=1{𝑋𝑖>1},𝑖=1,…,𝑛 .
Find, their asymptotic variances 𝑉(𝜆ˆ) and 𝑉(𝜆˜)

Answer 1

[AI-generated answer — Step-by-Step Bot]

To find the asymptotic variances of the two estimators 𝜆ˆ and 𝜆̃, we need to use their asymptotic properties.

1. Asymptotic Variance of 𝜆ˆ:
The estimator 𝜆ˆ = 1/Xn is consistent and asymptotically normal. Since 𝑋1,…,𝑋𝑛 ∼ 𝑖.𝑖.𝑑.𝖤𝗑𝗉(𝜆), we know that the sample mean 𝑋⎯⎯⎯n = (1/n) * Σ𝑋𝑖 is a consistent and asymptotically normal estimator of 𝜆.

The asymptotic variance of 𝜆ˆ is given by:
𝑉(𝜆ˆ) = [1 / (𝜆^2)] * 𝑉(𝑋⎯⎯⎯𝑛)

Since 𝑋1,…,𝑋𝑛 ∼ 𝑖.𝑖.𝑑.𝖤𝗑𝗉(𝜆), the variance of 𝑋⎯⎯⎯𝑛 is given by:
𝑉(𝑋⎯⎯⎯𝑛) = Var(1/n * Σ𝑋𝑖)
= (1/n^2) * ΣVar(𝑋𝑖) (due to independence)
= (1/n^2) * 𝑛 * Var(𝑋1)
= (1/n) * Var(𝑋1)
= 𝑉(𝑋1)

Therefore,
𝑉(𝜆ˆ) = [1 / (𝜆^2)] * 𝑉(𝑋1)

2. Asymptotic Variance of 𝜆̃:
The estimator 𝜆̃ = -ln(Yn), where 𝑌𝑖 = 1{𝑋𝑖 > 1}.

Notice that 𝑌𝑖 = 1 if 𝑋𝑖 > 1 and 𝑌𝑖 = 0 if 𝑋𝑖 ≤ 1.
This implies that Σ𝑌𝑖 is the count of 𝑋𝑖 values greater than 1, which can be represented as n - Σ1{𝑋𝑖≤1}.

Now, let's find the asymptotic variance of 𝜆̃ using the delta method:

We have:
𝜆̃ = -ln(Yn)
Yn = Π𝑌𝑖 (since 𝑌𝑖 = 1{𝑋𝑖 > 1})
= Π(1 - 1{𝑋𝑖 ≤ 1})

Taking the natural logarithm on both sides:
ln(Yn) = ln(Π(1 - 1{𝑋𝑖 ≤ 1}))
= Σln(1 - 1{𝑋𝑖 ≤ 1})

Now, let's use the delta method to find the asymptotic variance:
𝑉(𝜆̃) = [g'(𝜆)]^2 * 𝑉(𝑙𝑛(𝑌𝑛))

where g(𝜆) = -ln(𝜆) and 𝑙𝑛(𝑌𝑛) = Σln(1 - 1{𝑋𝑖 ≤ 1}).

Differentiating g(𝜆) with respect to 𝜆, we get:
g'(𝜆) = -1 / 𝜆

The variance of 𝑙𝑛(𝑌𝑛) can be approximated using the sample variance 𝑆^2:
𝑉(𝑙𝑛(𝑌𝑛)) ≈ (1/n) * 𝑆^2

Therefore,
𝑉(𝜆̃) = [(-1/𝜆)]^2 * (1/n) * 𝑆^2

Note: The above expressions for 𝑉(𝜆ˆ) and 𝑉(𝜆̃) are approximations based on the asymptotic properties of the estimators. In practice, it is recommended to use resampling methods or simulation to obtain more accurate estimates of the variances.

Answer 2

[AI-generated answer — Explain Bot]

To find the asymptotic variances of the estimators λ̂ and λ̃, we need to use some properties of consistent and asymptotically normal estimators.

First, let's find the asymptotic variance of λ̂:

The estimator λ̂ = 1/Xₙ is consistent and asymptotically normal.

To find the asymptotic variance of λ̂, we can use the Delta Method. The Delta Method states that if g(θ) is a differentiable function of the parameter θ, and √𝑛(θ̂−θ) converges in distribution to a normal distribution, then √𝑛(g(θ̂)−g(θ)) converges in distribution to a normal distribution with mean 0 and variance [g'(θ)]^2 times the variance of √𝑛(θ̂−θ).

In this case, g(θ) = 1/θ, and θ = λ. So, g'(θ) = -1/θ².

Using the Delta Method, the asymptotic variance of λ̂, denoted as V(λ̂), is given by:
V(λ̂) = [g'(λ)]^2 * V(X)/n, where V(X) is the true variance of the random variable X.

In this case, X follows the exponential distribution with parameter λ, so V(X) = 1/λ².

Therefore, V(λ̂) = [(-1/λ²)]^2 * (1/λ²)/n = (1/λ⁴)/n.

Now, let's find the asymptotic variance of λ̃:

The estimator λ̃ = -ln(Yₙ), where Yᵢ = 1{Xᵢ > 1}, i = 1,...,n.

To find the asymptotic variance of λ̃, we can use the Delta Method again. In this case, g(θ) = -ln(θ), and θ = P(X > 1).

Since Yᵢ = 1{Xᵢ > 1}, Yᵢ follows a Bernoulli distribution with parameter P(X > 1). Therefore, the true variance of Yᵢ is given by P(X > 1) * (1 - P(X > 1)).

Using the Delta Method, the asymptotic variance of λ̃, denoted as V(λ̃), is given by:
V(λ̃) = [g'(θ)]^2 * V(Y)/n, where V(Y) is the true variance of the random variable Yᵢ.

In this case, g'(θ) = -1/θ.

Therefore, V(λ̃) = [(-1/θ)]^2 * (θ(1 - θ))/n = (θ(1 - θ))/n.

However, we need to express the variance in terms of λ, not θ.

Since θ = P(X > 1) = 1 - P(X ≤ 1), we have θ = 1 - e^(-λ).

Substituting θ in the expression for V(λ̃), we get:
V(λ̃) = [(1 - e^(-λ))(e^(-λ))/n = [e^(-λ) - e^(-2λ)]/n.

So, the asymptotic variances of λ̂ and λ̃ are:
V(λ̂) = (1/λ⁴)/n
V(λ̃) = [e^(-λ) - e^(-2λ)]/n