Question

The Rayleigh density function is given by:

f(y) = (θ / (2y)) * e^(-θy²) for y > 0, and 0 elsewhere.
The quantity Y² has an exponential distribution with mean θ. If Y₁, Y₂, …, Yₙ denotes a random sample from a Rayleigh distribution, show that Wₙ = (1/n) * Σᵢ₌₁ⁿ Yᵢ² is a consistent estimator for θ.

Answer 1

Wₙ = (1/n) * Σᵢ₁ₙ Yᵢ² is an unbiased and consistent estimator for θ because its expected value is θ and its variance approaches zero as the sample size n approaches infinity.

Step-by-step explanation:

To show that Wₙ = (1/n) * Σᵢ₁ₙ Yᵢ² is a consistent estimator for θ, we must prove that Wₙ is both unbiased and that its variance goes to zero as n approaches infinity. An estimator is considered unbiased if the expected value of the estimator is equal to the parameter it is estimating. In this case, we need E(Wₙ) = θ. Furthermore, to prove that the variance of Wₙ approaches zero, we consider Var(Wₙ).

Since it is given that Y² follows an exponential distribution with mean θ, this implies that E(Y²) = θ. Because the Yᵢ's are independent and identically distributed (i.i.d.), we can say that E(Wₙ) = E((1/n)*Σᵢ₁ₙ Yᵢ²) = (1/n)*Σᵢ₁ₙ E(Yᵢ²) = (1/n)*n*θ = θ, making Wₙ an unbiased estimator of θ.

The variance of the mean of i.i.d. random variables is the variance of one random variable divided by n. Therefore, Var(Wₙ) = Var((1/n)*Σᵢ₁ₙYᵢ²) = (1/n²)*Σᵢ₁ₙVar(Yᵢ²). Since Y² is exponentially distributed, Var(Y²) = θ². Substituting this in, we get Var(Wₙ) = (1/n²)*n*θ² = θ²/n. As n approaches infinity, θ²/n approaches zero, indicating that Wₙ is a consistent estimator for θ.