Question

If Y1,Y2,…,Yn are random observations from a exponentail distribution with pdf fY(y)=e1e^−y/e,y>0, |use exisitng results for E(Yi) and Var(Yi) | (a) Show that θ^1=Y^ and θ^2=nYmin are unbiased estimators for θ. (b) Calculate the relative efficiencies of θ^2 with respect to θ^1, that is Var(θ^1)/Var(θ^2). Which one is more efficient?

Answer 1

The question involves showing that θ1 = Є and θ2 = nYmin are unbiased estimators for θ in an exponential distribution, and calculating the relative efficiency using their variances where the estimator with the lower variance is more efficient.

Step-by-step explanation:

Unbiased Estimators of θ and Relative Efficiency

The question involves finding unbiased estimators for a parameter θ in an exponential distribution and comparing their efficiencies. It is given that the random observations Y1, Y2, …, Yn are from an exponential distribution with a probability density function (pdf) given by fY(y) = (1/θ)e−y/θ, for y > 0.

An unbiased estimator is one where the expected value, E(−), is equal to the parameter it estimates. To prove that θ1 = Є and θ2 = nYmin are unbiased estimators for θ:

• Use the property that E(Yi) = θ for the exponential distribution to show that E(θ1) = E(Є) = θ.
• Show that E(θ2) = E(nYmin) = θ using the fact that Ymin is also an exponential distribution with mean θ/n.

To calculate the relative efficiency of θ2 with respect to θ1, use the formula Var(θ1)/Var(θ2). The estimator with the lower variance is more efficient because it leads to more consistent estimates of the parameter.