Suppose that Θ and W are independent, both with variance 1 , and that X=Θ+W . Furthermore, E[Θ]=1 and E[W]=2 . The LLMS estimator Θˆ=aX+b has
a=
b=
a=.5
b=-.5
=.5X-.5
To find the values of a and b for the LLMS estimator Θˆ=aX+b, we need to use the properties of expectation and variance.
1. Start by finding the expectation of X:
E[X] = E[Θ + W] = E[Θ] + E[W] = 1 + 2 = 3.
2. Next, find the variance of X:
Var[X] = Var[Θ + W] = Var[Θ] + Var[W] (since Θ and W are independent) = 1 + 1 = 2.
3. Now, let's find the value of a:
We know that the best linear unbiased estimator (BLUE) is obtained by a = Cov(Θ, X) / Var[X].
Cov(Θ, X) = E[ΘX] - E[Θ]E[X]
= E[Θ(Θ + W)] - 1⋅3 (using the linearity of expectation)
= E[Θ^2 + ΘW] - 3.
Since Θ and W are independent, and their variances are both 1, we have:
E[Θ^2 + ΘW] = E[Θ^2] + E[ΘW] = Var[Θ] + E[Θ]E[W] = 1 + 1⋅2 = 3.
Substituting this into the expression for Cov(Θ, X):
Cov(Θ, X) = 3 - 3 = 0.
Therefore, a = 0 / Var[X] = 0 / 2 = 0.
4. Finally, let's find the value of b:
b = E[Θ] - aE[X] = 1 - 0⋅3 = 1.
Thus, the LLMS estimator Θˆ = 0X + 1 = 1. Therefore, the values of a and b for the LLMS estimator are a = 0 and b = 1.