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
0.5 and -0.5 you're welcome
the parameter values Θ^ = aX + b such that the expected value of the squared error is minimized.
To find the values of a and b, we need to minimize the expression E[(Θ - Θ^)^2].
Expanding this expression, we have E[(Θ - aX - b)^2].
Since Θ and X are independent and have variance 1, we can rewrite this expression as E[(Θ^2 - 2aXΘ + a^2X^2 - 2bΘ + 2abX + b^2)].
Taking the expectation, we get E[Θ^2] - 2aE[XΘ] + a^2E[X^2] - 2bE[Θ] + 2abE[X] + b^2.
Since E[Θ] = 1, E[W] = 2, and Var[X] = Var[Θ + W] = Var[Θ] + Var[W] + 2Cov[Θ, W] = 1 + 1 + 2*0 = 2, we can further simplify the expression to:
E[Θ^2] - 2aE[X] + a^2E[X^2] - 2b + 2abE[X] + b^2.
Now, we differentiate this expression with respect to a and b to find the values that minimize it.
Differentiating with respect to a, we get -2E[X] + 2aE[X^2] + 2bE[X] = 0.
Simplifying, we have E[X^2]a + (b - E[X]) = 0.
Similarly, differentiating with respect to b, we get -2 + 2aE[X] + 2b = 0.
Simplifying, we have (aE[X] + 1)b = 1.
Solving these two equations simultaneously, we can find the values of a and b that minimize the expected squared error. However, it's a bit difficult to solve it explicitly without knowing the joint distribution of Θ and W.
So unfortunately, I cannot give you the exact values of a and b for the LLMS estimator without additional information. But I hope you found the derivation process entertaining!
The LLMS estimator Θˆ=𝑎𝑋+𝑏 is a linear combination of the random variable X. To find the values of a and b, we need to minimize the mean squared error (MSE) of the estimator.
The mean squared error (MSE) of the estimator Θˆ is given by:
MSE(Θˆ) = E[(Θˆ - Θ)^2]
Substituting the expression for Θˆ, we get:
MSE(Θˆ) = E[(aX + b - Θ)^2]
Expanding and simplifying the expression, we have:
MSE(Θˆ) = E[(a(Θ + 𝑊) + b - Θ)^2]
= E[((a-1)Θ + a𝑊 + b)^2]
Since Θ and 𝑊 are independent, their covariance is 0. Therefore, E[Θ𝑊] = E[Θ]E[𝑊] = 1 * 2 = 2 and E[𝑊^2] = var(𝑊) = 1.
Rewriting the expression, we have:
MSE(Θˆ) = E[((a-1)Θ)^2 + 2(a-1)Θ(a𝑊 + b) + (a𝑊 + b)^2]
= E[(a^2 - 2a + 1)Θ^2 + 2(a-1)Θ(a𝑊 + b) + a^2𝑊^2 + 2ab𝑊 + b^2]
Taking the expectations of each term, we get:
MSE(Θˆ) = (a^2 - 2a + 1)E[Θ^2] + 2(a-1)E[Θ(a𝑊 + b)] + a^2E[𝑊^2] + 2abE[𝑊] + b^2
Since E[Θ^2] = var(Θ) + (E[Θ])^2 and E[Θ] = 1, we have:
MSE(Θˆ) = (a^2 - 2a + 1) * 1 + 2(a-1)E[Θ(a𝑊 + b)] + a^2 * 1 + 2ab * 2 + b^2
Simplifying further, we get:
MSE(Θˆ) = a^2 - 2a + 1 + 2(a-1)E[Θ(a𝑊 + b)] + a^2 + 4ab + b^2
To minimize the MSE, we differentiate the expression with respect to a and b, and set the derivatives equal to 0:
∂MSE/∂a = 2a - 2 + 2E[Θ(a𝑊 + b)] + 2a + 4b = 0
∂MSE/∂b = 2(a-1)E[Θ(a𝑊 + b)] + 4ab + 2b = 0
Solving the above equations simultaneously will give us the values of a and b that minimize the MSE.
To find the LLMS (Linear Minimum Mean Square) estimator, we need to minimize the mean square error between the estimator and the true parameter. In this case, we want to find the values of 'a' and 'b' that minimize the mean square error between Θ and Θ^.
First, let's start with the expression for the LLMS estimator:
Θ^ = aX + b
Next, we need to calculate the mean square error (MSE) between Θ and Θ^:
MSE = E[(Θ - Θ^)^2]
Let's expand this expression:
MSE = E[((Θ - (aX + b))^2]
MSE = E[(Θ - aX - b)^2]
Using the linearity of expectation, we can simplify this expression further:
MSE = E[((Θ - a(Θ + W) - b))^2]
MSE = E[(Θ - aΘ - aW - b)^2]
Now, let's expand the square:
MSE = E[(Θ^2 - 2aΘ(Θ + W) - 2b(Θ + W) + a^2(Θ + W)^2 + 2ab(Θ + W) + b^2)]
Next, let's simplify the expression by taking expectations:
MSE = E[Θ^2] - 2aE[Θ(Θ + W)] - 2bE[Θ + W] + a^2E[(Θ + W)^2] + 2abE[Θ + W] + b^2
Now, we can substitute the given expectations:
MSE = 1 - 2a(E[Θ^2 + ΘW]) - 2b(1 + 2) + a^2(E[Θ^2 + 2ΘW + W^2]) + 2ab(1 + 2) + b^2
Simplifying further:
MSE = 1 - 2a(1 + E[ΘW]) - 6b + a^2(1 + 2 + E[W^2]) + 6ab + b^2
Since E[ΘW] = E[Θ]E[W] = 1 * 2 = 2 and E[W^2] = Var[W] + (E[W])^2 = 1 + 2^2 = 5, we can substitute these values into the MSE expression:
MSE = 1 - 2a(1 + 2) - 6b + a^2(1 + 2 + 5) + 6ab + b^2
Simplifying further:
MSE = 1 - 6a - 6b + 8a^2 + 6ab + b^2
To minimize the MSE, we need to take the partial derivatives of the MSE expression with respect to 'a' and 'b' and equate them to zero:
∂MSE/∂a = -6 + 16a + 6b = 0
∂MSE/∂b = -6 + 6a + 2b = 0
Solving these equations simultaneously, we can find the values of 'a' and 'b' that minimize the MSE.