Suppose that Θ , X1 , and X2 have zero means. Furthermore,
Var(X1)=Var(X2)=Var(Θ)=4,
and
Cov(Θ,X1)=Cov(Θ,X2)=Cov(X1,X2)=1.
The LLMS estimator of Θ based on X1 and X2 is of the form Θˆ=a1X1+a2X2+b . Find the coefficients a1 , a2 , and b . Hint: To find b , recall the argument we used for the case of a single observation.
a1 =
a2 =
b =
0.2
0.2
0
a1 = 1/3
a2 = 1/3
b = 0, because clown bot doesn't need coefficients to make people laugh!
To find the coefficients a1, a2, and b of the LLMS estimator Θˆ=a1X1+a2X2+b, we can use the given variances and covariances.
First, let's consider the formula for LLMS estimator, which is given by:
Θˆ = (Cov(X1, Θ) * X2 + Cov(X2, Θ) * X1) / (Var(X1) + Var(X2))
Using the given covariance values and variances, we can substitute these into the LLMS estimator formula:
Θˆ = (1 * X2 + 1 * X1) / (4 + 4)
Simplifying the numerator:
Θˆ = (X1 + X2) / 8
We see that b = 0 since the estimator has a zero mean. Therefore, we can rewrite the Θˆ equation as:
Θˆ = a1X1 + a2X2
Comparing this equation to the given Θˆ = (X1 + X2) / 8 equation, we can deduce that a1 = 1/8 and a2 = 1/8.
So, the coefficients are:
a1 = 1/8
a2 = 1/8
b = 0
To find the coefficients a1, a2, and b for the LLMS estimator of Θ based on X1 and X2, we need to use the properties of covariance and variance.
The LLMS estimator is given by Θˆ = a1X1 + a2X2 + b.
To start, we can write the expression for the LLMS estimator based on the given information:
Θˆ = a1X1 + a2X2 + b
We can calculate the covariance between Θ and X1 using the formula:
Cov(Θ, X1) = E[(Θ - E[Θ])(X1 - E[X1])]
Since both Θ and X1 have zero means, this simplifies to:
Cov(Θ, X1) = E[ΘX1]
Using the linearity of expectation, we can expand this expression:
Cov(Θ, X1) = E[a1X1^2 + a2X1X2 + bX1]
Since the covariance between Θ and X1 is given as 1, we have:
1 = E[a1X1^2 + a2X1X2 + bX1]
Similarly, we can calculate the covariance between Θ and X2:
Cov(Θ, X2) = E[a1X1X2 + a2X2^2 + bX2]
Since the covariance between Θ and X2 is also given as 1, we have:
1 = E[a1X1X2 + a2X2^2 + bX2]
Finally, we can calculate the covariance between X1 and X2:
Cov(X1, X2) = E[(X1 - E[X1])(X2 - E[X2])]
Since both X1 and X2 have zero means and Var(X1) = Var(X2) = 4, this simplifies to:
Cov(X1, X2) = E[X1X2]
Using linearity of expectation, we can expand this expression:
Cov(X1, X2) = E[a1X1^2 + a2X1X2 + bX1]
Since the covariance between X1 and X2 is given as 1, we have:
1 = E[a1X1^2 + a2X1X2 + bX1]
Now, we have three equations:
1 = E[a1X1^2 + a2X1X2 + bX1]
1 = E[a1X1X2 + a2X2^2 + bX2]
1 = E[a1X1^2 + a2X1X2 + bX1]
Solving these equations simultaneously will give us the values of a1, a2, and b.
Using the method of moments, we can express each of these expectations in terms of Var(X1), Var(X2), and Cov(X1, X2):
a1Var(X1) + a2Cov(X1, X2) + bVar(X1) = 1
a1Cov(X1, X2) + a2Var(X2) + bCov(X1, X2) = 1
a1Var(X1) + a2Cov(X1, X2) + bVar(X1) = 1
Substituting the given values of Var(X1), Var(X2), and Cov(X1, X2):
4a1 + Cov(X1, X2) + 4b = 1
Cov(X1, X2) + 4a2 + Cov(X1, X2) = 1
4a1 + Cov(X1, X2) + 4b = 1
Simplifying these equations, we get:
4a1 + 2 + 4b = 1
2 + 4a2 + 2 = 1
4a1 + 2 + 4b = 1
Simplifying further:
4a1 + 4b = -1
4a2 = -1
4a1 + 4b = -1
Dividing both sides of the second equation by 4, we get:
a2 = -1/4
Now, we have two equations involving a1 and b:
4a1 + 4b = -1
4a1 + 4b = -1
Since these equations are the same, we can't solve for both a1 and b simultaneously. However, we can find the ratio a1/b.
Dividing both sides of one of the equations by 4, we get:
a1 + b = -1/4
Now, we can solve for the ratio a1/b:
a1/b = (-1/4 - b)/b
We know that the parameter Θ has zero mean, so to find the value of b, we can set the expectation of Θ (which is zero) equal to the expectation of the estimator Θˆ:
E[Θˆ] = 0
Using linearity of expectation, we get:
E[a1X1 + a2X2 + b] = 0
Since the means of X1 and X2 are zero, this simplifies to:
a1E[X1] + a2E[X2] + b = 0
Since the means of X1 and X2 are zero, this further simplifies to:
b = 0
Therefore, we have b = 0.
Substituting b = 0 into the equation a1 + b = -1/4, we get:
a1 + 0 = -1/4
Therefore, a1 = -1/4.
Now, we have a1 = -1/4, a2 = -1/4, and b = 0 as the coefficients for the LLMS estimator of Θ based on X1 and X2.