Let X=U+W with E[U]=m, var(U)=v, E[W]=0, and var(W)=h. Assume that U and W are independent.
The LLMS estimator of U based on X is of the form U^=a+bX. Find a and b. Express your answers in terms of m, v, and h using standard notation.
a=- unanswered
b=- unanswered
Suppose we further assume that U and W are normal random variables and then construct U^LMS, the LMS estimator of U based on X, under this additional assumption. Would U^LMS be the identical to U^, the LLMS estimator developed without the additional normality assumption in part (1)?
Yes - answered
a=v/(v+h)
b=m*h/(v+h)
Yes
Can someone please answer?
a= m b= 1
The LMS estimator U^LMS would be identical to the LLMS estimator U^ developed without the additional normality assumption.
To find the values of a and b in the LLMS estimator, we need to minimize the mean squared error (MSE) between U and U^ = a + bX.
The MSE is given by:
MSE = E[(U - U^)^2]
Let's substitute U^ with a + bX:
MSE = E[(U - (a + bX))^2]
Expanding the square and using the linearity of the expectation operator, we get:
MSE = E[U^2 - 2U(a + bX) + (a + bX)^2]
= E[U^2 - 2aU - 2bUX + a^2 + 2abX + b^2X^2]
= E[U^2] - 2aE[U] - 2bE[UX] + a^2 + 2abE[X] + b^2E[X^2]
Since U and W are independent, E[UX] = E[U]E[X] = mE[X] and E[X] = E[U + W] = E[U] + E[W] = m.
Also, we know that E[X^2] = var(X) + E[X]^2.
X = U + W, so var(X) = var(U + W) = var(U) + var(W) = v + h.
Substituting these values, we have:
MSE = E[U^2] - 2am + 2abm + a^2 + 2abh + b^2(v + h)
To minimize the MSE, we differentiate it with respect to a and b and set the derivatives equal to zero.
∂MSE/∂a = -2m + 2bm = 0
∂MSE/∂b = 2am + 2bh = 0
From the first equation, we obtain a = m/b.
Substituting this value of a into the second equation, we get:
2m(b/b) + 2bh = 0
2m + 2bh = 0
bh = -m
Hence, b = -m/h.
Substituting the value of b back into the first equation, we find:
a = m/(-m/h) = -h.
Therefore, the LLMS estimator of U based on X is U^ = -h + (-m/h)X, or in simplified form, U^ = -h - mX/h.
Now, let's move on to part 2 of the question.
If we assume that U and W are normally distributed and construct the LMS estimator U^LMS of U based on X, the LMS estimator would be the same as the LLMS estimator U^ = -h - mX/h. This is because the LMS estimator does not rely on the normality assumption to derive its form.
So, U^LMS would be identical to U^, the LLMS estimator developed without the additional normality assumption.
a=m*h/(v+h)
b=v/(v+h)