Let X and W be independent and uniformly distributed on [−1,1]. We have given the following facts:
E[X]=E[X^3]=E[X^5]=0
E[X^2]=1/3
E[X^4]=1/5
Suppose that Y=X^3+W
Find the LMS estimate of Y, given that X=x.
(Notice we are trying to estimate Y from X, not the opposite direction. ) (Your answer should be a function of x.)
Y^LMS(x)= ?
Find the LLMS estimate for Y, given that X=x. (Your answer should be a function of x.)
Y^LLMS(x)= ?
can anyone please help here?
To find the LMS estimate of Y given X=x, we need to minimize the mean square error (MSE) between the estimated value and the true value of Y.
Let's start by finding the LMS estimate, denoted as Y^(LMS)(x).
Step 1: Let's find the expected value of the estimated value of Y, denoted as E[Y^(LMS)(x)].
Since X and W are independent, we have E[Y^(LMS)(x)] = E[X^3 + W].
Since E[W] = 0, we can simplify it to E[Y^(LMS)(x)] = E[X^3].
From the given information, we know that E[X^3] = 0. Therefore, E[Y^(LMS)(x)] = 0.
Step 2: Let's find the variance of the estimation error, denoted as Var[Y^(LMS)(x) - Y].
The estimation error is given by Y^(LMS)(x) - Y = X^3 + W - (X^3 + W) = 0.
Since the variance of a constant is zero, we have Var[Y^(LMS)(x) - Y] = Var[0] = 0.
Step 3: Since the variance is zero, the mean square error (MSE) is also zero.
So, the LMS estimate of Y given X=x is Y^(LMS)(x) = 0.
Now, let's find the LLMS estimate of Y given X=x.
The LLMS estimate is the least squares estimate, which can be found by minimizing the expected value of the square error between the estimated value and the true value of Y.
Step 1: Let's find the expected value of the squared error, denoted as E[(Y^(LLMS)(x) - Y)^2].
Consider Y^(LLMS)(x) = aX + b, where a and b are constants.
The squared error can be written as (Y^(LLMS)(x) - Y)^2 = (aX + b - (X^3 + W))^2.
Expanding the square, we get (aX + b - (X^3 + W))^2 = (aX + b - X^3 - W)^2.
Step 2: Let's find E[(Y^(LLMS)(x) - Y)^2] by taking the expectation of the squared error over X and W.
E[(Y^(LLMS)(x) - Y)^2] = E[(aX + b - X^3 - W)^2].
Next, we simplify the expression:
E[(aX + b - X^3 - W)^2] = E[(aX + b - X^3)^2 - 2(aX + b - X^3)W + W^2].
Since X and W are independent, E[W] = 0, and E[W^2] = Var[W].
E[(aX + b - X^3 - W)^2] = E[(aX + b - X^3)^2 - 2(aX + b - X^3)W + W^2]
= E[(aX + b - X^3)^2] - E[2(aX + b - X^3)W] + E[W^2]
= E[(aX + b - X^3)^2] - 0 + Var[W].
Step 3: We need to minimize E[(Y^(LLMS)(x) - Y)^2] with respect to a and b.
Minimizing E[(aX + b - X^3)^2] is equivalent to minimizing E[(X^3 - aX - b)^2].
To minimize this expression, we need E[X^3 - aX - b] = 0 and E[(X^3 - aX - b)X] = 0.
Taking expectations of those expressions, we have:
E[X^3 - aX - b] = E[X^3] - aE[X] - b = 0,
which simplifies to E[X^3] - b = 0. Since E[X^3] = 0, we have b = 0.
E[(X^3 - aX - b)X] = E[X^4 - aX^2] = E[X^4] - aE[X^2] = 0,
which simplifies to E[X^4] = aE[X^2]. Substituting the values, we have:
1/5 = a(1/3). Solving this equation, we get a = 3/5.
Therefore, the LLMS estimate of Y given X=x is Y^(LLMS)(x) = (3/5)X.
To summarize:
LMS estimate: Y^(LMS)(x) = 0
LLMS estimate: Y^(LLMS)(x) = (3/5)X.