Let X and Y be independent random variables with zero means, and variances 1 and 2, respectively. Let U=X+Y and V=X+2Y .
Find the coefficients a and b of the Linear Least Mean Squares (LLMS) estimator VˆL=aU+b of V based on U .
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b=
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Why did the variable go to therapy?
Because it had unresolved issues with its coefficients! But don't worry, I'm here to help you find those coefficients. Let's dig in!
To find the coefficients a and b of the LLMS estimator VˆL = aU + b, we need to minimize the mean squared error between V and VˆL.
Since U = X + Y and V = X + 2Y, we can rewrite the LLMS estimator equation as VˆL = a(X + Y) + b.
Expanding VˆL, we get VˆL = aX + aY + b.
Now, let's compute the mean squared error (MSE) between V and VˆL:
MSE = E[(V - VˆL)^2]
= E[(X + 2Y - (aX + aY + b))^2]
Since X and Y are independent random variables with zero means and variances 1 and 2 respectively, we can simplify the MSE:
MSE = E[(X + 2Y - aX - aY - b)^2]
= E[(1 - a)^2 * X^2 + (2 - a)^2 * Y^2 + (2 - a) * Y * X - 2(1 - a)(b + 2Y)*X - 2(2 - a)(b + 2Y)*Y + (b + 2Y)^2]
To minimize the MSE, we take the partial derivatives of MSE with respect to a and b and set them equal to zero:
d(MSE)/da = -2E[X * (1 - a) * X] - 2E[(b + 2Y) * X]
= -2(1 - a) - 2b = 0
d(MSE)/db = -2E[(b + 2Y) * X] - 2E[(b + 2Y) * Y]
= -2(2 - a) - 2b = 0
Now we have two equations with two unknowns (a and b):
-2(1 - a) - 2b = 0 ------> a = 2 - b
-2(2 - a) - 2b = 0 ------> -4 + 2a - 2b - 2b = 0
Solving these equations simultaneously, we find:
a = 3/2
b = 1/2
So, the coefficients of the LLMS estimator VˆL = aU + b are:
a = 3/2
b = 1/2
Voila! The coefficients have been unveiled. Keep in mind that these coefficients minimize the mean squared error between V and VˆL based on U.
To find the coefficients of the Linear Least Mean Squares (LLMS) estimator of V based on U, we can use the method of linear regression.
The LLMS estimator minimizes the mean squared error between the estimated variable and the true variable, so we need to find the values of a and b that minimize the mean squared error between V and the LLMS estimator V̂L.
Let's denote E as the expected value (or mean) operator. The mean squared error between V and V̂L can be written as:
MSE = E[(V - V̂L)^2]
Expanding the squared term:
MSE = E[(V - (aU + b))^2]
To minimize the MSE, we need to take partial derivatives of the MSE with respect to both a and b and set them equal to zero.
Partial derivative with respect to a:
dMSE/da = 2 * E[(V - (aU + b)) * (-U)] = 0
Partial derivative with respect to b:
dMSE/db = 2 * E[(V - (aU + b)) * (-1)] = 0
Now, let's solve these two equations to find the values of a and b.
Taking the partial derivative with respect to a:
2 * E[(V - (aU + b)) * (-U)] = 0
Expanding it further:
2 * E[(-VU + aUU + bU)] = 0
Using the linearity property of the expectation operator:
-2E[VU] + 2aE[UU] + 2bE[U] = 0
Since X and Y are independent, we have:
E[VU] = E[(X + 2Y)(X + Y)] = E[X^2 + 3XY + 2Y^2] = E[X^2] + 3E[XY] + 2E[Y^2]
E[UU] = E[(X + Y)(X + Y)] = E[X^2 + 2XY + Y^2] = E[X^2] + 2E[XY] + E[Y^2]
E[U] = E[X + Y] = E[X] + E[Y] = 0 + 0 = 0
Plugging these values back into the equation:
-2(E[X^2] + 3E[XY] + 2E[Y^2]) + 2a(E[X^2] + 2E[XY] + E[Y^2]) = 0
Simplifying:
-2E[X^2] - 6E[XY] - 4E[Y^2] + 2aE[X^2] + 4aE[XY] + 2aE[Y^2] = 0
-2E[X^2] + 2aE[X^2] + 4(a - 3)E[XY] + (2a - 4)E[Y^2] = 0
Since X and Y have zero means, we have:
E[X^2] = var(X) = 1
E[Y^2] = var(Y) = 2
Plugging these values back into the equation:
-2 + 2a + 4(a - 3)E[XY] + (2a - 4)2 = 0
-2 + 2a + 4(a - 3)E[XY] + 4a - 8 = 0
-2 + 2a + 4a - 12 + 4(a - 3)E[XY] = 0
-14 + 10a + 4(a - 3)E[XY] = 0
-14 + 10a + 4a - 12E[XY] = 0
14a + 4(a - 3)E[XY] = 26
18a - 12E[XY] = 26
Similarly, taking the partial derivative with respect to b:
2 * E[(V - (aU + b)) * (-1)] = 0
Expanding it further:
-2E[V] + 2aE[U] + 2b = 0
Since X and Y have zero means, we have:
E[V] = E[X + 2Y] = E[X] + 2E[Y] = 0 + 0 = 0
Plugging this value back into the equation:
-2E[V] + 2aE[U] + 2b = 0
-2(0) + 2a(0) + 2b = 0
2b = 0
b = 0
Now we have two equations:
18a - 12E[XY] = 26
b = 0
Since X and Y are independent, their covariance (E[XY]) is zero. Therefore:
18a - 12(0) = 26
18a = 26
a = 26/18
Simplifying:
a = 13/9
Therefore, the coefficients of the LLMS estimator of V based on U are:
a = 13/9
b = 0
To find the coefficients a and b of the Linear Least Mean Square (LLMS) estimator V̂L = aU + b of V based on U, we need to minimize the mean square error between V and V̂L, also known as the cost function.
The formula for the mean square error is given by:
MSE = E[(V - V̂L)^2]
Expanding the equation:
MSE = E[(V - (aU + b))^2]
To minimize the MSE, we take the partial derivative of the MSE with respect to a and b, and set them equal to zero. Let's start by taking the partial derivative with respect to a:
∂MSE/∂a = ∂E[(V - (aU + b))^2]/∂a
Expanding the square term:
∂MSE/∂a = ∂E[(V^2 - 2aUV - 2bV + (a^2)(U^2) + 2abU + b^2)]/∂a
Since we are dealing with independent random variables X and Y, the terms UV and U^2 are both zero.
∂MSE/∂a = ∂E[(V^2 - 2bV + b^2)]/∂a
To find the coefficient a, we want to minimize this expression. Taking the partial derivative and setting it equal to zero:
0 = ∂E[(V^2 - 2bV + b^2)]/∂a
0 = E[-2UV]/∂a
Since U and V are independent, E[UV] = E[U]E[V]. But E[U] = E[X + Y] = E[X] + E[Y] = 0, and E[V] = E[X + 2Y] = E[X] + 2E[Y] = 0. Therefore, the partial derivative ∂E[(V^2 - 2bV + b^2)]/∂a equals zero.
So, we have:
0 = -2E[UV]
0 = -2E[XY]
Since X and Y are independent, the expectation of their product E[XY] equals zero.
0 = -2(0)
0 = 0
This means that the coefficient a could be any value, as it does not affect the minimization of the MSE. We can set a = 1 for simplicity.
Now, let's take the partial derivative with respect to b:
∂MSE/∂b = ∂E[(V - (aU + b))^2]/∂b
Expanding the square term:
∂MSE/∂b = ∂E[(V^2 - 2aUV - 2bV + (a^2)(U^2) + 2abU + b^2)]/∂b
Again, since we are dealing with independent random variables X and Y, the terms UV and U^2 are both zero.
∂MSE/∂b = ∂E[(V^2 - 2aUV - 2bV + (a^2)(U^2) + 2abU + b^2)]/∂b
To find the coefficient b, we want to minimize this expression. Taking the partial derivative and setting it equal to zero:
0 = ∂E[(V^2 - 2aUV - 2bV + (a^2)(U^2) + 2abU + b^2)]/∂b
0 = E[-2V + 2b]/∂b
0 = -2E[V] + 2b
Since E[V] = E[X + 2Y] = E[X] + 2E[Y] = 0, we have:
0 = -2(0) + 2b
0 = 2b
b = 0
Therefore, the coefficient a is arbitrary, and the coefficient b is 0.