Let Θ be an unknown random variable that we wish to estimate. It has a prior distribution with mean 1 and variance 2. Let W be a noise term, another unknown random variable with mean 3 and variance 5. Assume that Θ and W are independent.
We have two different instruments that we can use to measure Θ. The first instrument yields a measurement of the form X1=Θ+W, and the second instrument yields a measurement of the form X2=2Θ+3W. We pick an instrument at random, with each instrument having probability 1/2 of being chosen. Assume that this choice of instrument is independent of everything else. Let X be the measurement that we observe, without knowing which instrument was used.
Give numerical answers for all parts below.
E[X]=
- unanswered
E[X2]=
- unanswered
The LLMS estimator of Θ given X is of the form aX+b. Give the numerical values of a and b.
a=
- unanswered
b=
- unanswered
1. 7.5
2. 97
3. 0.071
4. 0.467
Can someone explain me how to find E[X]? thanks in advance
To find the expected value of a random variable, we need to find the mean of its probability distribution.
1. E[X]:
Since X is a random variable, and we don't know which instrument was used to measure Θ, we need to find the expected value of X by considering both instruments.
For the first instrument:
E[X1] = E[Θ + W] (by definition of X1)
= E[Θ] + E[W] (since Θ and W are independent)
= E[Θ] + E[W] (since E[a+b] = E[a] + E[b] for independent variables)
Given that E[Θ] = 1 and E[W] = 3 (as stated in the question), we can substitute the values:
E[X1] = 1 + 3 = 4
For the second instrument:
E[X2] = E[2Θ + 3W] (by definition of X2)
= 2E[Θ] + 3E[W] (since Θ and W are independent)
Using the given values, E[Θ] = 1 and E[W] = 3:
E[X2] = 2(1) + 3(3) = 2 + 9 = 11
Therefore, E[X] = (E[X1] + E[X2]) / 2 = (4 + 11) / 2 = 15 / 2 = 7.5
2. E[X2]:
From the previous calculations, we found that E[X2] = 11.
3. The LLMS estimator of Θ given X is of the form aX + b. To find the numerical values of a and b, we need to use the formulas for LLMS estimation.
The LLMS estimator is given by:
Θ̂(X) = E[Θ] + Cov[Θ, X] / Var[X] * (X - E[X])
Given that Θ and X are independent, Cov[Θ, X] = 0. Also, Var[X] = Var[Θ + W] = Var[Θ] + Var[W] = 2 + 5 = 7.
Substituting these values into the LLMS estimator formula, we get:
Θ̂(X) = E[Θ] + 0 / 7 * (X - E[X])
= E[Θ] + 0
= 1
Therefore, the numerical values of a and b are:
a = 0
b = 1