Probability
posted by A .
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]= ?
E[X2]= ?
The LLMS estimator of Θ given X is of the form aX+b. Give the numerical values of a and b.
a= ?
b= ?

E[X]=7.5

E[X^2] = 97

a=?
b=? 
a and b ?

a and b ?????

How did you get 97?

If anybody knows how to get the E[X^2] can you enlighten us please? I am just really thrown off by the fact that it has 2 ways of measuring X.

E[X]=7.5
E[X^2]=98.5
a=0.071
b=0.467 
re: E[X^2]:
Remember that Var[X] = E[X^2](E[X])^2 and also that (in shorthand) Var[X] = E[Var] + Var[E]. Use this last identity to get Var[X]; we already have E[X] so we can easily get E[X^2]. 
correct answer for E[X^2] is 98.5
(NOT 97) 
b=0.471

Can someone explain me how to find E[X]? thanks in advance

you can find E[X] using the law of total expectation
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