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= ?

  1. Anonymus

    E[X]=7.5

  2. Anonymous

    E[X^2] = 97

  3. qwerty

    a=?
    b=?

  4. anonymous

    a and b ?

  5. Mary

    a and b ?????

  6. anonymous

    How did you get 97?

  7. Anonym

    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.

  8. Anonym

    E[X]=7.5
    E[X^2]=98.5
    a=0.071
    b=0.467

  9. Anon

    re: E[X^2]:
    Remember that Var[X] = E[X^2]-(E[X])^2 and also that (in short-hand) 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].

  10. Anonymous

    correct answer for E[X^2] is 98.5
    (NOT 97)

  11. Sam

    b=0.471

  12. cle

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

  13. Anonymous

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

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