Question

(a) The Bayesian setup: The posterior distribution 2 points possible (graded) Observe that if ??, ?? and ? are given, then each Y? is a gaussian: Y? | (??, ??, ?) ~ ? (?? + ??X?, 1/?). Therefore, the likelihood function of the vector (Y?, . . . , Y?) given (??, ??, ?) is of the form (1/?(2?/?))? exp (-?/2 ? (?? - ?? - ??X?)²) It turns out that the distribution of (??, ??) given ? and Y?, . . . , Y? is a 2-dimensional Gaussian. In terms of ?, Y and ?, what is its mean and covariance matrix? Hint: look ahead and see what part (b) is asking. What answer do you hope would come out, at least for one of these two things? (Type X for ?, trans(X) for the transpose ??, and X^(-1) for the inverse ??¹ of a matrix ?.) Mean: Covariance: (b) 1 point possible (graded) What is the Bayes estimator (??, ??)^Bayes for (??, ??)? Hint: Use your answer from part (a). However, as hinted: the answer here is guessable, even if you didn't solve the previous part. (Answer in terms of ?, Y and ?.) (Type X for ?, trans(X) for the transpose ?? of a matrix ?, and X^(-1) for the inverse ??¹ of a matrix ?.) (??, ??)^Bayes =

          (a) The Bayesian setup: The posterior distribution
2 points possible (graded)
Observe that if ??, ?? and ? are given, then each Y? is a gaussian: Y? | (??, ??, ?) ~ ? (?? + ??X?, 1/?).
Therefore, the likelihood function of the vector (Y?, . . . , Y?) given (??, ??, ?) is of the form
(1/?(2?/?))? exp (-?/2 ? (?? - ?? - ??X?)²)
It turns out that the distribution of (??, ??) given ? and Y?, . . . , Y? is a 2-dimensional Gaussian. In terms of ?, Y and ?, what is its mean and covariance matrix?
Hint: look ahead and see what part (b) is asking. What answer do you hope would come out, at least for one of these two things?
(Type X for ?, trans(X) for the transpose ??, and X^(-1) for the inverse ??¹ of a matrix ?.)
Mean:
Covariance:
(b)
1 point possible (graded)
What is the Bayes estimator (??, ??)^Bayes for (??, ??)?
Hint: Use your answer from part (a). However, as hinted: the answer here is guessable, even if you didn't solve the previous part.
(Answer in terms of ?, Y and ?.)
(Type X for ?, trans(X) for the transpose ?? of a matrix ?, and X^(-1) for the inverse ??¹ of a matrix ?.)
(??, ??)^Bayes =

(a) The Bayesian setup: The posterior distribution
2 points possible (graded)
Observe that if ??, ?? and ? are given, then each Y? is a gaussian: Y? | (??, ??, ?) ? (?? + ??X?, 1/?).
Therefore, the likelihood function of the vector (Y?, . . . , Y?) given (??, ??, ?) is of the form
(1/?(2?/?))? exp (-?/2 ? (?? - ?? - ??X?)²)
It turns out that the distribution of (??, ??) given ? and Y?, . . . , Y? is a 2-dimensional Gaussian. In terms of ?, Y and ?, what is its mean and covariance matrix?
Hint: look ahead and see what part (b) is asking. What answer do you hope would come out, at least for one of these two things?
(Type X for ?, trans(X) for the transpose ??, and X^(-1) for the inverse ??¹ of a matrix ?.)
Mean:
Covariance:
(b)
1 point possible (graded)
What is the Bayes estimator (??, ??)^Bayes for (??, ??)?
Hint: Use your answer from part (a). However, as hinted: the answer here is guessable, even if you didn't solve the previous part.
(Answer in terms of ?, Y and ?.)
(Type X for ?, trans(X) for the transpose ?? of a matrix ?, and X^(-1) for the inverse ??¹ of a matrix ?.)
(??, ??)^Bayes =

Added by Allison M.

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