{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw5_stat210a_solutions

# hw5_stat210a_solutions - UC Berkeley Department of...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 5 Fall 2006 Issued: Thursday, September 14, 2006 Due: Thursday, September 21, 2006 Graded exercises Problem 5.1 a) We want to show that R ( θ,δ ) = E θ ( ( X- θ ) θ ) does not depend on θ . To do that write: E θ ( ( X- θ ) 2 θ ) = E θ ( X- θ ) 2 θ = θ θ = 1 where the second equality follows from the fact that the variance of a Poi( θ ) random variable has variance θ . b) From item c below, we have that for the Gamma( a,b ) prior a Bayes estimator is: δ ( X ) = a + ∑ i X i- 1 n + b By constructing a sequence of priors with b k ↓ 0 and a k → 1, we have: δ k ( X ) = a k- 1 + ∑ i X i b k + n Hence: δ k ( X ) → ∑ i X i n and the sequence of priors approaches the “uniform” distribution on R + . c) We have that: π ( θ ) = exp[( a- 1)log( θ )- bθ + a log( b )- log Γ( a )] and p ( x | θ ) = exp "- nθ + X i x i log( θ ) # 1 Q i x i ! So: p ( θ | x ) ∝ exp " ( a + X i X i- 1)log( θ )- ( b + n ) θ # 1 which corresponds to a Γ( a + ∑ i X i ,b + n ) family (could also get this from Γ being conjugate prior to Poisson).conjugate prior to Poisson)....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

hw5_stat210a_solutions - UC Berkeley Department of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online