Dr. Hackney STA Solutions pg 108

# Dr. Hackney STA Solutions pg 108 - Second Edition 7-11 b...

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Second Edition 7-11 b. For one observation of X and θ the joint pdf is h ( x,θ | τ ) = f ( x | θ ) π ( θ | τ ) , and the marginal pdf of X is m ( x | τ ) = Z -∞ h ( x,θ | τ ) dθ. Thus, the joint pdf of X = ( X 1 ,...,X n ) and θ = ( θ 1 ,...,θ n ) is h ( x , θ | τ ) = Y i h ( x i i | τ ) , and the marginal pdf of X is m ( x | τ ) = Z -∞ ··· Z -∞ Y i h ( x i i | τ ) 1 ...dθ n = Z -∞ ··· ±Z -∞ h ( x 1 1 | τ ) 1 ² n Y i =2 h ( x i i | τ ) 2 ...dθ n . The 1 integral is just m ( x 1 | τ ), and this is not a function of θ 2 ,...,θ n . So, m ( x 1 | τ ) can be pulled out of the integrals. Doing each integral in turn yields the marginal pdf m ( x | τ ) = Y i m ( x i | τ ) . Because this marginal pdf factors, this shows that marginally X 1 ,...,X n are independent, and they each have the same marginal distribution, m ( x | τ ). 7.26 First write f ( x 1 ,...,x n | θ ) π ( θ ) e - n 2 σ 2 x - θ ) 2 -| θ | /a where the exponent can be written n 2 σ 2 x - θ ) 2 - | θ | a = n 2 σ 2 ( θ - δ ± ( x )) + n 2 σ 2 ( ¯ x 2 - δ 2 ± ( x ) ) with δ ± ( x ) = ¯ x ± σ 2 na , where we use the “+” if
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