# Partial solutions to assignment 10 Math 5080-1, Spring 2011 ```Partial solutions to assignment 10
Math 5080-1, Spring 2011
p. 224, #44. We know that
fX|Θ (x | θ) = θn exp −θ
n
X
!
xi
I{xi &gt; 0 for all i = 1, . . . , n},
i=1
and the prior pdf is
π(θ) = β exp(−βθ)I{θ &gt; 0}.
(a) Note that, viewed as a function of θ,
fX|Θ (x | θ)π(θ)
fX (x)
∝ fX|Θ (x | θ)π(θ)
fΘ|X (θ | x) =
n
= βθ exp −θ
n
X
!
xi − βθ
i=1
∝ θn exp (−θ [β + nx̄]) .
This is the pdf of GAMMA([β + nx̄]−1 , n + 1). And the posterior is this
pdf, after you plug in the data in the calculus variables; i.e., it is the
distribution GAMMA([β + nX̄]−1 , n + 1).
(b) We are asked to compute E[Θ | X]. But, according to the previous computation, this is the mean of a GAMMA([β + nX̄]−1 , n + 1); therefore,
we consult the table of our text to see that the Bayes estimator under
n+1
E[Θ | X] =
.
β + nX̄
(c) The answer is E[1/Θ | X]. Therefore, we seek to find first the expectation
of 1/Y when Y has a GAMMA pdf.
Suppose Y ∼ GAMMA(ζ , κ). Then,
Z ∞
Z ∞
1
−1
E(1/Y ) =
y fY (y) dy = κ
y κ−2 e−y/ζ dy
ζ Γ(κ) 0
0
Z ∞
1
= κ
wκ−2 ζ κ−2 e−w θ dw
[w := y/ζ]
ζ Γ(κ) 0
Z ∞
1
1
=
wκ−2 e−w dw =
&middot; Γ(κ − 1)
ζΓ(κ) 0
ζΓ(κ)
1
.
=
ζ(κ − 1)
1
Therefore, we plug in ζ = [β + nX̄]−1 and κ = n + 1 to see that
β + nX̄
β
= X̄ + .
n
n
E[1/Θ | X] =
This is the Bayes estimator for 1/θ.
(d) Under absolute loss the Bayes estimator of θ is the median [and not
the mean] of the posterior pdf. The posterior has been shown to be
GAMMA([β+nX̄]−1 , n+1). I other words, conditional on X = (X1 , . . . , Xn ),
the distribution of Θ is GAMMA([β + nX̄]−1 , n + 1). Therefore, conditional on X,
2Θ
∼ χ2 (2(n + 1));
[β + nX̄]−1
see Theorem 8.3.3 on page 269. The left-hand side is just 2Θ[β + nX̄].
We seek to find the point m [for median] such that
P {Θ ≤ m | X} =
1
.
2
Equivalently,
P
1
2Θ[β + nX̄] ≤ 2m[β + nX̄] X = .
2
Because the conditional pdf of 2Θ[β + nX̄] given X is χ2 (2(n + 1)), it
follows that m has to be chosen so that 2m[β + nX̄] is the 50th percentile
of χ2 (2(n + 1)) = χ2 (2n + 2); i.e.,
m=
1
χ2 (2n + 2).
2[β + nX̄] 0.5
(e) We are now asked to find a number m0 such that
1
1
0 P
≤m X = .
Θ
2
Equivalently,
Θ≥
1 1
X = .
m0 2
Θ&lt;
1 1
X = .
m0 2
P
Equivalently,
P
In other words, 1/m0 is the posterior median of Θ, which we called m;
that is,
2[β + nX̄]
m0 = 2
.
χ0.5 (2n + 2)
2
```