Eric Slud November 24, 2013

Solutions of HW6 Problems in Stat 700

(1).(a) As we saw in class, nX ∼ Poisson(nλ) and E{(nX)2 − nX} =n(nλ)2 +nλ−nλ = n2λ2, so that T = X2 − n−1 X is unbiased for λ2. SinceT is obviously a function of the complete sufficient statistic X, the Lehmann-Scheffe Theorem shows that T is the UMVUE.

(b) Writing the Poisson(λ) probability mass function in terms of the changedparameter θ = λ2, the Cramer-Rao bound is found directly as{

nE(∂2

θ2log(e−

√θ θX/2

X!)}

={n(− θ−3/2

4− E(

−X

2θ2)}−1

=4θ3/2

n=

4λ3

n

(c) We can calculate Var(T ) = ET 2−(ET )2 in terms of nX = X1+ · · ·+Xn as

= n−4 E[nX(nX−1)(nX−2)(nX−3)+4nX(nX−1)(nX−2)+ 2nX(nX−1)−λ4

= 4λ3/n+ 2λ2/n2. The bound is not achieved (note that this variance of T isstrictly larger than the C-R lower bound) because the parametric distributionin terms of θ is not a natural exponential family.

(2).(a) For x ≤ y < ϑ, we can calculate the probability (up to top-order termsin dx and dy) as

P (V1 ∈ [x, x+dx), V(n) ∈ [y, y+dy)) =2x

ϑ2

(xϑ

)2n−2

I[x=y] +2x

ϑ2dx

(2n− 2) y2n−3

ϑ2n−2I[x<y]

After dividing by the probability P (V(n) ∈ [y, y+dy)) = 2ny2n−1dy/ϑ2n, we findthat the conditional probability distribution of V1 at x given V(n) ∈ [y, y + dy)is the mixture with weights 1/n and (n− 1)/n respectively of a mass-point 1 aty with a density 2(x/y2) I[0<x<y], and the conditional distribution function for0 < y < ϑ is

FV1|V(n)(t|y) =

1

nI[x=y] +

n− 1

n

(min(y, t))2

y2I[t≥0]

Since V(n) is a sufficient statistic by the factorization theorm and is easily checkedto be complete, and since (3/2)V1 is obviously unbiased for ϑ, the Lehmann-cheffe Theorem impies that the UMVUE is

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2E(V1 |V(n)) =

3

2

{ 1

nV(n) +

n− 1

n

∫ V(n)

0

2y2

V 2(n)

dy =2n+ 1

2nV(n)

This final result could have beend optained directly by noticing that EV(n) =2nϑ/(2n+ 1).

(3). Let µj = 32, 36, 40 for j = 1, 2, 3, and denote θ = (µ, 1/σ2), with priorprobability the product of the probability mass function 0.3δµ1+0.4δµ2+0.3δµ3

1

(in µ) and the density 100 t, e−10t (which is Gamma(2, 10), in 1/σ2 at argumentt > 0). Then the marginal density for Y = (Y1, . . . , Yn), which we will need forthe denominator of our Bayes’-rule posterior density calculation, is

fY(y) = (2π)−n/2 (0.3)3∑

j=1

(4

3)I[j=2]

100Γ(n/2 + 2)

(10 + 0.5 [(n− 1)S2y + n(y − µj)2])n/2+2

where S2y = (n−1)−1

∑ni=1(yi−y)2, y = n−1

∑ni=1 yi. In getting this expression,

we integrated

fY|ϑ(y |m, t) = (2π)−n/2 tn/2 exp(− (t/2)[

n∑i=1

y2i − 2nm y + nm2])

= (2π)−n/2 tn/2 exp(− (t/2)[(n− 1)S2

y + n(Y −m)2])

Then the posterior mass-function/density for ϑ at (µj , t) (discrete in the 1stcoordinate, continuous in the 2nd) is

( 43 )I[j=2] tn/2+1 exp

(− (10 + 0.5[(n− 1)S2

y + n(y − µj)2) t

)∑3

k=1(43 )

I[k=2] (10 + 0.5[(n− 1)S2y + n(y − µk)2)−n/2−2

With the given numbers S2y = 6.25, y = 37.2 in this problem,

(n− 1)S2y + n (y − µj)

2 =

99(6.25) + 100(5.2)2 = 3322.7599(6.25) + 100(2.2)2 = 1102.7599(6.25) + 100(2.8)2 = 1402.75

and the ratio (10+0.5(1102.75))/(10+0.5(1402.75) = 1−.21086, which raised tothe power 52 is really negligible. Therefore the posterior is approximately den-erate at µ2 = 36 in the µ component, and in 1/(sigma2) = t is Gamma(100/2+2, 10 + 0.5(1102.75)) = Gamma(52, 561.38).

(4). Recall throughout that X and Y are discrete random variables, withpX(x, ϑ) =

∑y:h(y)=x pY (y, ϑ). Then we compute directly from the defini-

tion, freely passing the derivative with respect to ϑ across the possibly infinitesummation over all y such that h(y) = x:

IX(ϑ) = E([∂

∂ϑlog pX(X,ϑ)]2

)=

∑x

1

pX(x, ϑ)

( ∂

∂ϑpX(x, ϑ)

)2

=∑x

1

pX(x, ϑ)

( ∑y:h(y)=x

∂

∂ϑpY (y, ϑ)

)2

=∑x

1

pX(x, ϑ)

( ∑y:h(y)=x

√pY (y, ϑ) {

1√pY (y, ϑ)

∂

∂ϑpY (y, ϑ)}

)2

2

Applying the Cauchy-Schwarz inequality to the final (·)2, we find

IX(ϑ) ≤∑x

1

pX(x, ϑ)

∑y:h(y)=x

pY (y, ϑ) ·∑

y:h(y)=x

(∂

∂ϑpY (y, ϑ))

2/pY (y, ϑ)

=∑x

∑y:h(y)=x

(∂

∂ϑpY (y, ϑ))

2/pY (y, ϑ) = IY (ϑ)

(5).(a) First we note that by independence of sample elements, E(X1X2) = p2 isunbiased for p2 in this setting where the complete sufficient statistic is Sn = nX.So Lehmann-Scheffe implues that the unique UMVUE is the Rao-Blackwellizedestimator (when evaluated at m = Sn

E(X1 X2 |nX = m) =

1∑j=0

1∑k=0

j k P (X1 = j, X2 = k |, Sn = m)

=

1∑j=0

1∑k=0

jk p2(

n− 2

m− j − k

)pm−j−k (1−p)n−2−(m−j−k)

/[(n

m

)pm (1−p)n−m

]and this last expression is immediately checked to be equal to

(n−2m−2

)/(nm

)=

(m(m− 1)/(n(n− 1)). Thus, substituting Sn = nX for m, we find the UMVUE= X(nX − 1)/(n− 1).

(6). The posterior is proportional (as a function of the parameter λ) toγ e−γλ e−nλ λnY /(Y1! · · ·Yn!), and therefore is Gamma(nY +1, n+ γ). In gen-eral, for Z ∼ Gamma(α, β), we minimize the convex function E((a−Z)2/Z) =E(Z)− 2a+E(a2/Z) of the real variable a by setting the derivative equal to 0,i.e.,

a = 1/E(1/Z)) = 1/

∫ ∞

0

βα

Γ(α)tα−2 eβ t dt = 1

/(

β

α− 1) =

α− 1

β

Substituting α = nY + 1, β = n+ γ for our posterior density, we find that theBayes optimal estimator for this strange loss function is nY /(n+γ). Note thatthese last lines correct an error in my previous posted solution: the answer isdifferent from the optimal estimator (nY +1)/(n+γ) under the squared-errorloss function.

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