STAT 3023/3923
Semester 2 Statistical Inference 2021
Week 4 Tutorial Solutions
1. If X = (X, Y ) is bivariate normal with X and Y being standard normals with coefficient of correlation ρ. The joint density is
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f(x, y) = √1 e−(x2+y2−2ρxy)/[2(1−ρ2)]. 2π 1−ρ2
Let U = X/Y then
P(U ≤ u) = f(x,y)dx dy + f(x,y)dx dy.
0 {x≤uy} −∞ {x≥uy}
Differentiating with respect to u gives the density
= yf (uy, y)dy + −yf (uy, y)dy
= 2 ∞ √y exp[−y2(u2 +1−2ρu)/[2(1−ρ2)] dy
= 2π(u2 + 1 − 2ρu).
fU(u)= π(1−2ρu+u2).
0 2π 1−ρ2 2√1−ρ2
If X and Y are uncorrelated (ρ = 0) then U has a Cauchy distribution.
θxi (1 − θ)1−xi
= 1−θ (1−θ)n
= exp{ηT(x)−nlog(1+eη)}
xi log 1−θ +nlog(1−θ)}
E(T) = ψ′(θ) = nθ Var(T ) = ψ′′(θ) = n.
where T(X) = n X is a sufficient statistic for θ. Here η = log θ and
i=1 ψ(η) = n log(1 + eη) .
′ eη E(T)=ψ(η)=n1+eη =nθ.
Var(T)=ψ (η)=n(1+eη)2 =nθ(1−θ),
as T ∼ B(n, θ). (b)
n 1 − 1 ( x − θ ) 2 l(x,θ)= √e2i
= exp{θ( xi) − θ2} exp(−
x2i )(2π)−n/2 2 i=1
so T = ni=1 Xi is sufficient for θ and ψ(θ) = n2θ2.
l(x, θ) = Πni=1(2πθ)− 12 e−x2i /(2θ)
= (2πθ)−n/2e−(ni=1 x2i )/(2θ)
so here T (X) = ni=1 Xi2 is the sufficient statistic for θ by the fatorization theorem. 4.
l(x,α,β) = = =
i I(0
i=1 i=1 so T1(x) = ni=1 log xi and T2(x) = x(n).
The joint sufficient statistics are T1(X) = ni=1 log Xi and T2(X) = X(n).
This is in canonical form with T (x) = log(x) and ψ(α) = α log β + log Γ(α). (b)
f(x,α,β) = 1 xα−1e−x/βI(0 < x < ∞) βαΓ(α)
= exp{αlog(x)−βx−(αlogβ+logΓ(α))}.x−1I(0