Statistical Inference | Assignment

Statistical Inference
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Exercise 4.2
Question One
Consider a random sample of size n x1,x2,x3,.......xn from a density function f (x , θ), with
unknown parameter θ . Ones we observe the values of x1,x2,x3,...........xn, the likelihood
function is given by L(θ) = ∏
i=1
n
f ( xi θ)ie the likelihood function is the product of the marginal
distributions.
The score function S(θ∨x1 : n) is derived by taking the differential of the log likelihood fuction
with respect to θ. Ie dl(θ)
dθ . It worthy noting that, the L(θ) and In L(θ) have their maximum at the
same value of θ and therefore it is easier to find the maximum using the logarithms of the
likelihood function.
It follows that, given the pdf of x1,x2,x3......xn which are iid, which is
F ( x ,θ ) = e x−θ
1+ ex−θ x ∈ R, then the likelihood function is given by by; ∏
i=1
n
f (xi θ) where i = 1,2,
....n
f ( x , θ ) = d
dx F ( x ,θ ) = ex−θ
1+ ex−θ x ∈ R d
dx ex−θ (1+ex−θ )−1 = d
dx = ex−θ+1
f ( x , θ )=ex−θ x ∈ R,
e x1−θ
1+ ex 1−θ * ex 2−θ
1+ ex 2−θ * ex 3−θ
1+ ex 3−θ *......................* exn−θ
1+ exn−θ , taking the cumulative function
2 | P a g e

F ( x ,θ ) = e x−θ
1+ ex−θ x ∈ R, = f ( x , θ )=ex−θ (1+ ex−θ)−1 x ∈ R, = ex−θ(1+e− x+θ) = ex−θ+1
f ( x , θ )=ex−θ +1 x ∈ R
∏
i=1
n
f ( xi θ)=∏
i=1
n
e xi−θ
+ 1 x ∈ R = e∑
i=1
n
(x−θ)
+1
Log L f (xi θ) = ∑
i=1
n
( xi−θ ) log e +log 1 = L f (xi θ) = ∑
i=1
n
( xi−θ ) +log1
Question two
S(θ∨x1 : n) = dl(θ)
dθ L f ¿ = dl (θ)
dθ ∑
i=1
n
( xi−θ ) + log 1 = ∑
i=1
n
(−x 1+θ )
L(θ∨x1 : n) = Var(S(θ∨x1 : n)) Var(∑
i=1
n
(−x 1+θ )) = Var(∑
i=1
n
xi)+Var(n θ) (Smirnov, 2011)
Considering the Var(x1) = π 2
3 it follows that Var(∑
i=1
n
xi) = ∑
i=1
n
Var ¿ ¿) = 1
n2 ∑
i=1
n
var xi
= 1
n2 * n* π 2
3 = nπ2
3 n
J(θ∨x1 : n) is given by the negative expectation of the second differential of the log likelihood
function. Given as: J(θ∨x1 : n) = -E( d2 l(θ)
dθ ∑
i=1
n
( xi−θ ) +log 1) (Donnelly et al. 2016)
3 | P a g e