Question 1:

a. f(x|θ) = θx-1(1-θ) x=1,2,……. f(θ)=1, 0< θ<1

Assuming that a series of observation x=1, 2 ... are sample from f(x|θ) and that each observation is independent and is identically distributed (iid), the joint probability density function for x1,….., xn is the product of the individual pdfs: fx1,….,xnθ=i=1nfxiθ =i=1nθxi-1(1-θ) =(1-θ)nθι=1nxi - n

The maximum likelihood estimate of θ can be found by first finding posterior:
Posterior ∝ prior * likelihood
Since fθ=1 fθx∝fx|θf(θ) Posterior ∝ (1-θ)nθxi-n*1 ∝ (1-θ)nθxi-n
The posterior f(θ|x) is proportional to Beta distribution since 0<θ<1, i.e. Beta (α,β)
Beta distribution: fθα,β=Γ(α+β)Γα+Γ(β)θα-1(1-θ)β-1 Therefore by comparing them we have that α-1=i=1nxi-n and β-1=n

Hence α=i=1nxi-(n-1) and β=n+1

We can estimate θ by first taking log of the likelihood in order to maximisefθx:

lnfθxi=ln⁡(1-θnθxi-n)

lnfθx1,….,xn=nln1-θ+(i=1nxi-n)lnθ and then taking the derivative with respect to θ:

d ln(f(θ|xi))dθ∝- n1(1-θ)+(i=1nxi –n)1θ
Setting the derivative equal to 0 and solving for θ:

dlnf(θ|xi)dθ=n-11-θ+(i=1nxi-n)1θ=0 n11-θ=i=1nxi –n1θ 1-θθ=ni=1nxi -n 1θ=ni=1nxi-n+1 1θ=i=1nxii=1nxi-n θ=i=1nxi-ni=1nxi

⟹θ=1-ni=1nxi

b. f(x|θ) = θxe-θx! x=1,2,……. f(θ)=1, 0< θ<1

Assuming that a series of observation x=1, 2 ... are sample from f(x|θ) and that each observation is independent and is identically distributed (iid), the joint probability density function for x1,….., xn is the product