Please help - statistics question

Question

Suppose that Y₁, Y₂,...,Y_n form a random sample from a PARETO (α, β) distribution, where α and β are unknown. Show that the maximum Likelihood Estimates of α and β (i.e. α-hat and β-hat) can be found by solving the following set of equations:

1.  (n/α) + n ln(β-hat) = Σⁿ_i=1 ln[β-hat + Y_i]
2.  (nα/β-hat) = (α-hat+1) Σⁿ_i=1 [β-hat + Y_i]^-1

NOTE: The pdf of a PARETO (α, β) distribution is given by

f(y) = {(αβ^α)/(β+y)^α+1 if y>0

{0 if y≤0

Answer 1

We solve problems like this by maximizing the log likelihood rather than the likelihood, as this converts multiplications into sums.

As f(y) = αβ^α/(β+y)^α+1 , ln(f(y)) = ln(α) + α ln(β) - (α+1) ln(β+y), and the log likelihood is

n ln(α) + n α ln(β) - (α+1) ∑ ln(β+y_i)

We then maximize by setting the α and β derivatives equal to 0.

For the α derivative,

n/α + n ln(β) - ∑ ln(β+y_i) = 0, or n/α + n ln(β) = ∑ ln(β+y_i)

For the β derivative,

n α/β - (α+1) ∑1/(β+y_i) = 0, or n α/β = (α+1) ∑1/(β+y_i)

Yet, as the values of α and β thus calculated are estimates, not the true values, we can replace them with hat.

n/α-hat + n ln(β-hat) = ∑ ln(β-hat + y_i)

n α-hat/β-hat = (α-hat+1) ∑1/(β-hat + y_i)