ptolema said:
Homework Statement
The independent random variables X_1, ..., X_n have the common probability density function f(x|\alpha, \beta)=\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1} for 0\leq x\leq \beta. Find the maximum likelihood estimators of \alpha and \beta.
Homework Equations
log likelihood (LL) = n ln(α) - nα ln(β) + (α-1) ∑(ln xi)
The Attempt at a Solution
When I take the partial derivatives of log-likelihood (LL) with respect to α and β then set them equal to zero, I get:
(1) d(LL)/dα = n/α -n ln(β) + ∑(ln xi) = 0 and
(2) d(LL)/dβ = -nα/β = 0
I am unable to solve for α and β from this point, because I get α=0 from equation (2), but this clearly does not work when you substitute α=0 into equation (1). Can someone please help me figure out what I should be doing?
Your expression for LL is correct, but condition (2) is wrong. Your problem is
\max_{a,b} LL = n \ln(a) - n a \ln(b) + (a-1) \sum \ln(x_i) \\<br />
\text{subject to } b \geq m \equiv \max(x_1, x_2, \ldots ,x_n)
Here, I have written ##a,b## instead of ##\alpha, \beta##. The constraint on ##b## comes from your requirement ##0 \leq x_i \leq b \; \forall i##. When you have a bound constraint you cannot necesssarily set the derivative to zero; in fact, what replaces (2) is:
\partial LL/ \partial b \leq 0, \text{ and either } \partial LL/ \partial b = 0 \text{ or } b = m
For more on this type of condition, see, eg.,
http://en.wikipedia.org/wiki/Karush–Kuhn–Tucker_conditions
In the notation of the above link, you want to maximize a function ##f = LL##, subject to no equalities, and an inequality of the form ##g \equiv m - b \leq≤ 0##. The conditions stated in the above link are that
\partial LL/ \partial a = \mu \partial g / \partial a \equiv 0 \\<br />
\partial LL / \partial b = \mu \partial g \partial b \equiv - \mu
Here. ##\mu \geq 0## is a Lagrange multiplier associated with the inequality constraint, and the b-condition above reads as ##\partial LL / \partial b \leq 0##, as I already stated. Furthermore, the so-called "complementary slackness" condition is that either ##\mu = 0## or ##g = 0##, as already stated.
Note that if ##a/b \geq 0## you have already satisfied the b-condition, and if ##a/b > 0## you cannot have ##\partial LL / \partial b = 0##, so you must have ##b = m##