# Finding maximum likelihood estimator

1. Mar 22, 2014

### ptolema

1. The problem statement, all variables and given/known data

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 $\beta$.

2. Relevant equations

log likelihood (LL) = n ln(α) - nα ln(β) + (α-1) ∑(ln xi)

3. 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?

Last edited: Mar 22, 2014
2. Mar 22, 2014

### Mugged

So there might be some mistakes in the way you computed the log (LL) function. The term premultiplying log(β) should probably be reworked. Hint: log(β^y) = ylog(β). But what is y? It is not αn.

3. Mar 23, 2014

### Ray Vickson

Your expression for LL is correct, but condition (2) is wrong. Your problem is
$$\max_{a,b} LL = n \ln(a) - n a \ln(b) + (a-1) \sum \ln(x_i) \\ \text{subject to } b \geq m \equiv \max(x_1, x_2, \ldots ,x_n)$$
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
$$\partial LL/ \partial a = \mu \partial g / \partial a \equiv 0 \\ \partial LL / \partial b = \mu \partial g \partial b \equiv - \mu$$
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 $b = m$

4. Mar 23, 2014

### Mugged

Ray, log((b^a)^N) = a^N*log(b) ≠ aNlog(b)?

5. Mar 23, 2014

### Ray Vickson

We have $(b^a)^2 = b^a \cdot b^a = b^{2a},$ etc.

6. Mar 23, 2014

### Mugged

Ah..ok, my bad. This problem is harder than I thought...KKT coming in a statistics problem. Thanks.

7. Mar 23, 2014

### Ray Vickson

It's not that complicated in this case. For $a,b > 0$ the function $LL(a,b)$ is strictly decreasing in $b$, so for any $a > 0$ its maximum over $b \geq m \,(m > 0)$ lies at $b = m$. You don't even need calculus to conclude this.