# Maximum likelihood estimate for pdf

I am attempting a problem from Larsen and Marx, 4th edition that asks to find the maximum likelihood estimate for $\theta$ in the pdf: $$f(y; \theta) = \dfrac{2y}{1-\theta^{2}}, \theta \leq y \leq 1$$

It also states that a random sample of size 6 yielded measurements 0.70, 0.63, 0.92, 0.86, 0.43, and 0.21. I used the definition of the likelihood function to get: $$L(\theta) = \prod_{i=1}^n \dfrac{2y_{i}}{1-\theta^{2}} = 2^{n}(1-\theta^{2})^{-n} * \prod_{i=1}^n y_{i}$$

Because it is easier to deal with $\ln L(\theta)$ for the purpose of deriving and finding a $\theta$ that maximizes $L(\theta)$: $$\ln L(\theta) = -n \ln (1-\theta^{2}) + n \ln 2 + \ln \prod_{i=1}^n y_{i}$$ $$\dfrac{d \ln L(\theta)}{d\theta} = \dfrac{2n\theta}{(1-\theta^{2})}$$

However, setting this derivative to zero would mean $\theta_{e} = 0$ regardless of sample size and leaves me with a pdf of $f_{Y} = 2y$. Did I go wrong somewhere in my solution?

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The expression $f(y ; \theta)$ is only a pdf for $\theta=0$. Are you sure it is correctly written? –  Stefan Hansen Jul 6 '12 at 17:30
Your $f(\ ;\theta)$ is not a PDF. You probably mistyped the condition $\theta\leqslant y\leqslant1$ (and not $0\leqslant y\leqslant1$). You could try again to do the exercise with this (corrected) hypothesis. –  Did Jul 6 '12 at 17:31

Did I go wrong somewhere in my solution?

Apart for the typo (see did's comment), you made the most common error student make when computing ML: to just compute the derivative and set it to zero, before even asking about the domain of your function. Remember that your goal is to find the maximum of the function; now, you're supposed to know (Calculus I) that the global maximum of a function does not necessary happen on a critical point (null derivative), one should also look for the domain boundaries and non differentiable points (if they exist).

Always, when doing MLE, before computing the derivative, look at the function, ask which is the domain, if it's differentiable, perhaps try to graph it, see if it's easy to deduce where it's increasing/decreasing, etc.

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Here is a variant of (or an addendum to) @leonbloy's admonestation: never forget the range of the argument of a density or, better still, always write densities with their range included, for example using Iverson brackets. Here the density is $f(y;\theta)=2(1-\theta^2)^{-1}\cdot y\cdot [\theta\leqslant y\leqslant1]$ hence the likelihood $L(\theta)$ is not what you write. –  Did Jul 6 '12 at 18:31