# 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

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