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How to find maximum likelihood estimate for a parameter $a$ in distribution: $$f(x)=\begin{cases} \frac{\left| x\right| }{a ^{2} } &\text{for } x \in\left[ -a;a\right] \\0 &\text{for } x \not\in\left[ -a;a\right] \end{cases}$$

I did the following steps: $$L\left( a;x_{1},...,x_{n}\right) = \prod_{i=1}^{n} \frac{\left| x_{i}\right| }{a^{2}} = \frac{ \prod_{i=1}^{n}\left| x_{i}\right| }{a^{2n}}$$ $$\ln L = -2n \ln a + \sum_{i=1}^{n} \ln \left| x_{i}\right|$$ $$\frac{\partial \ln L}{\partial a} = \frac{-2n}{a}=0$$

What to do next?

Normally (in other examples) i got something like this: $c*=\frac{1}{\overline{X}}$ where $c*$ is some estimated parameter, but here I will get $a*=\infty$, which is wrong IMO.

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  • $\begingroup$ Is this for some subject, or for similar self-study from a text or similar? If so, could you please add the self-study tag? $\endgroup$
    – Glen_b
    May 27, 2013 at 3:30

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Your mistake is simply in taking the derivative and setting it equal to zero without thinking about what you're actually doing (not all maxima are turning points, he said, meaningfully). What you're actually identifying there would be a minimum.

Let

$$\sum_{i=1}^{n} \ln \left| x_{i}\right|=K\,,$$

say, and draw the log-likelihood function.

... but take care about what the smallest possible value for $a$ is.

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