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I am having trouble understanding how to calculate the Maximum Likelihood Estimator when x depends on $\theta$.

For example, to find the MLE of $$f(x) =\frac{2x}{\theta^2} \mbox{ where } 0\le x \leq \theta $$ we cannot use the traditional methods of finding Maximum Likelihood Estimators. Can someone please explain to me the steps I should take to find the MLE of this function?

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  • $\begingroup$ MLE finds the parameter $\theta$ for which your data has the largest likelihood. So you need to tell us what your data is and what your likelihood is. And what is $f(x)$? $\endgroup$
    – frank
    Commented Oct 5, 2022 at 4:12
  • $\begingroup$ My data is a random sample (X1, X2, ..., Xn) from X. f(x) is in the question. $\endgroup$
    – compscinewb
    Commented Oct 5, 2022 at 4:16
  • $\begingroup$ Your likelihood is thus$$L(\theta|x_!,\ldots,x_n)=\frac{2^n\prod x_i\mathbb I_{(0,\theta)}(x_i)}{\theta^{2n}}$$and you have to maximise this function of $\theta$. $\endgroup$
    – Xi'an
    Commented Oct 5, 2022 at 6:49

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Assuming independent values of the observations, the likelihood would be $$ 2^n\prod_{i=1}^nx_i/(\theta^{2n})$$ In order to maximize this, you would want $\theta$ to be as small as possible (the numerator is fixed once the sample is taken); but you are constrained by the fact that $x < \theta$, so your optimal choice would be $\hat\theta_{MLE} = \max\{x_1,\ldots,x_n\}$.

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