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I want to find the MLE of $X = \theta{Y}$, where $\theta > 0$ and $Y \sim \mathrm{Beta}(2,1)$. The density for $X$ is given by $$f_{\theta}(x) = \frac{2x}{\theta^{2}}$$

on $[0,\theta]$.

It has been a bit problematic finding the MLE here (for fixed $x \in [0, \theta]$, doesn't seem to be any value $\hat{\theta}$ that maximizes this function). I'd appreciate any help with this.

forgetfulF
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If you have just a single observation, $X$, then $\theta$ is necessarily $\ge X$, so the decreasing function $\theta\mapsto2x/\theta^2$ has as its domain the interval $[X,\infty)$. And there is indeed a value of $\theta$ in that domain where $\theta\mapsto2x/\theta^2$ assumes its maximum value.

If there are $n$ i.i.d. observations $X_1,\ldots,X_n$, then the domain is $\left[\max\{X_1,\ldots,X_n\},\infty\right)$.

Michael Hardy
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  • In the case of just a single observation, is there a way to get the functional representation of the maximum likelihood estimator? – forgetfulF Feb 02 '15 at 00:33
  • What do you mean by the functional representation? – Michael Hardy Feb 02 '15 at 00:34
  • I just meant $\hat{\theta}(x)$. The function twhich, evaluated at $x$ gives the maximizing value $\theta$ for fixed $x$. – forgetfulF Feb 02 '15 at 00:36
  • Since the likelihood function $L(\theta)$ is decreasing on the interval $[x,\infty)$ (or $[\max{x_1,\ldots,x_n}$ in the case of multiple observations), the maximum occurse when $\theta$ is at the left endpoint. – Michael Hardy Feb 02 '15 at 00:38