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In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection, so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ in which $y_{(n)}=\max \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ and this is maximized when $\theta=1/y_{(n)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(n)}} \, . $$

In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,\,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection (why?), so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and (why?) $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ in which $y_{(n)}=\max \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ and this is maximized (why? draw the graph for some fixed sample $y$) when $\theta=1/y_{(n)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(n)}} \, . $$

In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection, so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ in which $y_{(1)}=\min \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ and this is maximized when $\theta=1/y_{(1)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(1)}} \, . $$

In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection, so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ in which $y_{(n)}=\max \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(n)}]}(\theta) \, , $$ and this is maximized when $\theta=1/y_{(n)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(n)}} \, . $$

In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^n \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection, so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^n I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ in which $y_{(1)}=\min \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^n I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ and this is maximized when $\theta=1/y_{(1)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(1)}} \, . $$

In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that $$ f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, . $$ Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as $$ I_{[0,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, . $$ Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is $$ L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, . $$ But a product of indicators is the indicator of the intersection, so $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, , $$ and $$ I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ in which $y_{(1)}=\min \{y_1,\dots,y_n\}$. Hence, $$ L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(1)}]}(\theta) \, , $$ and this is maximized when $\theta=1/y_{(1)}$, yielding the maximum likelihood estimator $$ \hat{\Theta}_{ML}=\frac{1}{Y_{(1)}} \, . $$